ANALYSIS OF SMALL GROUPS

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1 ANALYSIS OF SMALL GROUPS 1. Big Enough Subgroups are Normal Proposition 1.1. Let G be a finite group, and let q be the smallest prime divisor of G. Let N G be a subgroup of index q. Then N is a normal subgroup of G. Proof. The group G acts on the coset space G/N by left translation, giving a map from G to the symmetric group on q letters, G Aut(G/N) S q. Let K denote the kernel of the map. Clearly K N since each g K must in particular left translate N back to itself. Thus, since q is the smallest prime divisor of G, q = G/N G/K = q (product of primes p q). On the other hand, the first isomorphism theorem says that G/K is isomorphic to the image of G in S q, a subgroup of S q. Thus G/K q!, so that G/K = q (product of primes p < q). Comparing the two displays shows that G/N = G/K, and so the containment K N now gives K = N. Thus N is normal because it is a kernel. 2. Semidirect Products First we discuss what it means for a given group to be a semidirect product of two of its subgroups. Let G be a group, let K be a normal subgroup, and let Q be a complementary subgroup. That is, G = KQ, K G, K Q = 1 G. Define a map σ : Q Aut(K), q ( σ q : k qkq 1). Thus σ qq = σ q σ q for all q, q Q, i.e., σ is a homomorphism. And of course σ q (kk ) = σ q (k)σ q (k ) for all k, k K since conjugation is an inner automorphism. Then the group-operation of G is kq k q = kq kq 1 q q = kσ q ( k) q q. This group structure describes G as a semidirect product of K and Q. notation is G = K σ Q. When Q acts trivially on K by conjugation, i.e., when all elements of K and Q commute, the semidirect product is simply the direct product. Regardless of whether the semidirect product is direct, we have a short exact sequence 1 K G Q 1, 1 The

2 2 ANALYSIS OF SMALL GROUPS where the first map is inclusion and the second is kq q. Furthermore, the sequence splits, in that the composite Q G Q, where the first map is inclusion, is the identity. The short exact sequence makes clear that K is so-named because it is the kernel group in the sequence, and similarly Q is so-named because it is the quotient group. Second we discuss ingredients that suffice to construct a semidirect product. Suppose that we have the data a group K, a group Q, a homomorphism σ : Q Aut(K). Define the following operation on the set G = K Q: (k, q)(k, q ) = (kσ q (k ), qq ). Note that the operation does not assume any sort of product between elements of K and elements of Q. The operation is associative, The identity is (1 K, 1 Q ), ((k, q)(k, q )) (k, q ) = (kσ q (k ), qq ) (k, q ) = (kσ q (k )σ qq (k ), (qq )q ) = (kσ q (k σ q (k )), q(q q )) = (k, q) (k σ q (k ), q q ) = (k, q) ((k, q )(k, q )). (k, q)(1 K, 1 Q ) = (kσ q (1 K ), q1 Q ) = (k, q), (1 K, 1 Q )(k, q) = (1 K σ 1Q (k), 1 Q q) = (k, q). And the inverse of (k, q) is (σ q 1(k 1 ), q 1 ), (k, q)(σ q 1(k 1 ), q 1 ) = (kσ q (σ q 1(k 1 )), qq 1 ) = (1 K, 1 Q ), (σ q 1(k 1 ), q 1 )(k, q) = (σ q 1(k 1 )σ q 1(k), q 1 q) = (1 K, 1 Q ). Thus G is a group. If we identify K with its embedded image K 1 Q in G and similarly identify Q with 1 K Q in G then the group operation becomes the semidirect product law, and K is normal, kq k q = kσ q (k )qq, (, q)(k, 1 Q )(, q) 1 = (, q)(k, 1 Q )(, q 1 ) = (, 1 Q ). Thus the data K, Q, and σ give a semidirect product construction G = K σ Q without assuming that K and Q are subgroups of a common group. 3. Groups of Order p 2 Let G = p 2 where p is prime, and let G act on itself by conjugation. The class formula gives p 2 = Z(G) + O x p 2 / G x.

3 ANALYSIS OF SMALL GROUPS 3 The sum is a multiple of p, and hence so is Z(G). We are done unless Z(G) = p. In this case, consider a noncentral element g. Its isotropy group contains Z(G) and g, making it all of G. This contradicts the noncentrality of g. Thus Z(G) = p 2, i.e., Z(G) = G, i.e., G is abelian. 4. Metacyclic Groups of Order pq Let p and q be primes with q > p. We seek nonabelian groups G of order pq. Any q-sylow subgroup of G, K = {1, a,, a q 1 }, is big enough to be normal. Thus, letting a p-sylow subgroup be Q = {1, b,, b p 1 }, we have G = KQ, K G, K Q = 1 G. That is, G is a semidirect product of K and Q. The only question is how Q acts on K by conjugation. The automorphisms of K are a a e for any nonzero e Z/qZ, and the composition of a a e and a a f is a a ef. That is, Aut(K) (Z/qZ). Elementary number theory (see below) shows that there is at least one generator g modulo q such that {1, g, g 2, g 3,, g q 2 } gives all the values 1, 2, 3,, q 1 modulo q. That is, g q 1 = 1 mod q is the first positive power of g that equals 1 modulo q, so that Aut(K) is cyclic of order q 1. Therefore, there are nontrivial maps σ : Q Aut(K) if and only if p q 1. In this case, the unique order p subgroup of Aut(K) is isomorphic to {1, g (q 1)/p, g 2(q 1)/p, g (p 1)(q 1)/p }, whose nontrivial elements are precisely the values i Z/qZ such that i 1 but i p = 1. Thus the nontrivial maps Q Aut(K) are b ( a a i), i 1 mod q but i p = 1 mod q. In sum, nonabelian groups of order pq exist only for p q 1, in which case they are a, b a q = b p = 1, bab 1 = a i where i 1 mod q but i p = 1 mod q. Especially, if p = 2 then the only possibility is i = 1 mod q, giving the dihedral group D q. For a given p and q with p q 1, different values of i give isomorphic groups. To see this, first note that any two such values i and i satisfy i = i e mod q for some e {1,, p 1}. Now let b = b e. Then the relations a q = b p = 1, bab 1 = a i become a q = b p = 1, ba b 1 = a i.

4 4 ANALYSIS OF SMALL GROUPS To see where the last relation comes from, compute b e ab e = b e 1 bab 1 b (e 1) = b e 1 a i b (e 1) = (b e 1 ab (e 1) ) i = (b e 2 ab (e 2) ) (i2 ) = = a (ie). Since b p = 1, this same calculation with p in place of e shows again why we need i p = 1 mod q, forcing p to divide q 1. In general, any semidirect product K σ Q where K and Q are cyclic (not necessarily of prime order) is called metacyclic A Brief Excursion into Elementary Number Theory. We have cited the following result. Proposition 4.1. Let q be prime. Then (Z/qZ) is cyclic, with ϕ(q 1) generators. An elementary proof is possible, and indeed it is standard. But we have the tools in hand to give a more sophisticated argument. First of all, if (Z/qZ) is cyclic then our analysis of cyclic groups has already shown that it has ϕ(q 1) generators. So only the cyclicity is in question. The proof begins with the observation that a polynomial over a field can not have more roots than its degree. Lemma 4.2. Let k be a field. Let the polynomial f k[x] have degree d 1. Then f has at most d roots in k. Naturally, the field that we have in mind here is k = Z/qZ. The lemma does require that k be a field, not merely a ring. For example, the quadratics polynomial X 2 1 over the ring Z/24Z has eight roots, {1, 5, 7, 11, 13, 17, 19, 23} = (Z/24Z). Proof. If f has no roots then we are done. Otherwise let a k be a root. The polynomial division algorithm gives f(x) = q(x)(x a) + r(x), deg(r) < 1 or r = 0. (Here the quotient polynomial q(x) is unrelated to the prime q in the ambient discussion.) Thus r(x) is a constant. Substitute a for X to see that in fact r = 0, and so f(x) = q(x)(x a). By induction, q has at most d 1 roots in k and we are done. Now, since (Z/qZ) is a finite abelian group, it takes the form (Z/qZ) Z/d 1 Z Z/d 2 Z Z/d k Z, 1 < d 1 d 2 d k. The additive description of the group shows that the equation d k X = 0 is solved by all d 1 d 2 d k group elements. Multiplicatively, the polynomial X d k 1 (Z/qZ)[X] has d 1 d 2 d k roots. Thus k = 1 and so (Z/qZ) is cyclic.

5 ANALYSIS OF SMALL GROUPS 5 5. Groups of Order 8 Let G be a nonabelian group of order 8. Then G must contain a subgroup a of order 4 but no element of order 8. The subgroup a is big enough to be normal. Suppose that G has no other subgroup of order 4. Consider an element b that does not lie in the subgroup generated by a. Then we have (since G is not abelian) a 4 = b 2 = 1, ba = a 3 b. The displayed conditions describe the dihedral group D 4. Otherwise G has a second subgroup b of order 4. The left cosets of a are itself and b a, so that b 2 a. Thus a 2 = b 2. Now we have and so To understand the group better, let Then so that, since c 1 = b 3 a 3 = ab 3 = a 3 b, a 4 = b 4 = 1, a 2 = b 2, ba = a 3 b, G = {1, a, a 2, a 3, b, ab, a 2 b, a 3 b}. c = ab. c 2 = ab ab = a 4 b 2 = b 2 = a 2, ab = c, ba = a 3 b = c 1, bc = bab = a 3 b 2 = a, cb = ab 2 = a 3 = a 1, ca = aba = b, ac = a 2 b = b 3 = b 1. We see that the G is the group of Hamiltonian quaternions. 6. Groups of Order 12 The left- Consider a nonabelian group G of order 12. Let K = {1, a, a 2 } be a 3-Sylow subgroup, so that G/K = 4. translation action of G on the coset space G/K gives a homomorphism σ : G Aut(G/K) S 4, g (σ g : γk gγk). (Note that Aut(G/K) is a group even though G/K may not be.) The kernel of σ is a subgroup of K since for any g in the kernel we must have gk = K. If σ has trivial kernel then G = A 4. (Recall that A n is the unique index-2 subgroup of S n. Indeed, if H S n doesn t contain some 3-cycle then there are at least three cosets. So an index-2 subgroup contains all 3-cycles, making it A n.) Otherwise, the kernel of σ is K, making K a normal subgroup of G. Let Q be a 2-Sylow subgroup of G. Since G = KQ, K G, K Q = 1 G, G is a semidirect product of K and Q. The only question is how Q acts on the generator a of K by conjugation.

6 6 ANALYSIS OF SMALL GROUPS The order-4 group Q is abelian. If its isomorphism type is C 2 C 2 then it takes the form Q = {1, b, c, bc} where b 2 = c 2 = 1 and cb = bc. In this case, the only nontrivial action of Q on K is, up to relabeling, bab = a, cac = a 2, and so G = a, b, c a 3 = b 2 = c 2 = 1, ba = ab, cb = bc, ca = a 2 c. Here the element ab has order 6, and its inverse is a 2 b, and its conjugate under the order 2 element c is its inverse, cabc = a 2 b. Thus G = D 6. On the other hand, if the isomorphism type of Q is C 4 then Q = {1, b, b 2, b 3 }. In this case, the only nontrivial action of Q on K is bab 1 = a 2, and so Alternatively, let ã = ab 2. Then also G = a, b a 3 = b 4 = 1, ba = a 2 b. G = ã, b ã 6 = 1, b 2 = ã 3, bã = ã 5 b. We don t yet know that such a group exists, but in fact it manifests itself as a subgroup of the cartesian product S 3 C 4, specifically the subgroup generated by ã = ((1 2 3), g 2 ), b = ((1 2), g), where g generates C 4.

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