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Let G be a nonabelian group of order $p^3$, where $p$ is a prime number. Prove that the center of $G$ is of order $p$.

Proof

Since $G$ is not abelian, the order of its center cannot be $p^3$. Since it is a $p$-group, the center cannot be trivial. So the order of $Z(G)$ is either $p^2$ or $p$.

Suppose, for contradiction, that $Z(G) = p^2$. Since $p$ is prime, we can assume that a subgroup $H= \langle p \rangle$ of order $p$ exists in $G$. We can also assume that $H$ and $Z(G)$ are disjoint. Otherwise, if there didn't exist a disjoint subgroup of order $p$, then the order of $G$ would be $p^2$. Since $Z(G)$ is the center, they commute with p. Since they commute with $p$, they must also commute with all powers of $p$. So $G = Z(G) \times H \implies$ G is abelian since $H$ and $Z(G)$ are abelian. So $|Z(G)|=p$.

Do you think my answer is correct?

Thanks in advance

azimut
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    You can use that if $G/Z(G)$ is cyclic then $G$ is abelian, and then $G/Z(G)$ would have order $p$ if $Z(G)=p^2$ and hence it would be cyclic implying $G$ is abelian . – user10444 Jul 14 '13 at 22:28
  • @user1044 Thanks a lot. That's a much shorter proof. But I'm just wondering...is my answer still correct? –  Jul 14 '13 at 22:35
  • Just a clarification when you write $H=\langle p \rangle $ do you mean $p$ is an element of $G$ (I am checking the details now) – user10444 Jul 14 '13 at 22:36
  • @user1044 Yes...I should have clarified that... –  Jul 14 '13 at 22:38
  • Your proof contains several things I don't understand. How do you conclude that $G\cong Z(G)\times H$? When you say $H$ and $Z(G)$ are disjoint, do you mean they have trivial intersection? Why is that the case? – Ittay Weiss Jul 14 '13 at 22:39
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    But how do you know $p \in G$ ? The only thing you know is that $G$ is non abelian, you can say that since $p \mid # G$ then by Sylow's first theorem there is an element $a$ of order $p$. – user10444 Jul 14 '13 at 22:43
  • @Artus I just noticed something $H$ is cyclic and hence abelian then $H \subset Z(G)$ they can't be disjoint. – user10444 Jul 14 '13 at 22:49
  • @user10444 Thanks for your comment about $p \in G$. Actually I meant that $p$ represents an element in $G$, not that the "number" $p$ is in $G$. Anyway, my notation was a bit weird, and it didn't make sense...so thanks for your comment. –  Jul 14 '13 at 23:01
  • @julien I mixed the up, I forgot the element has to commute with all the group. – user10444 Jul 14 '13 at 23:02
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    @IttayWeiss I'm not sure...maybe I'm wrong. –  Jul 14 '13 at 23:08
  • @Artus could you clarify on why $H$and $Z(G)$ are disjoint suppose I take $x\in H \cap Z(G) \not= e_G$ then what? – user10444 Jul 14 '13 at 23:10
  • @user10444 Yeah, that's why I'm probably wrong... –  Jul 14 '13 at 23:34
  • Mostly the same question, though not for proof verification – ccorn Jul 14 '16 at 17:16

3 Answers3

13

I have a question can somebody please clarify. Why is the possibility that $|Z(G)|=1$ not considered? Thanks!

UPDATE:

Here is my attempt at a solution.

Consider $Z(G)$ center of the group $G$. We know that $Z(G)\leq G$.

By Lagrange's Theorem $|Z(G)|$ must divide $|G|$.

Since $|G|=p^{3}$ the only possibilities are $1, p, p^{2}, p^{3}$.

$|Z(G)|\neq p^{3}$ because otherwise we will have $Z(G)=G$ but $G$ is non-abelian.

$|Z(G)|\neq p^{2}$ also because otherwise we will have the order of the factor group by the center as $|G/Z(G)|=|G|/|Z(G)|= p^{3}/p^{2} = p$.

We have:

$|G/Z(G)|=p \implies G/Z(G)$ is cyclic $\implies G$ is abelian. But $G$ is non-abelian.

Now $|Z(G)|\neq 1$ also because $G$ is a $p-group$ and $p-groups$ have non-trivial center.

(thanks to Tobias Kildetoft for pointing this out).

Thus the only possible order for $Z(G)$ is $p$.

$QED$

Jyrki Lahtonen
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chowching
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  • I realize you do not have the required rep to comment, but please don't post answers asking for clarifications. The reason $|Z(G)|=1$ is not considered is that it is a well-known result that the center of a $p$-group is non-trivial. – Tobias Kildetoft Sep 05 '13 at 08:24
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    Sorry about striking out the first sentence of your post. A flagger thinks it is not an answer, because you asked for clarification there. Granted, you did not answer the question Do you think my answer is correct?, so the flag is not without merit in that sense. – Jyrki Lahtonen Jun 21 '15 at 15:55
  • Sorry to comment on a 8 years old answer, but I don't quite get why $\vert Z(G) \vert = p^{2} $ have been discarded? – AKP2002 Dec 02 '21 at 11:07
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    @AKP2002 Because then anything outside the center would then commute with everything, so the group would be abelian, contrary to hypothesis. – Robert Shore Dec 16 '21 at 20:08
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I don't think the proof is correct. The following doesn't make sense to me.

We can also assume that $H$ and $Z(G)$ are disjoint. Otherwise, if there didn't exist a disjoint subgroup of order $p$, then the order of $G$ would be $p^2$.

As pointed out by user10444, typically you apply the statement

If G/Z(G) is cyclic, then $G$ is abelian.

to exclude the case $\lvert Z(G)\rvert = p^2$.

The statement is not hard to show: Let $gZ(G)$ be a generator of the cyclic group $G/Z(G)$. Then $G = \langle \{g\}\cup Z(G)\rangle$. Since each pair of generators of $G$ commutes, $G$ is abelian.

azimut
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2

Slightly different approach:

The conjugacy class equation $|G|=|Z|+\sum_{a\notin Z}\frac{|G|}{|N(a)|}\rightarrow p\mid|Z|$

$G$ non-abelian $\rightarrow |Z|=p$ or $p^2$

Let $a\in G,a\notin Z\rightarrow N(a)\neq G$

$a\notin Z \wedge a\in N(a)\rightarrow Z<N(a)<G$

$\rightarrow |N(a)|=p^2$ and $|Z|=p$

ZSMJ
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