Character Degrees of Extensions of PSL 2 (q) and SL 2 (q)

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1 Character Degrees of Extensons of PSL (q) and SL (q) Donald L. Whte Department of Mathematcal Scences Kent State Unversty, Kent, Oho 444 E-mal: July 7, 01 Abstract Denote by S the projectve specal lnear group PSL (q) over the feld of q elements. We determne, for all values of q > 3, the degrees of the rreducble complex characters of every group H such that S H Aut(S). We also determne the character degrees of certan extensons of the specal lnear group SL (q). Explct knowledge of the character tables of SL (q), GL (q), PSL (q), and PGL (q) s used along wth standard Clfford theory to obtan the degrees. 1 Introducton In a seres of recent artcles, M. L. Lews and the author have studed varous propertes of the set of degrees of rreducble complex characters of nonsolvable groups. Ths always requres detaled nformaton on the character degrees of fnte smple groups and, n order to extend results to general nonsolvable groups, often requres nformaton on the degrees of almost smple groups; that s, groups H such that S H Aut(S) for some smple group S. The most nterestng cases tend to be groups nvolvng the smaller smple groups wth few character degrees, n partcular the -dmensonal projectve specal lnear groups PSL (q), whch we wll denote by L (q) n Atlas [] notaton. Several of these studes requred ncreasngly detaled nformaton about the character degrees of groups H wth L (q) H Aut(L (q)) (see [5, 6, 7, 8]). Of course, the character table of L (q) s well-known, as s the automorphsm group, and so the character degrees of H are known n prncple. Snce L (q) s a normal subgroup of such a group H, each character degree of H wll be χ(1) j for some rreducble character χ of L (q) and some dvsor j of H : L (q). Determnng the values of j for whch χ(1) j s a character degree of H for a specfc group H and character χ of L (q) s not theoretcally dffcult. However, gven the more detaled nformaton requred n recent work and the number of dfferent possbltes for H, χ, and q, t has become much more convenent to have a complete answer coverng all cases than to derve the necessary nformaton on a case-by-case bass. For more recent work, t has also become useful to have nformaton about the degrees of extensons of the quassmple groups SL (q) when q s odd. In Theorem A, we gve the lst of character degrees of H, where L (q) H Aut(L (q)), n all cases. In Theorem B and Corollary C, we gve the character degrees of all extensons of SL (q) of a certan type. Let q = p f for some prme p. The outer automorphsm group of L (q) s generated by a feld automorphsm ϕ of order f and, f p s odd, a dagonal automorphsm δ of order. If p =, then δ s an nner automorphsm. The dagonal automorphsm δ s nduced by conjugaton on SL (q) by a dagonal matrx δ GL (q) of order q 1. We have L (q) δ = PGL (q) and SL (q) δ = GL (q). The character tables of PGL (q) and GL (q) are also known. In 3, we descrbe explctly the actons of the automorphsms on the conjugacy classes of SL (q), GL (q), L (q), and PGL (q), and n 4, we descrbe the actons of the automorphsms on the rreducble characters. 1

2 If q = p f > 5 s odd, the character degree set of L (q) s cd(l (q)) = {1, q, (q + ε)/, q 1, q + 1}, where ε = ( 1) (q 1)/, and the character degree set of L (q) for even q or PGL (q) for odd q s cd(l (q)) = {1, q, q 1, q + 1}. The characters of degrees 1 and q are nvarant n Aut(L (q)) and n fact extend to rreducble characters of H for any L (q) H Aut(L (q)). The two characters of degree (q + ε)/ are nvarant under ϕ and are nterchanged by δ, so are easly handled. The characters of degrees q 1 and q + 1 belong to parametrzed famles and are nvarant under δ, but ther stablzers n ϕ depend on the parameters. In 5, we determne the subgroups of Aut(L (q)) that are stablzers of characters of degree q 1 or q + 1 of L (q) or PGL (q). We also determne the subgroups of ϕ that are stablzers of characters of SL (q) of degree q 1 or q + 1, and show that for each such subgroup there s a character of SL (q) and an extenson of the character to GL (q) wth the same stablzer. In 6, we show that for odd q, f H s any subgroup of Aut(L (q)) contanng L (q) but not contanng PGL (q), then H/ L (q) s cyclc. Hence, n any case, f L (q) H Aut(L (q)), then ether H/ L (q) s cyclc or H/ PGL (q) s cyclc. A character of PGL (q) or, f PGL (q) H, of L (q), wll therefore extend to ts stablzer n H and then the extensons wll nduce rreducbly to H by Clfford s Theorem. We are then able to determne the character degrees of H usng our knowledge of whch subgroups of H appear as stablzers. In 7, we consder subgroups of A = (SL (q) δ ) ϕ = GL (q) ϕ of the form H = (SL (q) δ α ) ϕ β. We determne the characters of H 0 = SL (q) δ α lyng over characters of SL (q) of each degree. We then fnd ther stablzers n ϕ β n order to determne the degrees of the characters of H. Notaton and Man Theorems If G s any fnte group, Irr(G) wll denote the set of rreducble complex characters of G. We denote by cd(g) = {χ(1) χ Irr(G)} the set of character degrees of G. We set q = p f, where p s prme and f s a postve nteger. We wll always assume p f > 3 because the character degrees of the automorphsm groups of the non-smple groups L () = S 3 and L (3) = A 4 are well-known and the general results we prove do not always apply n these cases. It wll be most convenent to work wth the conjugacy classes and characters of SL (q) and GL (q), whch are descrbed n [3] and [9], respectvely. We wll use the notaton of those sources for the classes and characters. The outer automorphsm group of L (q) s of order (q 1, ) f, and s generated by a feld automorphsm ϕ of order f and, f p s odd, a dagonal automorphsm δ of order (see [1] or []). Observe that PGL (q) = L (q) δ. If p =, then δ s an nner automorphsm and the center of SL (q) s trval, so that L (q) = PGL (q) = SL (q). We wll denote by H a subgroup of Aut(L (q)) satsfyng L (q) H Aut(L (q)). The followng theorem descrbes the set of character degrees of H for any such subgroup of Aut(L (q)) for any q. Theorem A follows drectly from Corollary 6.3 and Theorems 6.4, 6.5, 6.6, and 6.7. Theorem A. Let S = L (q), where q = p f > 3 for a prme p, A = Aut(S), and let S H A. Set G = PGL (q) f δ H and G = S f δ H, and let H : G = d = a m, m odd. If p s odd, let ε = ( 1) (q 1)/. The set of rreducble character degrees of H s wth the followng exceptons: cd(h) = {1, q, (q + ε)/} {(q 1) a : m} {(q + 1)j : j d},

3 . If p s odd wth H S ϕ or f p =, then (q + ε)/ s not a degree of H.. If f s odd, p = 3, and H = S ϕ, then 1.. If f s odd, p = 3, and H = A, then j 1. v. If f s odd, p =, 3, or 5, and H = S ϕ, then j 1. v. If f (mod 4), p = or 3, and H = S ϕ or H = S δϕ, then j. If q s odd, the dagonal automorphsm δ of L (q) s nduced by conjugaton on SL (q) by a dagonal matrx δ n GL (q) of order q 1, and GL (q) = SL (q) δ. Conjugaton of SL (q) by δ s an nner automorphsm. The feld automorphsm ϕ of L (q) of order f also acts as an outer automorphsm of order f on SL (q) and GL (q). We determne the character degrees of certan subgroups of A = (L (q) δ ) ϕ = GL (q) ϕ contanng SL (q). However, n ths case A/ SL (q) = F q Gal(F q /F p ), where F q s the feld of q elements and F q ts multplcatve group. The structure of ths group s too complcated to allow us to determne the degrees for every subgroup, and so we restrct our attenton to subgroups of the form H = (S δ α ) ϕ β, where α q 1 and β f. The next theorem s the general result for such a subgroup H. Theorem B follows drectly from Theorems 7.3, 7.6, and 7.7. For coprme ntegers x and y, we denote by O y (x) the order of x modulo y. Theorem B. Let A = (SL (q) δ ) ϕ = GL (q) ϕ, where q = p f > 3 for an odd prme p, and let H = (SL (q) δ α ) ϕ β, where α q 1 and β f. Set f/β = d = a m wth m odd, α = α/(, α), and l = (q 1)/(α ). The set of rreducble character degrees of H s cd(h) = {k : k d and k = O v (p β ) for some v (q 1)/α} {qk : k d and k = O v (p β ) for some v (q 1)/α} { } 1 (α,) (q 1)k : k d and k = O v(p β ) for some v l wth l/v odd { } 1 (α,) (q + 1)k : k d and k = O v(p β ) for some v l {(q 1) a : m} {(q + 1)j : j d}, wth the excepton that f p = 3, β = 1, and f s odd, then j 1. Fnally, we consder the specal cases where H = A or H s a partcular subgroup of A of ndex. Corollary C follows from Theorem B. Observe that the characters of A of degree q + 1 le over the two characters of SL (q) of degree (q + 1)/ when p = 3 and f s odd. There s no character of A of degree q + 1 lyng over a character of SL (q) of degree q + 1 n ths case. Corollary C. Let A = (SL (q) δ ) ϕ, where q = p f > 3 for an odd prme p, and A 0 = (SL (q) δ ) ϕ. Let f = a m wth m odd. The set of rreducble character degrees of A s cd(a) = {k, qk,(q 1) a,(q + 1)k : k f, m}. The set of rreducble character degrees of A 0 s cd(a 0 ) = {k, qk, 1 (q 1)a,(q 1) a, 1 (q + 1)k,(q + 1)k : k f, m}, wth the excepton that f p = 3 and f s odd, there s no rreducble character of A 0 of degree q +1. 3

4 3 Conjugacy Classes and Automorphsms Let q = p f, where p s a prme. We are consderng characters of extensons of L (q) and PGL (q), as well as of SL (q) and GL (q). The conjugacy classes, characters, and automorphsms are more easly descrbed for SL (q) and GL (q). For SL (q), we wll use the notaton and character tables of [3, 38], and for GL (q) we wll use the character table of [9]. Let ν be a generator of F q, the multplcatve group of the feld F q of q elements, τ a generator of F q, and γ = τ q 1. We denote 1 = [ ] [ 1 0, z = 0 1 ] [ 1 0, c = 1 1 ] [ 1 0, d = ν 1 ] [ ν 0, a = 0 ν 1 ] [ γ 0, b = 0 γ 1 If q s odd, then every element of SL (q) s conjugate to one of 1, z, c, cz, d, dz, a l for 1 l (q 3)/, or b m for 1 m (q 1)/. If q s even, then every element of SL (q) = L (q) s conjugate to one of 1, c, a l for 1 l (q )/, or b m for 1 m q/. Observe that n ether case, 1 l [(q )/] and 1 m [q/], where [x] denotes the greatest nteger less than or equal to x. We need to consder GL (q) and PGL (q) only when q s odd. We denote A 1 (l) = [ ν l 0 0 ν l ], A (l) = [ ν l 0 1 ν l ], A 3 (l 1, l ) = [ ν l ν l ], B 1 (l) = [ τ l 0 0 τ ql Every element of GL (q) s conjugate to one of A 1 (l) or A (l) for 1 l q 1, A 3 (l 1, l ) for 1 l 1 q 1, 1 l 1 q 1, and l 1 l, or B 1 (l) for 1 l q 1 and (q + 1) l. The outer automorphsm group of L (q), q = p f, s of order df, where d = (, q 1). It s generated by a dagonal automorphsm δ and a feld automorphsm ϕ. The dagonal automorphsm s nduced by conjugaton on SL (q) by the matrx [ ] ν 0 δ = 0 1 ]. ]. and these automorphsms act on elements of SL (q) by [ ] δ [ a b a ν = 1 ] b and c d νc d [ a b c d ] ϕ [ a p b = p c p d p ]. We have GL (q) = SL (q) δ, and the feld automorphsm acts on GL (q) n the same way. Moreover, δ acts as an nner automorphsm on SL (q). Observe that δ = F q and the acton of ϕ on δ s nduced by the acton of the generator of the Galos group of F q over F p. Hence (GL (q) ϕ )/ SL (q) = δ ϕ = F q Gal(F q /F p ). As the center Z of SL (q) s nvarant under both δ and ϕ, these maps nduce automorphsms δ and ϕ on L (q) = SL (q)/z by (gz) δ = g δ Z and (gz) ϕ = g ϕ Z, as usual. Denotng the nduced automorphsm ϕ on L (q) by ϕ as well, we have L (q) δ = PGL (q) and Aut(L (q)) = L (q) δ, ϕ = PGL (q) ϕ. Smlarly, the center of GL (q) s nvarant under ϕ and ϕ nduces an automorphsm ϕ on PGL (q), whch we wll also denote by ϕ. If q s even, then δ s an nner automorphsm and Aut(L (q)) = L (q) ϕ, and f q s odd, then δ s an outer automorphsm but δ s nner. Hence δ s of order d = (, q 1) modulo nner automorphsms. Snce entres n elements of SL (q) are from the feld of q = p f elements, ϕ s of order f. Moreover, f q s odd, then δ and ϕ commute modulo nner automorphsms, so that Aut(L (q))/ L (q) = δ ϕ, hence ther actons on conjugacy classes or characters wll commute. We now descrbe the actons of δ and ϕ on conjugacy classes. These lemmas follow from straghtforward calculatons. 4

5 Lemma 3.1. Let q be odd and assume notaton as above. In SL (q), the dagonal automorphsm δ nterchanges the conjugacy classes of c and d, nterchanges the conjugacy classes of cz and dz, and fxes all other conjugacy classes. Lemma 3.. Assume notaton as above and let 1 k < f. In SL (q), the automorphsm ϕ k sends. the conjugacy class of a l to the class of a r, where 1 r [(q )/] and r ±lp k (mod q 1),. the conjugacy class of b m to the class of b s, where 1 s [q/] and s ±mp k (mod q + 1), and fxes all other conjugacy classes. Lemma 3.3. Assume notaton as above wth q odd and let 1 k < f. In GL (q), the automorphsm ϕ k sends. the conjugacy class of A 1 (l) to the class of A 1 (r), where 1 r q 1 and r lp k (mod q 1),. the conjugacy class of A (l) to the class of A (r), where 1 r q 1 and r lp k (mod q 1),. the conjugacy class of A 3 (l 1, l ) to the class of A 3 (r 1, r ), where 1 r 1 q 1, 1 r q 1, r 1 l 1 p k (mod q 1), and r l p k (mod q 1), and v. the conjugacy class of B 1 (l) to the class of B 1 (t), where 1 t q 1, q + 1 t, and t lp k (mod q 1). 4 Characters and Automorphsms In order to determne whch subgroups of Aut(L (q)) and (SL (q) δ ) ϕ appear as stablzers of rreducble characters of SL (q), GL (q), L (q), or PGL (q), we frst determne condtons under whch these characters are nvarant under the acton of powers of δ or ϕ. It wll be more convenent to work wth characters and conjugacy class of SL (q) or GL (q). Let G be ether SL (q) or GL (q) and let Z = Z(G), so that G/Z s L (q) or PGL (q), respectvely. An automorphsm σ of G nduces an automorphsm σ on G/Z defned by (gz) σ = g σ Z. Smlarly, the rreducble characters of G/Z are precsely those defned by χ(gz) = χ(g), where χ Irr(G) and Z ker χ. It s straghtforward to check that χ σ = χ f and only f χ σ = χ. Hence the rreducble characters of G/Z nvarant under a partcular automorphsm σ are those characters of G nvarant under σ wth kernel contanng Z. We frst determne the characters of SL (q) and GL (q) whose kernels contan the center. Of course, f q s even, then the center of SL (q) s trval and the dagonal automorphsm s an nner automorphsm, and so SL (q) = L (q) = PGL (q). In the notaton of [3, 38], when q s even, SL (q) has rreducble characters. 1 G of degree 1,. ψ of degree q (= St, the Stenberg character),. χ, 1 (q )/, of degree q + 1, and v. θ j, 1 j q/, of degree q 1. We wll assume q s odd n the followng. For SL (q), we use the notaton and character table n [3, 38] and for GL (q), we use [9, ]. The followng results are easly obtaned from the respectve character tables. Lemma 4.1. Let G = SL (q) wth q odd. The rreducble characters of G wth kernel contanng Z(G) are as follows: 5

6 . 1 G of degree 1;. ψ of degree q (= St, the Stenberg character);. χ, for 1 (q 3)/ and even, of degree q + 1; v. θ j for 1 j (q 1)/ and j even, of degree q 1; v. ξ 1 and ξ of degree (q + 1)/, f q 1(mod 4); v. η 1 and η of degree (q 1)/, f q 1(mod 4). Lemma 4.. Let G = GL (q) wth q odd. The rreducble characters of G wth kernel contanng Z(G) are as follows:. χ (n) 1 and χ (n) q for n = (q 1)/ and n = q 1,. χ (m,n) q+1 for 1 n (q 3)/ and m = (q 1) n,. χ ((q 1)n) q 1 for 1 n (q 1)/. Notaton 4.3. We wll wrte χ (n) q+1 for the character χ(m,n) We wll wrte θ (n) q 1 q+1 wth 1 n (q 3)/ and m = (q 1) n. for the character χ((q 1)n) wth 1 n (q 1)/. We frst consder the acton of δ on the rreducble characters of SL (q) for odd q and relate the characters of SL (q) to those of GL (q) and the characters of L (q) to those of PGL (q). Lemma 4.4. Let q be odd. All characters of SL (q) of degrees 1, q, q + 1, and q 1 are nvarant under δ and each extends to q 1 dstnct rreducble characters of GL (q). Each of {ξ 1, ξ } and {η 1, η } s a sngle orbt under the acton of δ. Each of these characters extends to (q 1)/ rreducble characters of ts stablzer SL (q) δ n GL (q) and each extenson then nduces to an rreducble character of GL (q). Proof. Observe frst that GL (q) = SL (q) δ, so that GL (q)/ SL (q) s cyclc of order q 1. Hence by Gallagher s Theorem ([4, 6.17]), an nvarant character of SL (q) extends to q 1 dstnct rreducble characters of GL (q). The characters of degrees 1, q, q + 1, and q 1 have the same value on the classes of c and d, and so also on the classes of cz and dz. Hence these characters are nvarant under δ by Lemma 3.1. The pars of characters {ξ 1, ξ } and {η 1, η } are equal on all classes except the classes of c, d, cz, and dz. We have ξ 1 (c) = ξ (d), ξ 1 (d) = ξ (c), ξ 1 (cz) = ξ (dz), and ξ 1 (dz) = ξ (cz). Therefore, ξ δ 1 = ξ and ξ δ 1 = ξ by Lemma 3.1, and smlarly for η 1 and η. Recall that δ nduces an nner automorphsm on SL (q), and so the stablzer n GL (q) of each of these characters s SL (q) δ. The result then follows from Clfford s Theorem ([4, 6.11]) and Gallagher s Theorem. For the followng lemma, let ε = ( 1) (q 1)/ and denote by µ 1, µ the rreducble characters of SL (q) of degree (q + ε)/. Thus µ = ξ f q 1(mod 4) and µ = η f q 1(mod 4), so that µ 1, µ Irr(L (q)) by Lemma 4.1. Lemma 4.5. Let q be odd. All characters of L (q) of degrees 1, q, q + 1, and q 1 are nvarant under δ and each extends to two dstnct rreducble characters of PGL (q). The characters µ 1 and µ of L (q) of degree (q + ε)/ form a sngle orbt under the acton of δ and nduce to a sngle rreducble character of PGL (q) of degree q + ε. Proof. Ths follows from Lemma 4.4 and the fact that L (q) s of ndex n PGL (q) = L (q) δ. 6

7 We now consder the acton of the feld automorphsm ϕ on the rreducble characters of SL (q), L (q), and PGL (q). The stuaton for GL (q) s more complcated and wll be consdered n 7. Unless stated otherwse, q may be ether even or odd. Lemma 4.6. If χ s an rreducble character of SL (q), L (q), or PGL (q) of degree 1, q, (q 1)/, or (q + 1)/, then χ s nvarant under the acton of ϕ. Proof. It s clear from the character tables that the characters of SL (q), and hence of L (q), of degree 1 and q are nvarant under ϕ, as are the prncpal character and Stenberg character of PGL (q). For odd q, the remanng characters of degree 1, q of PGL (q) have value ( 1) k+l on the class of A 3 (k, l), and ether ( 1) l or ( 1) l on the class of B 1 (l). By Lemma 3.3, ϕ sends A 3 (k, l) to A 3 (kp, lp) and B 1 (l) to B 1 (lp). Snce p s odd n ths case, ( 1) k+l = ( 1) kp+lp and ( 1) l = ( 1) lp, and so these characters are also nvarant under ϕ. The characters of degree (q ± 1)/ occur only for SL (q) and L (q) wth q odd. By Lemma 3., the only conjugacy classes moved by ϕ are those of a l and b l. The values of the µ on these classes are 0, ( 1) l or ( 1) l. The class of a l, b l s sent to a ±lp, b ±lp, respectvely. Agan, snce p s odd, ( 1) l = ( 1) ±lp and these characters are nvarant under ϕ. We next determne condtons under whch a gven character of degree q +1 or q 1 s nvarant under a power of ϕ. The followng general result wll be very useful. Lemma 4.7. If ǫ s a complex kth root of unty and, j are ntegers, then ǫ + ǫ = ǫ j + ǫ j f and only f ±j(mod k). Proof. Observe that ǫ + ǫ = ǫ j + ǫ j f and only f ǫ ǫ j = ǫ j ǫ = ǫ ǫ j ǫ +j. Ths holds f and only f ether ǫ = ǫ j, n whch case j(mod k), or ǫ +j = 1, n whch case j(mod k). The next result also apples to characters of L (q), of course, as these are characters of SL (q). Lemma 4.8. Let q = p f for a prme p and f, and let 1 k f.. The character χ n of SL (q) or χ (n) q+1 of PGL (q) of degree q + 1 s nvarant under ϕ k f and only f p f 1 (p k 1)n or p f 1 (p k + 1)n.. The character θ n of SL (q) or θ (n) q 1 of PGL (q) of degree q 1 s nvarant under ϕ k f and only f p f + 1 (p k 1)n or p f + 1 (p k + 1)n. Proof. Dstnct characters of SL (q) or PGL (q) of degree q + 1 dffer only on the classes of a l of SL (q) or A 3 (l 1, l ) of GL (q). Smlarly, dstnct characters of degree q 1 dffer only on the classes of b m of SL (q) or B 1 (l) of GL (q). Let ρ be a complex prmtve (q 1)th root of unty, so that χ n (a l ) = ρ nl + ρ nl for the character χ n of SL (q) of degree q + 1. The character χ n s then nvarant under ϕ k f and only f χ n (a l ) = χ n ((a l ) ϕk ) = χ n (a lpk ) for all l, 1 l [(q )/], by Lemma 3., hence f and only f ρ nl + ρ nl = ρ nlpk + ρ nlpk 7

8 for all l. By Lemma 4.7, ths holds f and only f nlp k ±nl(mod q 1) for all l. Observng that q = p f and that f the congruence holds for l = 1 then t holds for all l, we see that χ n s nvarant under ϕ k f and only f np k ±n(mod p f 1) as clamed. Smlarly, for the character χ (n) q+1 of PGL (q), we have χ (n) q+1 (A 3(l 1, l )) = ρ n(l l 1 ) + ρ n(l l 1 ) and, by Lemma 3.3, χ (n) q+1 s nvarant under ϕk f and only f ρ n(l l 1 ) + ρ n(l l 1 ) = ρ n(l l 1 )p k + ρ n(l l 1 )p k for all l 1 l, 1 l 1 q 1, 1 l q 1. As before, ths s equvalent to n(l l 1 )p k ±n(l l 1 )(mod q 1) for all l 1, l. Snce q 4, ths must hold n partcular for l 1 = 1, l =, and so holds for all l 1, l f and only f np k ±n(mod p f 1) as clamed. Let σ be a complex prmtve (q + 1)th root of unty, so that θ n (b m ) = (σ nm + σ nm ) for the character θ n of SL (q) of degree q 1. The character θ n s then nvarant under ϕ k f and only f θ n (b m ) = θ n ((b m ) ϕk ) = θ n (b mpk ) for all m, 1 m [q/], by Lemma 3., hence f and only f σ nm + σ nm = σ nmpk + σ nmpk for all m. By Lemma 4.7, ths holds f and only f nmp k ±nm(mod q + 1) for all m. Agan, q = p f and f the congruence holds for m = 1 then t holds for all m, hence θ n s nvarant under ϕ k f and only f np k ±n(mod p f + 1) as clamed. For the character θ (n) q 1 of PGL (q), we have θ (n) q 1 (B 1(l)) = (σ nl + σ nlq ) = (σ nl + σ nl ), snce σ q = σ 1. By Lemma 3.3, θ (n) q 1 s nvarant under ϕk f and only f θ nl + θ nl = θ nlpk + θ nlpk for all l, 1 l q 1 and q + 1 l. As before, ths s equvalent to nlp k ±nl(mod q + 1), whch holds for all l f and only f np k ±n(mod p f + 1). The followng number-theoretc result wll be helpful n applyng Lemma 4.8. Lemma 4.9. If p s a prme and f, k are postve ntegers such that k f, then. (p f 1, p k 1) = p k 1, { (p 1, ) f f/k s odd. (p f 1, p k + 1) = p k + 1 f f/k s even,. (p f + 1, p k 1) = (p 1, ), 8

9 { p v. (p f + 1, p k + 1) = k + 1 f f/k s odd (p 1, ) f f/k s even. Proof. Observe frst that p k 1(mod p k 1) and p k 1(mod p k + 1), hence and p f (p k ) f/k (1) f/k 1(mod p k 1) p f (p k ) f/k ( 1) f/k (mod p k + 1). It follows that p k 1 p f 1, so that () holds. Smlarly, f f/k s even, then p k + 1 p f 1 and (p f 1, p k + 1) = p k + 1, whereas f f/k s odd, then p k + 1 p f + 1 and (p f + 1, p k + 1) = p k + 1. Snce p k 1 p f 1, we have that (p f + 1, p k 1) must dvde. Smlarly, f f/k s odd, then p k + 1 p f + 1 and so (p f 1, p k + 1) dvdes, whle f f/k s even, then p k + 1 p f 1 and (p f + 1, p k + 1) dvdes. If p =, then all of p k ± 1 and p f ± 1 are odd, but f p s odd then all of these ntegers are even. Hence these greatest common dvsors are 1 when p = and when p s odd, hence are equal to (p 1, ) as clamed. 5 Stablzers of Characters of Degree q 1 or q + 1 By Lemmas 4.4 and 4.5, the rreducble characters of SL (q) and L (q) of degree q 1 or q + 1 are nvarant n GL (q) and PGL (q), respectvely. Also, recall that f q s even, then δ s an nner automorphsm and SL (q) = PGL (q) = L (q). We now determne the subgroups of ϕ that occur as stablzers (n ϕ ) of characters of SL (q), L (q), GL (q), or PGL (q) of degree q 1 or q + 1. Throughout ths secton, we denote q = p f for some prme p and nteger f, and K wll denote a subgroup of ϕ of the form K = ϕ k for some postve dvsor k of f. We frst determne when K s a stablzer of a character of degree q 1. Lemma 5.1. Let q = p f, k f and K = ϕ k, and set n = (p f + 1)/(p k + 1).. If f/k s even, then K does not stablze any rreducble character of SL (q) or PGL (q) of degree q 1.. If f/k s odd, then K s the stablzer n ϕ of θ n Irr(SL (q)), θ (n) q 1 Irr(PGL (q)), and an extenson ˆθ n of θ n to GL (q). Proof. All characters of SL (q) of degree q 1 are θ j for some 1 j [q/] and for odd q, characters of PGL (q) of degree q 1 are θ (j) q 1 for 1 j (q 1)/. In partcular, note that 1 j < (q + 1)/ = (p f + 1)/ n any case. By Lemma 4.8, θ j and θ (j) q 1 are fxed by ϕk f and only f If f/k s even, then p f + 1 (p k 1)j or p f + 1 (p k + 1)j. (p f + 1, p k 1) = (p f + 1, p k + 1) = (, p 1) by Lemma 4.9. Thus f ϕ k stablzes θ j or θ (j) q 1, then (pf + 1)/(, p 1) must dvde j, whch s mpossble as 1 j < (p f + 1)/. Hence f f/k s even, ϕ k does not stablze any character of degree q 1. Assume now that f/k s odd, so by Lemma 4.9, p k +1 p f +1. Set n = (p f +1)/(p k +1). Snce p k + 1 3, we have 1 n = pf + 1 p k + 1 pf + 1 < pf + 1 = q + 1 3, 9

10 hence n s an nteger and n < (q + 1)/. If q s odd, ths mples n (q 1)/ and f q s even, ths mples n q/. Hence θ n s a character of SL (q) for any q and θ (n) q 1 s a character of PGL (q) for odd q. Moreover, we have p f + 1 (p k + 1)n, so that by Lemma 4.8, θ n and θ (n) q 1 are stablzed by K. If the stablzer, T, of θ n or θ (n) q 1 n ϕ properly contans K, then T = ϕt for some dvsor t of k wth 1 t < k. Snce ϕ t stablzes θ n or θ (n) q 1, we have pf +1 (p t +1)n or p f +1 (p t 1)n. Hence one of (p t + 1)/(p k + 1) or (p t 1)/(p k + 1) s an nteger. In any case, ths mples p k + 1 p t + 1, contradctng 1 t < k. Therefore, K s the stablzer n ϕ of θ n and θ (n) q 1. Fnally, we show that K s also the stablzer n ϕ of some extenson ˆθ n of θ n to GL (q). We assume q s odd as the result s trval f q s even. It s straghtforward to check that for any j, one extenson of θ j to GL (q) s χ (j) q 1. The rreducble characters of the cyclc group GL (q)/ SL (q) are precsely the characters χ () 1 of GL (q), for 1 q 1, and so by Gallagher s Theorem all of the extensons of θ j are the characters χ (j) q 1 χ() 1. We clam that the extenson ˆθ n = χ (n) q 1 χ() 1 where n = pf + 1 p k + 1 and = pf p k p k 1 = pk (p f k 1) p k, 1 has stablzer K n ϕ. Frst, observe that snce f/k s odd, f/k 1 s even, and so k dvdes (f/k 1)k = f k. Hence p k 1 p f k 1 and s an nteger wth 1 p f 1 = q 1. Thus ˆθ n s an extenson of θ n, and so the stablzer of ˆθ n s contaned n K. It remans to show that ˆθ n s nvarant under ϕ k. The values of the character ˆθ n and ts mage under ϕ k are as follows: ˆθ n (ˆθ n ) ϕ k A 1 (l) (q 1)ρ (n+)l (q 1)ρ (n+)pk l A (l) ρ (n+)l ρ (n+)pk l A 3 (l 1, l ) 0 0 B 1 (l) ǫ [n+(q+1)]l ǫ [nq+(q+1)]l ǫ [n+(q+1)]pkl ǫ [nq+(q+1)]pk l where ǫ s a complex prmtve (q 1)th root of unty and ρ = ǫ q+1 s a complex prmtve (q 1)th root of unty. For n and as defned above, we have n + = pf 1 p k 1 n + (q + 1) = n pf 1 p k 1 = q 1 p k 1 nq + (q + 1) = np k pf 1 p k 1 = [n + (q + 1)]pk. Thus t s clear that ǫ [nq+(q+1)]l = ǫ [n+(q+1)]pkl for all l. Also, q 1 [n + (q + 1)](p k 1), so that [nq + (q + 1)]p k l = [n + (q + 1)]p k l [n + (q + 1)]l (mod q 1), and hence ǫ [n+(q+1)]l = ǫ [nq+(q+1)]pkl. It follows that ˆθ n and (ˆθ n ) ϕ k have the same values on the classes B 1 (l). Fnally, q 1 (n + )(p k 1), so that (n + )l (n + )p k l (mod q 1) for all l. Therefore, ˆθ n and (ˆθ n ) ϕ k have the same values on all classes and ˆθ n s nvarant under ϕ k, and hence also under ϕ k. 10

11 Lemma 5.. Let q = p f be odd, k f, K = ϕ k, and n = (p f + 1)/(p k + 1).. If f/k s even or p k = 3, then K does not stablze any rreducble character of L (q) of degree q 1.. If f/k s odd and p k 3, then K s the stablzer n ϕ of θ n Irr(L (q)). Proof. The characters of L (q) of degree q 1 are θ j for some even nteger j wth 1 j (q 1)/. By Lemma 4.8, θ j s fxed by ϕ k f and only f p f + 1 (p k 1)j or p f + 1 (p k + 1)j. By the same argument as n Lemma 5.1, f f/k s even, then ϕ k does not stablze any character of degree q 1. Suppose now that p = 3 and k = 1, so that K = ϕ, and f 3 s odd. By Lemma 4.1, the characters of L (3 f ) of degree q 1 are the θ j wth 1 j (3 f 1)/ and j even. By Lemma 4.8, θ j s fxed by ϕ f and only f 3 f +1 j or 3 f +1 4j. Thus f K = ϕ stablzes θ j, then 3 f +1 4j. Snce f s odd, 3 f + 1 4(mod 8), and so (3 f + 1)/4 s odd and dvdes j. Snce j s even, ths mples (3 f + 1)/ dvdes j, contradctng 1 j (3 f 1)/. Therefore, f p = 3 and k = 1, then K = ϕ does not stablze any character of L (q) of degree q 1. We now assume p k > 3 and f/k s odd. Thus, by Lemma 4.9, p k + 1 p f + 1, and so n = (p f + 1)/(p k + 1) s an even nteger. Snce p k > 3, we have 1 n = pf + 1 p k + 1 < pf + 1 = pf + 1 = q + 1 4, hence n s a postve nteger and n < (q + 1)/, whch mples n (q 1)/. Therefore, θ n s a character of L (q) of degree q 1, and snce p f + 1 (p k + 1)n, θ n s stablzed by ϕ k. If the stablzer of θ n properly contans K, then there s a dvsor t of k wth 1 t < k such that ϕ t stablzes θ n. Hence p f + 1 (p t + 1)n or p f + 1 (p t 1)n, and so one of (p t + 1)/(p k + 1) or (p t 1)/(p k +1) s an nteger. In partcular, we must have p k +1 (p t +1); that s, p k p t +1. But p 3 and 1 t < k, and so p k p t+1 3p t = p t + p t > p t + 1, a contradcton. Therefore, K s the stablzer n ϕ of the character θ n of L (q). We next determne when K = ϕ k s the stablzer n ϕ of a character of degree q + 1. Lemma 5.3. Let q = p f be odd, k f, and K = ϕ k.. If p k 3, then K s the stablzer n ϕ of χ n Irr(SL (q)), χ (n) q+1 Irr(PGL (q)), and an extenson ˆχ n of χ n to GL (q), for n = (p f 1)/(p k 1).. If p = 3, k = 1, and f s even, then K = ϕ s the stablzer n ϕ of χ n Irr(SL (q)), χ (n) q+1 Irr(PGL (q)), and an extenson ˆχ n of χ n to GL (q), for n = (3 f 1)/4.. If p = 3, k = 1, and f s odd, then K = ϕ does not stablze any rreducble character of SL (q) or PGL (q) of degree q + 1. Proof. Characters of SL (q), PGL (q) of degree q+1 are χ j, χ (j) q+1 respectvely, for 1 j (q 3)/. By Lemma 4.8, χ j and χ (j) q+1 are fxed by ϕk f and only f p f 1 (p k 1)j or p f 1 (p k + 1)j. 11

12 Assume frst that p k 5 and let n = (p f 1)/(p k 1). We then have p k 1 >, so that n < (p f 1)/ = (q 1)/. Therefore, n q 1 1 = q 3, and so χ n Irr(SL (q)) and χ (n) q+1 Irr(PGL (q)). Moreover, (p k 1)n = p f 1, hence p f 1 dvdes (p k 1)n and χ n and χ (n) q+1 are nvarant under ϕk. Therefore, K s contaned n the stablzer, T, of χ n and χ (n) q+1 n ϕ. We have K = ϕ k ϕ t = T for some dvsor t of k. Snce χ n and χ (n) q+1 are nvarant under ϕ t, we have that p f 1 dvdes ether (p t 1)n or (p t + 1)n, that s, p f 1 (p t 1) pf 1 p k 1 or pf 1 (p t + 1) pf 1 p k 1. Hence ether p k 1 p t 1 or p k 1 p t + 1. Snce t k, we have p t 1 p k 1, and so f p k 1 p t +1, then p t 1 p t +1 = (p t 1)+. Therefore, p t 1 and snce p s odd, ths means p t = 3. In partcular, p k 1 = 3 k 1 dvdes p t + 1 = 4, and hence k = 1 and p k = 3, contradctng p k 5. Therefore, p k 1 p t 1, so that k t, and snce t k, we have k = t and so K = T. Now let p k = 3 so that p k 1 = and p k + 1 = 4. It follows that f ϕ k = ϕ stablzes χ n or χ (n) q+1 for some n, then 3f 1 n or 3 f 1 4n. Hence 3 f 1 4n n any case, and f f s odd, then (3 f 1)/ s odd and dvdes n. Hence (q 1)/ dvdes n, contradctng 1 n (q 3)/. Therefore, f ϕ stablzes χ n or χ (n) q+1 for some n, then f s even. Conversely, f f s even, then 4 3 f 1. Settng n = (3 f 1)/4 = (q 1)/4, we have n < (q 1)/, so that n (q 3)/ and χ n Irr(SL (q)) and χ (n) q+1 Irr(PGL (q)). Moreover, 3 f 1 4n so that χ n and χ (n) q+1 are nvarant under ϕ. Hence K = ϕ s the stablzer of χ n and χ (n) q+1. Fnally, we show that unless f s odd and p k = 3, K s also the stablzer n ϕ of some extenson ˆχ n of χ n to GL (q). It s straghtforward to check that for any j, one extenson of χ j to GL (q) s χ (j,q 1) q+1. The rreducble characters of the cyclc group GL (q)/ SL (q) are precsely the characters χ () 1 of GL (q), for 1 q 1, and so by Gallagher s Theorem all of the extensons of χ j are the characters χ (j,q 1) q+1 χ () 1. We clam that the extenson ˆχ n = χ (n,q 1) q+1 χ () 1 has stablzer K n ϕ for some choce of. Snce ˆχ n s an extenson of χ n, the stablzer of ˆχ n s contaned n K. We need to show that ˆχ n s nvarant under ϕ k. The values of the character ˆχ n and ts mage under ϕ k are as follows: ˆχ n (ˆχ n ) ϕ k A 1 (l) (q + 1)ρ (n+)l (q + 1)ρ (n+)pk l A (l) ρ (n+)l ρ (n+)pk l A 3 (l 1, l ) (ρ nl 1 + ρ nl )ρ (l 1+l ) (ρ nl 1p k + ρ nl p k )ρ (l 1+l )p k B 1 (l) 0 0 where ρ s a complex prmtve (q 1)th root of unty. Frst assume p k 3 and let = q 1, so that ˆχ n = χ (n,q 1) q+1 χ (q 1) 1 = χ (n,q 1) q+1 and ρ = 1. We have q 1 n(p k 1), hence np k n (mod q 1) and ρ n = ρ npk. It follows from the table of character values above that ˆχ n s nvarant under ϕ k, and so also under ϕ k. Now suppose p = 3, k = 1 (so K = ϕ ), and f s even. In ths case, we have n = (3 f 1)/4 and we take = (3 f 1)/8. Snce f s even, s an nteger and ˆχ n = χ (n,q 1) q+1 χ () 1 s an extenson 1

13 of χ n. We have n+ = (q 1)/ and ρ n+ = 1. As p k = 3 s odd, t follows that ˆχ n and (ˆχ n ) ϕ 1 agree on the classes A 1 (l) and A (l). We also have n = and p k = 3, and so and ˆχ n (A 3 (l 1, l )) = ρ (3l 1+l ) + ρ (l 1+3l ) (ˆχ n ) ϕ 1 (A 3 (l 1, l )) = ρ (3l 1+l )3 + ρ (l 1+3l )3. Fnally, 9 = (q 1) +, hence ρ 9 = ρ, and so ˆχ n and (ˆχ n ) ϕ 1 agree on the classes A 3 (l 1, l ). Therefore, ˆχ n s nvarant under ϕ. Lemma 5.4. Let q = p f, k f, K = ϕ k, and n = (p f 1)/(p k 1).. If p k {, 3,, 5, 3 }, then K s the stablzer n ϕ of χ n Irr(L (q)).. For p k {, 3,, 5, 3 }, K s the stablzer n ϕ of an rreducble character of L (q) of degree q + 1 f and only f f/k even.. If p k = or 3 and f/ s odd, then there s an rreducble character of L (q) of degree q +1 whose stablzer s ϕ and hence s nvarant under K = ϕ. Proof. Characters of L (q) of degree q + 1 are χ j for 1 j (q )/ f q s even and 1 j (q 3)/, j even, f q s odd. By Lemma 4.8, χ j s fxed by ϕ k f and only f p f 1 (p k 1)j or p f 1 (p k + 1)j. Assume frst that K s the stablzer of χ j for some j, and suppose p k {, 3,, 5, 3 }. We also assume f/k s odd and work for a contradcton. If p k =, 3, or 5, so k = 1, and K stablzes χ j, then p f 1 (p 1)j or p f 1 (p + 1)j. Snce f = f/k s odd, (p f 1, p + 1) by Lemma 4.9, and so f p f 1 (p + 1)j, then p f 1 j,.e., q 1 j. Ths contradcts the fact that j < (q 1)/ n all cases. Hence p f 1 (p 1)j, and so (q 1)/(p 1) j. For p = or 3, ths agan contradcts j < (q 1)/. Fnally, for p = 5, we have (5 f 1)/4 j and snce f s odd, (5 f 1)/4 s odd. Recallng that j must be even, we then have (5 f 1)/4 j, and so (q 1)/ j, agan contradctng j < (q 1)/. Now suppose p k = or 3, so k =, and assume χ j s nvarant under K, hence under ϕ, for some j. In ths case we have p f 1 (p 1)j or p f 1 (p + 1)j. Snce f/k = f/ s odd, we have (p f 1, p + 1) by Lemma 4.9, and as before, ths mples that f p f 1 (p + 1)j, then q 1 j, a contradcton. Hence we must have p f 1 (p 1)j. If p =, ths means f 1 3j. Hence n fact f 1 ( 1 + 1)j, and so χ j s nvarant under ϕ. If p = 3, we have 3 f 1 8j. Snce f/ s odd, f s not dvsble by 4, and t follows that (3 f 1)/8 s odd and dvdes j, whch s even. Hence (3 f 1)/8 j, and so 3 f 1 4j = ( )j and agan ths mples χ j s nvarant under ϕ. Therefore, n ether case K = ϕ s not the full stablzer of χ j n ϕ. On the other hand, f k = and f/ s odd, then both ( f 1)/3 and (3 f 1)/4 are ntegers, ( f 1)/3 ( f )/ and (3 f 1)/4 (3 f 3)/, and (3 f 1)/4 s even. Hence, settng j = ( f 1)/3 when p = and j = (3 f 1)/4 when p = 3, we have that χ j Irr(L (q)). Snce f 1 = 3j and 3 f 1 = 4j, we obtan f 1 ( 1 + 1)j and 3 f 1 ( )j, and so χ j s nvarant under ϕ n ether case. We now have that f K s the stablzer n ϕ of an rreducble character of L (q) of degree q+1 and p k {, 3,, 5, 3 }, then f/k s even. We next consder the converse, and so let K = ϕ k and f p k {, 3,, 5, 3 }, we assume f/k s even. We wll fnd a j n each case such that K s the stablzer of χ j. We frst suppose p k {, 3, 5}, so that k = 1 and f s even. If p =, ths mples 3 f 1 and we set j = ( f 1)/3, so that j < ( f 1)/, so j (q )/ and χ j Irr(L (q)). If p = 3 or 13

14 5, then f even mples 8 p f 1 and we set j = (p f 1)/4. Thus j s even and strctly less than (q 1)/, hence j (q 3)/, and so χ j Irr(L (q)). In all three cases, we have j = (p f 1)/(p±1) and so p f 1 (p ± 1)j, whch mples that χ j s nvarant under ϕ. Hence the stablzer of χ j n ϕ s K = ϕ. Next, let p k = or 3, so that k = and f/k = f/ s even. By Lemma 4.9, p +1 p f 1 and we set j = (p f 1)/(p +1). For p =, we have j = ( f 1)/5 < ( f 1)/, hence j (q )/ and χ j Irr(L (q)). For p = 3, we have j = (3 f 1)/10 < (3 f 1)/, hence j (q 3)/. Also, snce f s even, 8 3 f 1 and so j s even. Thus agan χ j Irr(L (q)). In both cases, p f 1 (p + 1)j, and so χ j s stablzed by ϕ = ϕ k. Also p±1 < p +1, so (p±1)j < p f 1 and hence p f 1 dvdes nether (p + 1)j nor (p 1)j. It follows that χ j s not nvarant under ϕ, and so the stablzer of χ j n ϕ s K = ϕ, as clamed. We now suppose p k {, 3,, 5, 3 }. Set n = (p f 1)/(p k 1), whch s even as p k 1 p f 1. Snce p k 7, we have q 1 n = p k 1 q 1 < q 1 6. Therefore, f q s even, then n (q )/ and f q s odd, then n s even and n (q 3)/, so that χ n Irr(L (q)) n any case. Moreover, p f 1 (p k 1)n, and so χ n s nvarant under ϕ k and K s contaned n the stablzer T of χ n n ϕ. We know that T s of the form T = ϕ t for some postve dvsor t of k. In partcular, observe that t k mples p t 1 p k 1. Also, snce ϕ t stablzes χ n, we have p f 1 (p t ± 1)n; that s p f 1 (p t ± 1) pf 1 p k 1, whch mples that ether p k 1 (p t 1) or p k 1 (p t + 1). Frst, we show that p k 1 (p t + 1) cannot occur f p k {, 3,, 5, 3 }. If p k 1 (p t + 1), then t k mples p t 1 (p t + 1) = (p t 1) + 4. Hence p t 1 4 and p t =, 3, or 5, so t = 1 and we have p k 1 (p + 1) wth p =, 3, or 5. If p =, then k 1 6, hence k = 1 or and p k = or. If p = 3, then 3 k 1 8, hence k = 1 or and p k = 3 or 3. If p = 5, then 5 k 1 1, hence k = 1 and p k = 5. Therefore, f p k {, 3,, 5, 3 }, then p k 1 (p t + 1). Fnally, suppose p k 1 (p t 1). If p =, ths mples p k 1 (p t 1), hence k t and so t = k and T = K. If p s odd, ths mples (p k 1)/(p t 1) dvdes. Suppose k > t, so that (p k 1)/(p t 1) = and p k 1 = (p t 1). We have p k p p t, hence (p t 1) = p k 1 p p t 1 = p(p t 1) + (p 1) > (p t 1), a contradcton. Hence t = k and T = K n ths case as well. Therefore, f p k {, 3,, 5, 3 }, then K s the stablzer n ϕ of χ n. 6 Subgroups of Aut(L (q)) and Ther Degrees Our goal n ths secton s to determne the character degrees of every group H wth L (q) H Aut(L (q)). Recall that f q s even, then δ s an nner automorphsm, PGL (q) = L (q), and Aut(L (q)) = L (q) ϕ. Hence, f L (q) < H Aut(L (q)) wth q even, then H = L (q) ϕ k for some k f wth 1 k < f, and H/ L (q) s cyclc. If q s odd, then Aut(L (q)) = L (q) δ, ϕ and the outer automorphsm group of L (q) s Aut(L (q))/ L (q) = δ ϕ, where δ s of order and ϕ s of order f. The followng elementary lemma wll be useful n descrbng the subgroups of Aut(L (q)) n ths case. Lemma 6.1. If A = x y, where x = and y = f, then any subgroup of A that does not contan x s cyclc. 14

15 Proof. If f s odd, then A s cyclc, so we may assume f s even. In ths case, A contans exactly three elements of order, as does any noncyclc subgroup of A. Thus a subgroup of A not contanng the element x of order cannot contan three elements of order, hence s cyclc. Corollary 6.. If L (q) < H Aut(L (q)) wth q = p f, p an odd prme, then one of the followng occurs:. δ H so that PGL (q) H and H = PGL (q) ϕ k for some k f wth 1 k f;. H = L (q) ϕ k for some k f wth 1 k < f;. H = L (q) δϕ k for some k f wth 1 k < f and f/k even. Proof. If δ H, then H contans L (q) δ = PGL (q). If δ s not n H, then H/ L (q) s cyclc by the lemma, hence H = L (q) σ for some outer automorphsm σ. If H s not a subgroup of L (q) ϕ, then σ = δϕ k for some k f. Fnally, f/k s the order of ϕ k, and so f f/k s odd, then δ s n H. Corollary 6.3. Let L (q) H Aut(L (q)) and set G = PGL (q) f δ H and G = L (q) f δ H. If ˆχ Irr(H), then ˆχ(1) = χ(1) H : I H (χ), where χ s a consttuent of ˆχ G and I H (χ) s the stablzer of χ n H. Proof. We have that G H and H/G s cyclc. Hence a character χ Irr(G) extends to ts stablzer I H (χ) n H and each extenson nduces rreducbly to H by Theorem 6.11 of [4]. Every character of H lyng over χ wll therefore have degree χ(1) H : I H (χ). We frst determne the stablzers of characters of L (q) of degrees 1, q, and (q + ε)/, and the degrees of the characters of H lyng over these. Theorem 6.4. Let S = L (q) and let S H Aut(S). If χ Irr(S) has degree 1 or q, then every rreducble character of H lyng over χ has degree χ(1). Proof. By Lemmas 4.5 and 4.6, χ s nvarant n H. If H does not contan δ, then by Corollary 6., H/S s cyclc and χ extends to H. The result then follows from Gallagher s Theorem (see [4, 6.17]). If δ H, then by Lemma 4.5, χ extends to two rreducble characters of PGL (q). By Lemma 4.6, both of these are nvarant n H. Snce H/ PGL (q) s cyclc, these characters extend to H and the result agan follows from Gallagher s Theorem. Theorem 6.5. Let S = L (q) wth q odd and let S H Aut(S). Let µ Irr(S) wth µ(1) = (q + ε)/.. If H S ϕ, then µ s nvarant n H and every rreducble character of H lyng over µ s of degree (q + ε)/.. If H S ϕ, then the stablzer of µ n H s I H (µ) = H (S ϕ ) and H : I H (µ) =. Every rreducble character of H lyng over µ s of degree q + ε. Proof. If H S ϕ, then µ s nvarant n H by Lemma 4.6. Snce H/S s cyclc, µ extends to H and () follows from Gallagher s Theorem ([4, 6.17]). Assume now that H S ϕ, so δϕ k H for some nteger k. By Lemma 4.5, µ s not fxed by ths automorphsm, and so I H (µ) < H. By Lemma 4.6, µ s nvarant n H (S ϕ ) and we have H (S ϕ ) I H (µ) < H. Snce δ s an nner automorphsm, H : H (S ϕ ) =, and hence I H (µ) = H (S ϕ ). We have that I H (µ) S ϕ, and so I H (µ)/s s cyclc. Thus µ extends to I H (µ) and, by Gallagher s Theorem, each extenson has degree (q + ε)/. Fnally, by Clfford s Theorem, each extenson nduces to an rreducble character of H of degree H : I H (µ) (q + ε)/ = q + ε. 15

16 Theorem 6.6. Let S = L (q), where q = p f > 3 for a prme p, A = Aut(S), and let S H A. Let G = PGL (q) f δ H and G = S f δ H, and set H : G = d = a m, m odd. The degrees of the rreducble characters of H lyng over characters of G of degree q 1 are precsely (q 1) a l, where l s a postve dvsor of m, wth the excepton that f p = 3, f s odd, and H = L (q) ϕ, then l 1. Proof. Frst suppose δ H, so that H = PGL (q) ϕ f/d. (We nclude here the case where q s even.) By Corollary 6.3, for j d, there s a character of H of degree (q 1)j lyng over a character θ of PGL (q) of degree q 1 f and only f H : I H (θ) = j, hence f and only f I H (θ) = PGL (q) ϕ k, where k = (f/d)j. By Lemma 5.1, such a character θ of PGL (q) exsts f and only f f/k = d/j s odd, that s, j = a l for some l m. Suppose now that q s odd and δ H. By Corollary 6., H = L (q) δ c ϕ f/d, where c {0, 1} and f d s odd, then c = 0 snce δ H. Agan by Corollary 6.3, there s a character of H of degree (q 1)j lyng over a character θ of L (q) of degree q 1 f and only f I H (θ) = L (q) δ cj ϕ k, where k = (f/d)j. By Lemma 5., ths mples f/k = d/j must be odd, that s, j = a l for some l m. Conversely, f d/j s odd, then such a character θ of L (q) exsts except when p = 3 and k = 1. Snce d f, k = (f/d)j = 1 f and only f d = f (so f/d = 1) and j = 1. Observe that snce d = f s odd n ths case, c = 0 and H = L (q) ϕ. Theorem 6.7. Let S = L (q), where q = p f > 3 for a prme p, A = Aut(S), and let S H A. Let G = PGL (q) f δ H and G = S f δ H, and set H : G = d. The degrees of the rreducble characters of H lyng over characters of G of degree q + 1 are precsely (q + 1)j, where j s a postve dvsor of d, wth the followng exceptons:. f f s odd, p = 3, and H = A, then j 1;. f f s odd, p =, 3, or 5, and H = S ϕ, then j 1;. f f (mod 4), p = or 3, and H = S ϕ or H = S δϕ, then j. Proof. Frst, suppose q s odd and δ H, so that H = PGL (q) ϕ f/d. By Corollary 6.3, for j d, there s a character of H of degree (q + 1)j lyng over a character χ of PGL (q) of degree q + 1 f and only f H : I H (χ) = j, hence f and only f I H (χ) = PGL (q) ϕ k, where k = (f/d)j. By Lemma 5.3, such a character χ of L (q) exsts except when f s odd, p = 3, and k = 1. Snce d f, k = (f/d)j = 1 f and only f f = d (hence K = A) and j = 1, whch s excepton () n the statement of the theorem. Suppose now that ether q s odd and δ H or q s even. By Corollary 6., H = L (q) δ c ϕ f/d, where c {0, 1} and f d s odd, then c = 0 snce δ H. Agan, Corollary 6.3 mples that f j d, there s a character of H of degree (q + 1)j lyng over a character χ of L (q) of degree q + 1 f and only f I H (χ) = L (q) δ cj ϕ k, where k = (f/d)j. Lemma 5.4 mples that f p k {, 3,, 5, 3 } or f p k {, 3,, 5, 3 } and f/k s even, then such a character χ of L (q) exsts and H has a character of degree (q + 1)j. It therefore remans to consder the cases where p k {, 3,, 5, 3 } and f/k s odd. Suppose p =, 3, or 5, k = (f/d)j = 1, and f s odd. Snce d f, k = 1 mples that j = 1 and d = f, whch s odd, so that H = L (q) ϕ. In ths case, Lemma 5.4 mples there s no character of L (q) of degree q + 1 stablzed by H, hence H has no character of degree (q + 1)j for j = 1, whch s excepton () n the statement of the theorem. Fnally, suppose p = or 3, k = (f/d)j =, and f/ s odd, that s, f (mod 4). In ths case, k = (f/d)j = mples that ether j = 1 and d = f/, or j = and d = f. If j = 1 and d = f/, we have H = L (q) ϕ. By Lemma 5.4, there s a character χ of L (q) of degree q + 1 nvarant under ϕ, hence under H, and so H has a character of degree (q + 1)j wth j = 1. 16

17 If j = and d = f, then ether H = L (q) ϕ or H = L (q) δϕ, and the subgroup of H of ndex j = s K = L (q) ϕ. However, by Lemma 5.4, K s not the stablzer n H of any character of L (q) of degree q + 1 (any such character nvarant under ϕ s also nvarant under ϕ). Hence H does not have a character of degree (q + 1)j for j =, whch s excepton () n the statement of the theorem. Fnally, we observe that Theorem A now follows from Corollary 6.3 and Theorems 6.4, 6.5, 6.6, and Subgroups of (SL (q) δ ) ϕ and Ther Degrees Throughout ths secton we wll denote A = (SL (q) δ ) ϕ = GL (q) ϕ, where q = p f s odd. Our goal s to determne the character degrees of all subgroups of A of the form H = (SL (q) δ α ) ϕ β, where α q 1 and β f. We wll frst determne the characters of H 0 = SL (q) δ α lyng over the rreducble characters of SL (q) of degree 1, q, (q 1)/, and (q+1)/ and then determne the stablzers of these characters n H. As before, snce H/H 0 s cyclc, each character of H 0 wll extend to ts stablzer n H and then nduce rreducbly to H. For the characters of SL (q) of degrees q 1 and q+1, we have shown that each of the characters of degree q 1 or q + 1 from Lemmas 5.1 and 5.3 has an extenson to GL (q), hence also to H 0, wth the same stablzer n ϕ. The followng lemma wll be necessary for computng restrctons of characters of GL (q) to H 0. It s straghtforward to verfy the lemma usng the lst of conjugacy class representatves n [9]. Lemma 7.1. Let α q 1, H 0 = SL (q) δ α, and α = α/(, α). An element X of GL (q) s n H 0 f and only f the determnant of X s a power of ν α. Representatves of the the conjugacy classes of GL (q) that are contaned n H 0 are as follows:. A 1 (jα ), where 1 j (q 1)/α ;. A (jα ), where 1 j (q 1)/α ;. A 3 (l 1, l ), where 1 l q 1, l 1 l, and l 1 + l = jα for some nteger j; v. B 1 (jα), where 1 j (q 1)/α, q + 1 jα. Observe that f α s odd, each conjugacy class of GL (q) contaned n H 0 s a sngle conjugacy class of H 0, but f α s even, then the class of A (l) n GL (q) splts nto two conjugacy classes A (l) and A (l) of H Characters of H Lyng Over 1 SL and ψ The rreducble characters of SL (q) of degree 1 and q are the prncpal character 1 SL and the Stenberg character ψ, respectvely. These extend to the prncpal and Stenberg character of GL (q), denoted n [9] by 1 GL = χ (q 1) 1 and St = χ (q 1) q, respectvely. By Gallagher s Theorem, the remanng rreducble characters of GL (q) lyng over 1 SL and ψ are the products of these characters wth the q 1 rreducble characters of GL (q)/ SL (q), whch s cyclc of order q 1. Of course, the characters of ths quotent are precsely the extensons of 1 SL, whch are denoted χ (n) 1 for n = 1,,...,q 1 n [9], and the extensons of the Stenberg character are χ (n) q = St χ (n) 1 = χ (q 1) q χ (n) 1. The rreducble characters of H 0 lyng over 1 SL and ψ are therefore the restrctons of χ (n) 1 and χ (n) q = St χ (n) 1 to H 0. As the Stenberg character St s nvarant under ϕ, the stablzer n ϕ of the restrcton of χ (n) q to H 0 s equal to the stablzer of the restrcton of χ (n) 1. 17

18 Lemma 7.. Let q = p f be odd, H 0 = SL (q) δ α for α q 1, and l = (q 1)/α = H 0 : SL (q). Let β f. For n = 1,,...,q 1, the stablzer n ϕ β of (χ (n) and (χ (n) q ) H0 s ϕ βk, where k s the order of p β modulo l/(l, n). Proof. As noted above, we only need to consder the restrcton of χ (n) 1 to H 0. By Lemma 7.1, Lemma 3.3, and the values of χ (n) 1 gven n [9], the values of (χ (n) and ts mage under ϕ r for any r f are as follows: (χ (n) (χ (n) 1 )ϕ r H 0 A 1 (jα ) ρ njα ρ njα p r A (jα ) ρ njα ρ njα p r A 3 (l 1, l ) ρ njα ρ njαpr B 1 (jα) ρ njα ρ njαpr where ρ s a complex prmtve (q 1)th root of unty. Note that snce α q 1, we know that q + 1 α and so B 1 (α) s a conjugacy class of H 0. Therefore, f (χ (n) s nvarant under ϕ r (and hence under ϕ r ) we must have ρ nα = ρ nαpr, and thus nα nαp r (mod q 1). Conversely, f ths congruence holds, then, snce α s ether α or α, t s clear that (χ (n) s nvarant under ϕ r. We therefore have that (χ (n) s nvarant under ϕ r f and only f p f 1 nα(p r 1). Suppose now that ϕ r s n ϕ β, and let r = βk for a postve nteger k. Replacng α wth (q 1)/l, we have that (χ (n) s nvarant under ϕ βk f and only f p f 1 n((p f 1)/l)(p βk 1), hence f and only f l n(p βk 1), whch s equvalent to l/(l, n) (p βk 1). Therefore, f k s the order of p β modulo l/(l, n), then by defnton ϕ βk stablzes (χ (n), but no smaller postve power of ϕ β does, and so the stablzer of (χ (n) n ϕ β s ϕ βk. Theorem 7.3. Let q = p f be odd and set H = (SL (q) δ α ) ϕ β and H 0 = SL (q) δ α, where α q 1 and β f. Denote d = f/β = H : H 0 and l = (q 1)/α = H 0 : SL (q). The set of degrees of rreducble characters of H lyng over the prncpal character of SL (q) s {k : k d and k = O v (p β ) for some v l} and the set of degrees of rreducble characters of H lyng over the Stenberg character of SL (q) of degree q s {qk : k d and k = O v (p β ) for some v l}. Proof. By the remarks above, the rreducble characters of H lyng over 1 SL and ψ are precsely those lyng over the restrctons of χ (n) 1 and χ (n) q to H 0. As H/H 0 = ϕ β s cyclc, each rreducble character χ of H 0 wll extend to ts stablzer I n H and then nduce to H. Hence the degree of each character of H lyng over χ s H : I χ(1). Therefore, k s the degree of an rreducble character of H lyng over the prncpal character of SL (q) f and only f I k = H 0 ϕ βk s the stablzer n H of (χ (n) for some 1 n q 1. By Lemma 7., f I k s the stablzer n H of (χ (n), then k s the order of p β modulo the dvsor v = l/(l, n) of l. Conversely, suppose v l and k s the order of p β modulo v. Settng n = l/v, we have 1 n l q 1 and (l, n) = n = l/v. Hence k s the order of p β modulo v = l/(l, n), and I k s the stablzer n H of (χ (n). Therefore, the degrees of characters of H lyng over the prncpal character of SL (q) are as clamed. As the stablzers of (χ (n) and (χ (n) q ) H0 n H are the same, the second concluson n the lemma follows from the frst. 18