4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ According to GAP, SmallGroup(32,22) has a centre of order 8, and the derived subgroup has order 2. $\endgroup$ Sep 17, 2011 at 21:06
  • $\begingroup$ 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 $|G:\text{Z}(G)|=p^2$. $\endgroup$
    – yakov
    Jun 24, 2016 at 0:35

4 Answers 4

4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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! $\endgroup$
    – Steve D
    Sep 19, 2011 at 11:35
  • $\begingroup$ 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! $\endgroup$
    – Ralph
    Sep 19, 2011 at 19:50
0
$\begingroup$

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 $(C_{p^2}\times C_p)\rtimes C_{p^2}$.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 $|G:\text{Z}(G)|=p^2$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.