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I am trying to prove the following:

Let $|G|=p^n$ for $n\geq 1$ and $p$ prime. Prove that $|\mathcal{Z} (G)|\neq p^{n-1}$.

Here is what I have so far:

Suppose, to the contrary, that $|\mathcal{Z} (G)|=p^{n-1}$. Then there is a $g\in G$ such that $g\notin \mathcal{Z} (G)$. Let $H=\left\{g^nz:n\in \mathbb{Z}, z\in \mathcal{Z} (G)\right\}$. We show that $H\leq G$ by showing that if $a,b \in H$, then $ab^{-1}\in H$.

If $a,b \in H$, then $a=g^nz_1$ and $b=g^mz_2$ for some $n,m \in \mathbb{Z}$ and $z_1, z_2 \in \mathcal{Z} (G)$. Then since elements of $\mathcal{Z} (G)$ commute with all elements of $G$, and $\mathcal{Z} (G) \leq G$, we have $$ab^{-1}=(g^nz_1)(z_2^{-1}g^{-m})=g^n(z_1z_2^{-1})g^{-m}=g^{n-m}(z_1z_2^{-1}).$$ The element $g^{n-m}(z_1z_2^{-1})\in H$, since $(n-m)\in \mathbb{Z}$ and $z_1z_2^{-1}\in \mathcal{Z} (G)$ (since $\mathcal{Z} (G) \leq G$). Hence, $H \leq G$. Furthermore, $H$ is abelian since $$ab=(g^nz_1)(g^mz_2)=g^n(z_1g^m)z_2=g^n(g^mz_1)z_2=(g^{n+m})(z_1z_2)=g^{m+n}(z_2z_1)=g^m(g^nz_2)z_1 =g^m(z_2g^n)z_1=(g^mz_2)(g^nz_1)=ba.$$ Also, $\mathcal{Z} (G) \leq H$, since if $|g|=k$ for some $k\in \mathbb{Z}$ (we know such a $k$ exists because $G$ is finite) then $g^kz=1z=z\in H$ for all $z\in \mathcal{Z} (G)$.

Okay, here is where I run into a problem. I want to come to a contradiction by arguing that the order of $H$ is greater than the order of $\mathcal{Z} (G)$ since for example, $gz \notin \mathcal{Z}(G)$ for any $z\in \mathcal{Z}(G)$. Someone told me I would come to the conclusion that $H=G$, but how?? I can't see why $|H|=|G|=p^n$, and even if I did, how does that give me a contradiction. I'm confused! Some help would be greatly appreciated. Thanks!

Sarah
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    You could try to prove more generally that if $G/Z(G)$ is cyclic, then $G$ is abelian. That's a rather common exercise, and implies the desired result directly. – Daniel Fischer May 13 '14 at 15:28
  • And what @DanielFischer suggests is one of m.se's favorite questions: see https://math.stackexchange.com/questions/63087/if-g-zg-is-cyclic-then-g-is-abelian and various linked threads. – darij grinberg May 05 '23 at 22:47

2 Answers2

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Suppose $|Z(G)|=p^{n-1}$. Then, for $g\in G\setminus Z(G)$: $$Z(G)\lneq C_G(g)\lneq G$$ Contradiction, because there isn't any power of $p$ between $p^{n-1}$ and $p^n$, as requested by Lagrange's theorem.

citadel
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Sarah, despite your efforts, are you familiar with the fact that if $|G/Z(G)|$ is a prime number then $G$ is cyclic? So in your case you would get $p^n=p^{n-1}$, which is absurd.

Nicky Hekster
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  • I was not familiar with this fact. I'll have to see if it is in the Dummit & Foote book as a theorem or an exercise. Otherwise, I will have to prove it myself. Thanks. – Sarah May 13 '14 at 15:33
  • Your "proof" above is already a good step in that direction. Try to do it yourself! – Nicky Hekster May 13 '14 at 15:35
  • I'm being silly. I already knew that if a group has prime order then it is cyclic. – Sarah May 13 '14 at 15:37
  • Be careful - yes, $G/Z(G)$ is cyclic (generated by for example $gZ(G)$, with $g \notin Z(G))$ and try to conclude that $G$ must be abelian. – Nicky Hekster May 13 '14 at 15:44