CZ-GROUPS. Kristijan Tabak and Mario Osvin Pavčević Rochester Institute of Technology and University of Zagreb, Croatia

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1 GLASNIK MATEMATIČKI Vol. 51(71)(2016), CZ-GROUPS Kristijan Tabak and Mario Osvin Pavčević Rochester Institute of Technology and University of Zagreb, Croatia Abstract. We describe some aspects of the structure of nonabelian p-groups G for which every nonabelian subgroup has a trivial centralizer in G, i.e. only it s center. We call such groups CZ-groups. The problem of describing the structure of all CZ-groups was posted as one of the first research problems in the open problems list in Yakov Berkovich s book Groups of prime power order Vol 1 ([1]). Among other features of such groups, we prove that a minimal CZ-group must contain at least p 5 elements. The structure of maximal abelian subgroups of these groups is described as well. 1. Introduction and definitions Throughout the entire paper we will think of G as a finite p-group. We assume that every nontrivial nonabelian subgroup has a trivial centralizer in G, i.e. C G (H) = Z(H) for every nonabelian H G. We start by a formal definition of our main object to be investigated. Definition 1.1. A finite p-group G is called a CZ-group if for every nonabelian subgroup M < G, the centralizer of M in G equals to Z(M). We will shortly write G CZ p. We start our analysis with one quite trivial observation about the center of any G CZ p. Proposition 1.2. Let G CZ p. Then Z(G) A, for every nonabelian A G Mathematics Subject Classification. 20D15, 20D25. Key words and phrases. p-group, center, centralizer, Frattini subgroup, minimal nonabelian subgroup. This work has been fully supported by Croatian Science Foundation under the project

2 346 K. TABAK AND M.-O. PAVČEVIĆ Proof. Let s assume the opposite. Take some nonabelian group A < G. If g Z(G)\A, then g C G (A) Z(A). Thus, we have a contradiction with G CZ p. Now, we would like to describe the case when Z(G) is not contained in some abelian subgroup. In that case we prove that such an abelian subgroup can t be a maximal subgroup of some nonabelian subgroup. To be more precise, we have: Lemma 1.3. Let G CZ p. Let A G be an abelian subgroup such that Z(G) A. If M > A is nonabelian, then [M : A] p 2. Proof. Take g Z(G) \ A, where A is an abelian subgroup of G. Let M G be a nonabelian group such that [M : A] = p. Then A M. Notice that, from Z(G) M we conclude that there is a g Z(G) M such that [ g,a : A] p and g,a M, because of which, M = g,a is abelian. That is a contradiction, hence [M : A] p 2. In some sense, it is a natural question to ask could CZ structure be inherited from G to some smaller subgroup. The following result provides the answer. Proposition 1.4. Let G CZ p. If N < M < G, where N is nonabelian, then M CZ p. Proof. Since N is nonabelian, clearly M 1. So C G (N) = Z(N) and C G (M) = Z(M). Assume that C M (N) > Z(N). Then there is some g C M (N)\Z(N).SinceC M (N) C G (N), wehaveg C G (N)\Z(N).Thus, C G (N) Z(N) which is a clear contradiction. Therefore, C M (N) = Z(N), thus M CZ p. 2. Center of a nonabelian subgroup of a CZ-group Next topic that we cover is the question of the center of a CZ-group G. To be more specific, we will provide properties of the center of a maximal nonabelian subgroup of G and compare the center of G with centers of some of its subgroups. Lemma 2.1. Let G CZ p and M G is nonabelian. Then Z(G) Z(M). Proof. Assume that g Z(G) \ Z(M). Then g M, otherwise g Z(M). Hence, g C G (M) \ Z(M). This is a contradiction with C G (M) = Z(M). Next result deals with the center of a maximal nonabelian subgroup. Theorem 2.2. Let G CZ p and let M < G be a maximal nonabelian subgroup. Then one of the following is true:

3 CZ-GROUPS Z(M) = Z(G), 2. Z(M) > Z(G) and [G : M] = p. Proof. Take M < G, where M is maximal nonabelian. If Z(M) > Z(G), take g Z(M) \ Z(G). Then there is some x G \ M such that [x,g] 1 and g M. Notice that M,x is nonabelian. Assume that x p M. Then x p Z(M) = C G (M), so M,x p is also nonabelian. If M,x p = G, then x p is a generator. On the other hand x p Φ(G). Therefore, x is not a generator. Hence, M < M,x p < G, which is a contradiction with the assumption that M is maximal nonabelian. So, x p M and herewith we have proved [G : M] = p. Lemma 2.3. Let G CZ p and A < G be abelian of index p. Then for every x G\A, there is a nonabelian M < G, such that x p Z(M). Proof. Take x G\A. We know that G/A = xa. Take y A, such that [x,y] 1 (there is always such an y, otherwise G would be abelian). Take M = x,y. Then C G (M) = Z(M). Notice that x p A, so [x p,y] = 1. Now, it is clear that x p Z(M). 3. Minimal CZ-groups In this section we deal with CZ-groups which don t possess any nontrivial CZ-subgroup. We shall name such groups minimal CZ-groups. We start with a definition of a minimal CZ-group. Definition 3.1. A group G CZ p is called a minimal CZ-group if it doesn t possess a nontrivial CZ-subgroup. By Proposition 1.4, it is straightforward to see that if G is a minimal CZ-group, then every proper nonabelian subgroup is a minimal nonabelian group. For a CZ-group G which is determined to be minimal in this sense, we shall write G CZm p. For the sake of completeness, we repeat here the known result that classifies minimal nonabelian p-groups. Theorem 3.2. Let G be a minimal nonabelian p-group. Then G = p and G/G is abelian of rank 2. G is isomorphic to one of the following groups: 1. G = a,b a pm = b pn = 1, a b = a 1+pm 1, m 2, n 1 and G = p m+n, 2. G = a,b, a pm = b pn = c p = 1, [a,b] = c, [a,c] = [b,c] = 1, where G = p m+n+1, and if p = 2, then m+n > 2 and G is maximal cyclic normal subgroup, 3. G = Q 8. Our next result answers the question on the number of generators of a given G CZm p.

4 348 K. TABAK AND M.-O. PAVČEVIĆ Theorem 3.3. If G CZm p, then [G : Φ(G)] p 3. Proof. Let A < G be some maximal abelian subgroup (meaning that there is no abelian subgroup B such that A < B < G). Then A < x,a G for some x G \ A. It is clear that there is an a A such that [a,x] 1 (otherwiseawouldn tbemaximalabelian). TakeM = a,x.clearlym > 1. If M = G, then [G : Φ(G)] = p 2. If [G : M] p 2, then there is some N < G such that M < N < G. Therefore N CZ p which contradicts to G CZm p. If [G : M] = p, then it is clear that G = a,x,y, for some y G\M. Hence [G : Φ(G)] p 3. Notice that if G CZm p, then it is natural to assume G p 4. Otherwise, any proper subgroup would be of order at most p 2, thus abelian. In orderto deliveradescription ofaminimal CZ-group,we need toprovidesome information about automorphisms of minimal nonabelian groups, since such groups, as we have seen above, are main ingredients of minimal CZ-groups. Thus, we will start our analysis with groups of order p 4. For that, we need some technical results regarding the modular group of order p 3, which may be a minimal nonabelian subgroup of a putative CZ-group G. Throughout this paper we will denote that modular group and its generators and relations by M p 3 = a,b a p2 = b p = 1, a b = a 1+p. Another option is that a nonabelian subgroup of order p 3 is given by N = a,b, a p = b p = c p = 1, [a,b] = c, [a,c] = [b,c] = 1 and this notation of the group N will be kept throughout the paper as well. It is easy to see that for M p 3 the following holds: b j a i = a i(1 p)j b j, and (a i b j ) k = a i[1+(1 p)j +(1 p) 2j + +(1 p) (k 1)j] b kj for k 2. On the other hand, o(a i b j ) = p 2 for any i 0,p and i [p 2 1]\ {0,p},whileo(a p b j ) = p.finally, it isalsoeasyto seethat(a i b j ) b = (a i b j ) p+1. Also, because of a M p 3 we may assume that any automorphism of M p 3 is of the form a α b β a αk (a pi b j ) β for some integers k, i and j. The next result gives a description of such automorphisms. Lemma 3.4. Let M p 3 = a,b a p2 = b p = 1, a b = a 1+p. Let ϕ ij : M p 3 M p 3 be maps defined by ϕ ij (a α b β ) = a α (a pi b j ) β. Then 1. ϕ ij (a α b β a γ b δ ) = a α+γ(1 p)β +pi (1 p)(β+δ)j 1 (1 p) j 1 b j(β+δ), 2. ϕ ij (a α b β )ϕ ij (a γ b δ ) = a α+pi (1 p)βj 1 (1 p) j 1 +γ(1 p)jβ +pi(1 p) βj (1 p)δj 1 (1 p) j 1 b j(β+δ). Proof. Notice that a α b β a γ b δ = a α (b β a γ )b δ = a α a γ(1 p)β b β b δ. Therefore, we have ϕ ij (a α b β a γ b δ ) = ϕ ij (a α+γ(1 p)β b β+γ ) = a α+γ(1 p)β (a pi b j ) β+δ.

5 CZ-GROUPS 349 But, on the other hand (a pi b j ) β+δ = a pi[1+(1 p)j +(1 p) 2j + +(1 p) j(β+δ 1)j] b j(β+δ), hence ϕ ij (a α b β a γ b δ ) = a α+γ(1 p)β +pi (1 p) (β+δ)j 1 (1 p) j 1 b j(β+δ). Furthermore, we have ϕ ij (a α b β )ϕ ij (a γ b δ ) = a α (a pi b j ) β a γ (a pi b j ) δ Let us introduce shortcuts = a α a pi[1+(1 p)j +(1 p) 2j + +(1 p) j(β 1)] b jβ a γ a pi[1+(1 p)j +(1 p) 2j + +(1 p) j(δ 1)] b jδ. A = 1+(1 p) j +(1 p) 2j + +(1 p) j(β 1) = (1 p)βj 1 (1 p) j 1, B = 1+(1 p) j +(1 p) 2j + +(1 p) j(δ 1) = (1 p)δj 1 (1 p) j 1. We get ϕ ij (a α b β )ϕ ij (a γ b δ ) = a α a pia b jβ a γ a pib b jδ = a α+pia (b jβ a γ+pib )b jδ = {since b j a i = a i(1 p)j b j } = a α+pia a (γ+pib)(1 p)jβ b jβ b jδ = a α+pi (1 p)βj 1 (1 p) j 1 +γ(1 p)jβ +pi(1 p) βj (1 p)δj 1 (1 p) j 1 b j(β+δ). Throughoutthe comingresultswe will deal with the assumption that M p 3 is a normal subgroup of G. Proposition 3.5. Let G be a p-group and M p 3 = a,b G. Let d G\M be such that a d = a and b d b. Then [b,d] = 1. Proof. Since M p 3 G, the action via conjugation is an inner automorphism of M p 3. Let us use the notation (a α b β ) d = ϕ ij (a α b β ) = a α (a pi b j ) β. Then we would have ϕ ij (a α ) = a α = (a α ) d, ϕ ij (b β ) = (a pi b j ) β. Let us assume that b d = b j. Then ϕ ij (b) = a pi b j = b d, so pi 0 mod p 2, hence i {0,p}. We proved that ϕ ij (a α b β a γ b δ ) = a α+γ(1 p)β b j(β+δ). On the other hand ϕ ij (a α b β )ϕ ij (a γ b δ ) = a α a 0 b jβ a γ a 0 b jδ = a α (a jβ a γ )b jδ = a α a γ(1 p)jβ b jβ b jδ = a α+γ(1 p)jβ b j(β+δ). Therefore, it is necessary that a γ(1 p)β = a γ(1 p)jβ, so γ(1 p) β γ(1 p) jβ mod p 2. Now, we must have γ [ (1 p) jβ (1 p) β] 0 mod p 2. From here

6 350 K. TABAK AND M.-O. PAVČEVIĆ we get β(1 j) 0 mod p. Since this must be true for any β, we conclude that 1 j 0 mod p, so b d = b. Proposition 3.6. Let G be a p-group and M p 3 G. Let g G \ M p 3 such that a d = a and b d a p b. Then b d = a p b. Proof. Notice that (a α b β ) d = a α (a p b j ) β = ϕ 1j (a α b β ). It s necessary that ϕ 1j (a α b β )ϕ 1j (a γ b δ ) = ϕ 1j (a α b β a γ b δ ). Using Lema 3.4 we get On the other hand ϕ 1j (a α b β a γ b δ ) = a α+γ(1 p)β +p (1 p) (β+δ)j 1 (1 p) j 1 b j(β+δ). βj ϕ 1j (a α b β )ϕ 1j (a γ b δ 1 ) = a α+p(1 p) (1 p) j 1 +γ(1 p)βj +p(1 p) βj (1 p)δj 1 (1 p) j 1 b j(β+δ). Let us use abbreviations Λ = α+γ(1 p) β +p (1 p)(β+δ)j 1 (1 p) j, 1 Π = α+p (1 p)βj 1 (1 p) j 1 +γ(1 p)βj +p(1 p) βj(1 p)δj 1 (1 p) j 1. So, it is necessary that Λ Π mod p 2. We see that [ p 1+(1 p) j +(1 p) 2j + +(1 p) (β 1)j] pβ mod p 2. Similarly, we get and Hence, p (1 p)(β+δ)j 1 (1 p) j 1 (β +δ)p mod p 2 p (1 p)δj 1 (1 p) j 1 δp mod p2. a Λ = a α+γ(1 p)β +(β+γ)p = a Π = a α+βp+δp+γ(1 p)jβ. Then, γ(1 p) β γ(1 p) βj mod p 2. Thus we get only one possibility: j 1 mod p. Therefore, b d = a p b. Now we will focus our analysis to groups of order p 4. The main goal is to determine all minimal CZ-groups of order p 4. Proposition 3.7. Let G = a,b,c be a group of order p 4, p 2, where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If [a,c] = 1 and b c b, then also [b,c] = 1.

7 CZ-GROUPS 351 Proof. If [a,c] = 1, then a c = a. If b c b, then b c = b j for some j [p]. Notice that (a α b β a γ b δ ) c = a α+γ(1 p)β b (β+δ)j, while on the other hand we have (a α b β ) c (a γ b δ ) c = a α b jβ a γ b jδ = a α+γ(1 p)jβ b jβ+jδ. This gives us α+γ(1 p) jβ α+γ(1 p) β mod p 2. Since α, γ [p 2 ] and β [p], we get γ(1 p) jβ γ(1 p) β mod p. Now, it is easy to see that j 1 mod p, thus b c = b. Proposition 3.8. Let G = a,b,c be a group of order p 4, p 2, where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If a c = a and b c a p b, then b c = a p b. Proof. Put b c = a p b j. We use an idea that is similar to the previous proof. Firstly, notice that because of b j a i = a i(1 p)j b j we get (a α b β a γ b δ ) c = a α+pβ+pδ a γ(1 p)βj b βj+δj. On the other hand, after we use the automorphism property, we get (a α b β a γ b δ ) c = a α+pβ+pδ a γ(1 p)β b βj+δj. This gives us j 1 mod p. Notice that if G = a,b,c is of order p 4 and p > 2, where M p 3 = a,b, then from the assumption G CZ we get o(c) p 2. Otherwise, o(c) = p 3, implying c G to be maximal abelian. Then G would be M p 4, hence G is not a CZ-group. A contradiction. It can be shown that if G = a,b,c CZ p and G = p 4, p > 2 and M p 3 = a,b, then the assumption [a,c] = 1 yields b c = a p b and o(c) = p 2, with an additional property a p = c p. The alternative possibility is o(c) = p. But the next result shows that this is not possible. Theorem 3.9. Let G = M p 3,c be of order p 4, where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If a c = a, b c = a p b and o(c) = p, then G CZ p. Proof. Let A = a,c = C p 2 C p. Then A is a maximal abelian subgroup. Take φ : A A where φ(x) = [x,b]. We know that φ is a homomorphism and Im(φ) = G = a p. Since a ac = a, c ac = c, b ac = b, then ac Z(G). Notice that o(ac) = p 2 and G = p 4 = p Z(G) G. Thus Z(G) = p 2, hence Z(G) = ac. Clearly ac M p 3. Therefore ac C G (M p 3) \ Z(M p 3). So, M p 3 doesn t have a trivial centralizer, hence G CZ p. Lemma Let G = M p 3,c CZ p be of order p 4, where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If a c a and b c b, then [b,c] = 1. Proof. Take a c = a i, b c = b j where i [p 2 1], j [p 1]. Then (a α b β a γ b δ ) c = (a α a γ(1 p)β b β+δ ) c = a αi+γ(1 p)βi b (β+δ)j. On the other hand we have (a α b β a γ b δ ) c = a αi b βj a γi b δj = a α+γi(1 p)βj b (β+δ)j. This leads us to γi[(1 p) βj (1 p) β ] 0 mod p 2. Since γ is any integer, we get p (1 p) βj (1 p) β. Hence j = 1.

8 352 K. TABAK AND M.-O. PAVČEVIĆ Theorem Let G = M p 3,c be a group of order p 4, p > 2 where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If o(c) = p 2, then a c M p 3 \ a or a c > 1. Proof. Assume a c = a i and a c = 1. Then G = a,c and c p M \ a. Therefore c p = a pi b j where i {1,p} and j [p 1]. Notice that because of Z(M p 3) = a p we have (c p ) k = (a pi ) k b jk. Furthermore, a cp = a api b j = a bj = a (1+p)j = a 1+pj. For every k N we have a (api b j ) k = a bjk = a (1+pj)k. If we assume that a (1+pj)k = a, then (1 + pj) k 1 0 mod p 2. Therefore, p(kj 1) 0 mod p 2 and kj 1 mod p. Notice that such k always exists (and is not divisible by p) since C p is a field. Therefore, without losing generality we can take c p = b. Take ϕ Aut( a ) = Aut(C p 2) = C p(p 1) (here we need the assumption p > 2). Put ϕ(a) = a i. Then ϕ p (a) = a ip = a cp = a b = a 1+p. On the other hand ϕ p(p 1) (a) = a, hence (p+1) p mod p 2. This gives us p 0 mod p 2, which is an obvious contradiction. Therefore, a c M \ a or a c > 1. Proposition Let G = M p 3,c be a CZ-group of order p 4, p > 2 where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If o(c) = p 2 and a c a, then [b,c] 1. Proof. Assume that the claim is not true. That means [b,c] = 1. Notice that c C G (M p 3)\M p 3, otherwisegwouldn t be acz-group. Since a c a, then by the previous theorem a c > 1. Notice that a c = a p. We can write a p = c p. Take a c = a 1+i. Then a a 1+i is an automorphism of order p, thus a cp = a (1+i)p = a 1+pi = a, hence pi 0 mod p 2. So without losing generality we may write a c = a 1+p. Now, look at the element cb p 1. For it we have [c,cb p 1 ] = [b,cb p 1 ] = 1. On the other hand a cbp 1 = (a 1+p ) bp 1 = (a (p+1)p 1 ) p+1 = a (p+1)p = a 1+p2 = a. Hence, cb p 1 C G (M p 3)\M p 3, which is a contradiction. Thus, [b,c] 1. Lemma Let G = M p 3,c be a CZ-group of order p 4, p > 2 where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If a c = a 1+p, then b c = a p b j where j 0. Proof. Since M p 3 G, then b c M p 3. If b c a, then o(b) = p implies that we can write b c = a p. Therefore b c Φ(M p 3)char M p 3, where Φ(M p 3) stands for the Frattini subgroup, which is characteristic. Thus b Φ(M p 3) and so not a generator of M p 3, contradiction. Therefore b c = b j or b c = a p b j, where j 0. If b c = b j, then b c b j is an automorphism of order p 1 (since Aut(C p ) = C p 1 ). Therefore, b cp 1 = b, so [b,c p 1 ] = 1. On the other hand a cp 1 a and c p 1 is clearly a generator for G. But this contradicts Proposition 3.12.

9 CZ-GROUPS 353 Theorem Let G = M p 3,c be a CZ-group of order p 4, p > 2 where M p 3 = a,b a p2 = b p = 1, a b = a 1+p. If o(c) = p 2 and a c = a 1+p, then b c = a p b. Proof. Because of Theorem 3.11 we have a c > 1, thus without losinggeneralitywemaywrite a p = c p. BecauseofM p 3 G,we haveb c M p 3. As we have proved in the previous Lemma, we have b c = a p b j a p. From Theorem 2.2 we have Z(M p 3) = a p Z(G) > 1. Thus Z(G) = a p. Put z = a p. Then c 1 bc = zb j. From here we get bc = czb j and b 1 cb = zb 2 cb j+1 = zb p 2 cb j+1. Using this, we get c b = zb p 3 (bc)b j+1 = zb p 3 (czb j )b j+1 = z 2 b p 3 cb 2j+1 = = z k b p k 1 cb kj+1. Put k = p 1. Then c b = z p 1 cb j(p 1)+1 = z p 1 cb 1 j. Now, let us take a group N = c,b. Since c N, we get c b c. From here we get b 1 j c. If b 1 j b, then b 1 j = b c, so G = a,c is of order p 3. Thus, the only option is b 1 j = 1, hence b j = b. Proposition Let M p 3 = a,b a p2 = b p = 1, a b = a 1+p. Let G = M p 3,c be a CZ-group of order p 4 where o(c) = p 2. Then a G. Proof. Let us assume the opposite. Since M p 3 G, then a c = a i b j, where o(a i ) = p 2 and b j 1. Take N = a,c. Because a c p we have N = p 3, therefore N G. So a c N. This gives us a c = a i b j N, hence b j N. If b j 1, then b N and N = G, which gives us a contradiction. So, the only case is b j = 1 and a c a, hence a G. Now, we have only one candidate G = a,b,c a p2 = c p2 = b p = 1, a b = a c = az, b c = bz, z = a p for a CZ-group of order p 4 that contains M p 3. The next result will provide an answer regarding the status of such group. Theorem The group G = a,b,c a p2 = c p2 = b p = 1, a b = a c = az, b c = bz, z = a p is not a CZ-group. Proof. Notice that M p 3 = a,b a p2 = b p = 1, a b = a 1+p G. Let us assume that G is a CZ-group. Then, by previous results we have Z(G) = Z(M p 3) = a p = z. Take x = ab p 1 c. Then x M p 3, otherwise c M p 3 and G = p 4. Then a x = a abp 1c = a bp 1c = (a b ) bp 2c = (az) bp 2 c = za bp 2c = z(a b ) bp 3c = z 2 a bp 2c = = z p 1 a c = z p 1 za = a, which gives us x C G (a). On the other hand it is clear that from b 1 ab = az we get ab = zba, thus ba = z 1 ab. Using this, we get b a = a 1 ba = a 1 (z 1 ab) = z 1 b. Therefore b x = b abp 1c = (z 1 b) bp 1c = (z 1 b) c = z 1 b c = z 1 bz = b,

10 354 K. TABAK AND M.-O. PAVČEVIĆ thus x C G (b). So, x C G (M p 3)\Z(M p 3), which is a contradiction with the assumption that G is a CZ-group. In other words, we have proved the following result: Theorem If G CZm p is of order p 4, p > 2, then M p 3 G. It is easy to check the properties of an exponent of a minimal CZ-group of order p 4. Lemma Let G CZm p and G = p 4. Then exp(g) p 2. Proof. Let exp(g) > p 2. If exp(g) = p 4 then G = C p 4 CZ. If exp(g) = p 3, then there is some d G such that d G. Hence G = M p 4, which is not a CZ-group. Theorem 3.2 motivates us to deal with the possible minimal CZ-group that contains minimal nonabelian subgroup different than modular. Proposition Let G CZm p be of order p 4. Let N = a,b,c a p = b p = c p = 1, a b = ac, [a,c] = [b,c] = 1 G. Then there is a T N, such that T = C p C p and T G. Proof. We know that N = p 3 and N is minimal nonabelian group. Also N = c is maximal cyclic normal subgroup in N. Thus every maximal subgroup in N is isomorphic to C p C p. Take Γ 1 = {T N [N : T] = p}. As we ve seen, T = C p C p for every T Γ 1. We also know that Γ 1 1 mod p. Take d G\N such that G = N,d (such d always exists). Since d p N, we know that if d acts on Γ 1 nontrivially (via conjugation), then the orbits are of order p or 1. Therefore, there is some T Γ 1 which is fixed by conjugation with d. Hence T d = T, therefore T G. Now, we will use the previous result to describe any minimal CZ-group of order p 4 that contains a subgroup of order p 3 isomorphic to N. Proposition Let G CZm p be of order p 4 and N = a 1,b 1,c 1 a p 1 = bp 1 = cp 1 = 1, ab1 1 = a 1c 1, [a 1,c 1 ] = [b 1,c 1 ] = 1 G. Then there is a T N such that T = a,b = C p C p and T G. Additionally, N = T,b and [a,b] = c Z(N). Proof. We know that Z(N) = c 1. Then also C p = Z(N) Z(G) > 1. Thus, Z(N) = Z(G). Also, by previous result, we know that there is some T = a 2,c 2 = C p C p that is normal in G. Take some b 2 N \T. Since N = T,b 2, we must have a b2 2 a 2. Otherwise, N would be abelian. Since T G, then a b2 2 = ai 2 cj 2 for some i,j. Notice that if cj 2 1, then a 2,c 2 = a 2,c j 2. Therefore, we can write a b2 2 = ai 2c 2. Another option is a b2 2 = ai 2. Since T G we have T Z(G) > 1. If a b2 2 = ai 2, then we can write c 2 = T Z(N). Then N = E p 3 (elementary abelian group) which is a contradiction. So, the only

11 CZ-GROUPS 355 option is a b2 2 = ai 2 c 2. Thus, [a 2,b 2 ] = a i 1 2 c 2, thus a i 1 2 c 2 = c 2 = Z(N) = N. Therefore i = 1. Now, identify a 2 = a, b 2 = b, c 2 = c. Theorem Let G CZm p be of order p 4. Then G has no subgroup isomorphic to the minimal nonabelian group N = a,b,c a p = b p = c p = 1, a b = ac, [a,c] = [b,c] = 1. Proof. Let us assume the opposite. Let N G, where N = a,b,c a p = b p = c p = 1, a b = ac, [a,c] = [b,c] = 1. By our previous result, without losing generality, we can write T = a,c G. Since G/T = p 2, it is abelian, hence T G. Thus G p 2. Since G Φ(G) then also Φ(G) p 2. If Φ(G) = p 3, then dim(g) = 1 and G = C p 4, which is clearly a contradiction. Hence, the only option is Φ(G) = p 2 and Φ(G) = G = T. Therefore, G has 2 generators. Put G = x,y. Then it is clear that G = [x,y] = C p 2, which is a clear contradiction with G = T = C p C p. Using Theorems 3.17 and 3.21 we have reached one of the main results of this paper. We establish now the lower bound for the order of minimal CZ-groups. Theorem Let G CZm p. Then G p 5. Proof. If G CZm p CZ, then G has some nonabelian subgroup S < G such that C G (S) = Z(S). It is clear that S > p 3, thus G p 4. If G = p 4, then there is some minimal nonabelian S < G. So S = M p 3 or N. Both cases were eliminated by Theorems 3.17 and Maximal normal abelian subgroup of G CZ p This section collects another type of results, it deals with CZ-groups and repercussions on its maximal abelian subgroups. First, we provide a slightly different proof of Lemma from Berkovich s and Janko s book Groups of Prime Power Order, Vol. 2 ([2]). We will use notation A p B if A is a subgroup of B whose index is p. Similarly, we will write A p B if A B and [B : A] = p. Lemma 4.1. Let G be a p-group and A G its maximal abelian subgroup. Then for any x G\A there is some a A such that [x,a] 1 and [x,a] p = 1. Furthermore, [a, x, x] = 1, thus x, a is minimal nonabelian, i.e. every p- group is generated by minimal nonabelian subgroups. Proof. Take C A (x) for some x G\A. Clearly C A (x) < A, since otherwise x,a would be abelian and would contain A, which is a contradiction. Take x C A (x). It is a group because of x C A (x) = C A (x) x. One can see that x C A (x) < x A. Take B x A such that x C A (x) p B. Notice that x C A (x) A p A. Clearly x C A (x) A = C A (x). ThereforeC A (x) p A B.

12 356 K. TABAK AND M.-O. PAVČEVIĆ On the other hand, take b = x i a B (where a A.) Take g C A (x). Then (because g A) we have b g = (x i a) g = (x g ) i a g = x i a = b. Thus C A (x) Z(B). Now take a (A B) \C A (x) such that a p C A (x). Clearly [a,x] 1. Since [x,a] B Φ(B), [x,a] C A (x) x p B. Let us assume that [x,a] C A (x). Then [x,a] = x i a 1 for some a 1 C A (x). Then x 1 a 1 xa = x i a 1 and x a = x i+1 a 1. On the other hand, x xa = x xi+1 a 1 = x a1 = x, thus x a C A (x). Then (x a ) a 1 C A (x) a 1 = C A (x), so x C A (x) a contradiction! Therefore, [x,a] C A (x), so B C A (x). Knowing that for any finite group it holds [x,yz] = [x,z][x,y] z, we have 1 = [x,a p ] = [x,a p 1 a] = [x,a][x,a p 1 ] a. Because of [x,a p 1 ] B C A (x), we get [x,a p ] = [x,a][x,a p 1 ]. This gives us [x,a] p = 1. Finally, because of [x,a] C A (x), we get [a,x,x] = 1. Now we want to characterize the centralizer of a maximal abelian subgroup of a CZ-group because it certainly could be used for further describing the structure of a CZ-group. Lemma 4.2. Let G CZ p and let A G be a maximal abelian subgroup. Let x G\A. Then C A (x) Z(N) for every nonabelian N x A. Proof. Take x G \ A. If C A (x) = A, then x,a > A is abelian contradictingtotheassumptionforabeingmaximal. Thus, C A (x) < A.Take N x A, where N is nonabelian. Then, G CZ p implies C G (N) = Z(N). Take g C A (x) and y N. Then y = x j a 1, a 1 A. Hence, (x j a 1 ) g = (x g ) j a g 1 = xj a 1, because [g,x] = 1. Also, g A, so [g,a 1 ] = 1. Therefore, g C G (N) = Z(N), hence C A (x) Z(N). Corollary 4.3. Let G CZ p and let A G be a maximal abelian subgroup. Let x G \ A. Then for every nonabelian N x A, it holds Z(N)\C A (x) x C A (x). Proof. We know that C A (x) Z(N) < N x A. Take g = x j a 1 Z(N) \ C A (x). Then [g,x] = 1 and x g = x xj a 1 = x a1 = x which yields a 1 C A (x). Thus, Z(N)\C A (x) x C A (x). Corollary 4.4. Let G CZ p and let A G be maximal abelian subgroup. Let x G\A. Then for every nonabelian N x A, it holds C A (x) Z(N) x C A (x). Proof. Since C A (x) Z(N) and Z(N)\C A (x) x C A (x) we have [Z(N)\C A (x)] C A (x) x C A (x) C A (x). Because C A (x) x C A (x), we have Z(N) x C A (x). Hence Z(N) x C A (x).

13 CZ-GROUPS 357 Theorem 4.5. Let G CZ p. and let A Gbe a maximal abelian subgroup. Let x G\A such that x p A. Then Z( x A) = C A (x) = C G ( x A). Proof. Take x G \ A and x p A. Then, by previous result, we have C A (x) Z( x A) x AC A (x). Clearly [ x C A (x) : C A (x)] = p. If Z( x A) = x C A (x), then x Z( x A), hence [x,a] = 1. A contradiction with maximalityofa. ThereforeZ( x A) = C A (x) = C G ( x A). Lastequality is true due to G CZ p. Corollary 4.6. Let G CZ p and let A G be a maximal abelian subgroup. Then C A (T \A) Z(T) < A for any T G such that A < T. Theorem 4.7. Let G CZ p and let A G be a maximal abelian subgroup. Then for any x G\A there is some a A such that C = a,x is minimal nonabelian and C A (x) = Z(C) A C. Proof. We already know that for any x G \ A, there is some a A such that C = a,x is minimal nonabelian. Take t C A (x) \ C. Then t A and [t,a] = [t,x] = 1. Hence t C G (C) = Z(C) (due to G CZ), which is an obvious contradiction. Therefore, t C and C A (x) C. Now, take g C A (x). Then [g,x] = [g,a] = 1, so g Z(C) A. So far we have C A (x) Z(C) A. Now, take s Z(C) A but s C A (x). Then [s,x] = 1, so s C A (x). Again a contradiction. This gives us Z(C) A C A (x). Finally, we present our second main result, the full description of the centralizer of a generator that lies outside of a maximal abelian subgroup. Theorem 4.8. Let G CZ p and let A G be a maximal abelian subgroup. If x G \ A such that x p A, then Z(B) = C A (x) where B = x,a. Furthermore, there is a minimal nonabelian group M = x,a, where a A such that Z(M) = C A (x) A and M A p M. Proof. Take x G \ A such that x p A. Put B = x,a. Clearly A p B. Take some g C A (x). Then g Z(B) since [g,x] = [g,a 1 ] = 1 for any a 1 A. Hence C A (x) Z(B). Now, take h Z(B)\C A (x). If h A, then h C A (x). A contradiction. If h A, then h B \ A and [h,a] = 1. Therefore A,h > A is abelian, contradiction with the choice of A. Thus, Z(B) C A (x). We know that there is some a A such that M = x,a is minimal nonabelian. We already know that Z(M) A = C A (x). Additionally, we have B = MA. Then B = M A. This gives us [B : A] = [M : M A] = p. M A Thus M A p M. Since M is minimal nonabelian, we have Z(M) p M A p M.

14 358 K. TABAK AND M.-O. PAVČEVIĆ References [1] Y. Berkovich Groups of Prime Power Order Vol. 1., Walter de Gruyter GmbH & Co. KG, Berlin, [2] Y. Berkovich and Z. Janko, Groups of Prime Power Order Vol. 2., Walter de Gruyter GmbH & Co. KG, Berlin, [3] Y. Berkovich and Z. Janko, Groups of Prime Power Order Vol. 3., Walter de Gruyter GmbH & Co. KG, Berlin, K. Tabak Rochester Institute of Technology, Zagreb Campus D.T. Gavrana 15, Zagreb Croatia M.-O. Pavčević Department of applied mathematics Faculty of Electrical Engineering and Computing University of Zagreb Zagreb Croatia Received: