# Maximum value of the number of conjugacy classes of nonabelian p-groups with an abelian subgroup of index p

It is known that if $G$ is a nonabelian $p$-group of order $p^n$, with an abelian subgroup of index $p$, then the number $k(G)$ of conjugacy classes of $G$ can be as large as $p^{n-1} + p^{n-2} - p^{n-3}$, with equality if and only if $|G'| = p$, where $G'$ is the commutator subgroup of $G$. For $n = 5$, we have $k(G) \leq p^4 + p^3 - p^2$, and this upper limit is reached if and only if $|G'| = p$.

My question is, if for the case $n = 5$, $|G'| = p$, how large can the centre $Z(G)$ be? Of course, $|Z(G)| \leq p^3$, but we can get any better information?

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According to GAP, SmallGroup(32,22) has a centre of order 8, and the derived subgroup has order 2. – Sonia Balagopalan Sep 17 '11 at 21:06
If a nonabelian $p$-group $G$ has an abelian sugroup of index $p$, then $|G|=p|G'||\text{Z}(G)$. Therefore, if $|G'|=p$, then $|G:\text{Z}(G)|=p^2$. – yakov Jun 24 at 0:35

I just realized, that the arguments below work in greater generality:

Let $G$ be a $p$-group with an abelian subgroup $A$ of index $p$. Then there is a short exact sequence $$1 \to Z(G) \to A \to G' \to 1.$$ In particular, $|Z(G)|\cdot |G'| = |A| = |G|/p$.

Proof: Let $g \in G$ such that $gA$ generates $G/A$. Then $G = \lt A,g \gt$. Since $A$ is abelian, the map $$f: A \to G', a \mapsto [g,a]$$ is a homomorphism of groups. Let $B \le A$ be the kernel of $f$. The elements of $B$ commute with elements of $A$ and with $g$. Thus $B \le Z(G)$.

Next, let's show $Z(G) \le A$: Since $G$ is not abelian, $G'$ is non-trivial. As the intersection of the center of a $p$-group with a non-trivial normal subgroup is non-trivial, we find $a_0 \in Z(G) \cap G'$ of order $p$. Let's assume $f(A) = G'$ has already been shown. Then, there is $a_1 \in A$ such that $f(a_1) = a_0$, i.e. $ga_1g^{-1} = a_0a_1$. Now, if $z$ is any element of $Z(G)$, write $z=ag^i$ ($a \in A$). Hence $$1 = [z,a_1] = ag^ia_1g^{-i}a^{-1}a_1^{-1} = (g^ia_1g^{-i})a_1^{-1} = ...= (a_0^ia_1)a_1^{-1} = a_0^i.$$ Since $a_0$ has order $p$, $i$ is a multiple of $p$ and from $g^p \in A$, $z \in A$ follows.

Thus, we can apply $f$ to $Z(G)$ and $f(Z(G))= \lbrace 1 \rbrace$ yields $Z(G) \le B$, whence $Z(G) = B$ and the sequence $1 \to Z(G) \to A \xrightarrow[]{f} G' \to 1$ is exact.

Finally, let's show $f(A) = G'$. From definition, $[g,a] \in f(A)$ for all $a \in A$. Suppose $[g^i,a] \in f(A)$ for all $a \in A$. Then, $$[g^{i+1},a] = g(g^iag^{-i})g^{-1}a^{-1} = g[g^i,a]ag^{-1}a^{-1}=g[g^i,a]g^{-1}(gag^{-1}a^{-1})$$ $$=[g^i,gag^{-1}][g,a] \in f(A).\hspace{125pt}$$ If $x=ag^i, y=bg^j \in G$, then a simple computation, using that $A$ is abelian, shows $[x,y] = [g^i,b][g^j,a^{-1}] \in f(A)$, finishing the proof. qed.

Old version:

$|G'| = p$ is a strong condition that severely limits the possibilities for $G$. In fact, it can be shown:

If $G$ is a non-abelian $p$-group with $|G'| = p$, which has an abelian subgroup $A$ of index $p$, then $Z(G)$ is an index-$p$ subgroup of $A$. In particular, $|Z(G)| = |G| / p^2$.

Remark 1: $G$ can be of any order $p^n$

Remark 2: From the proof below, it should not be to hard, to classify all groups $G$ from the statement.

Proof: Since $G/A$ is abelian, $G' \le A$. Let $G' = \lt a_0 \gt$. Since the center of a $p$-group non-trivially intersects every non-trivial normal subgroup, $a_0 \in Z(G)$ follows. Let $g \in G$ represent a generator of $G/A$ (note: $G = \lt A,g \gt$) and let $$f: A \to G', a \mapsto [g,a].$$ Since $A$ is abelian and $G$ non-abelian and $G'$ cyclic of prime order, $f$ is a surjective homomorphism. Let $B$ be the kernel of $f$ and let $a_1 \in A$ with $[g,a_1] = a_0$, i.e. $ga_1g^{-1}=a_0a_1$. Because elements from $B \le A$ commute with $g$ and $A$, one conlcudes $B \le Z(G)$.

Next I show $Z(G) \le A$: Let $z=ag^i \in Z(G)$. Thus $$1 = [z,a_1] = ag^ia_1g^{-i}a^{-1}a_1^{-1}= (g^ia_1g^{-i})a_1^{-1}=...=(a_0^ia_1)a_1^{-1}=a_0^i$$ and since $a_0$ has order $p$, $i$ is a multiple of $p$, implying $z \in A$ (note: $g^p \in A$).

Hence $B \le Z(G) \lvertneqq A$. Finally, surjectivity of $f$ implies $(A:B) = p$, whence $Z(G) =B$ and $Z(G) = |G|/p^2.\quad$qed.

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I would just like to mention that (1) this a great answer and (2) a somewhat more general version of your theorem can be found as Lemma 4.6 in Isaacs's Finite Group Theory. The fact that $Z(G)\le A$ can also be proved by contradiction: if not, then $G=AZ(G)$ and that easily implies $G$ is abelian! – Steve D Sep 19 '11 at 11:35
I'm pleased to hear that you like the answer. Thanks for your proof of $Z(G) \le A$ by contradiction: That's much better than my lengthy considerations! – Ralph Sep 19 '11 at 19:50

A GAP computations gives the following:

Let $p=2,3,5,7,11,13,17$, and let $G$ be a group of order $p^5$ such that $|G'|=p$. Then $|Z(G)|=p$ or $p^3$.

Further, for these primes $p$ it seems that there are only two groups such that its center has $p$ elements. These groups are semi-direct products of the form $(C_{p^2}\times C_p)\rtimes C_{p^2}$.

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Answer on the question. Must be |Z(G)|\in{p,p^3} only. Asssume, however, that |Z(G)|=p^2. If A\le G is minimal nonabelian, then A\ne G (otherwise, |Z(G)|=p^3). In that case G=A*C_G(A), and it is easily seen that the order of the center of such G is not p^2.

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If a nonabelian $p$-group $G$ has an abelian subgroup of index $p$, then $|G|=p|G '||\text{Z}(G)|$. Therefore, if $|G'|=p$, then $|G:\text{Z}(G)|=p^2$.

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