Prove/disprove: Let $G$ s.t $|G| = p^3 \implies |G'| \leq p$.

Question

Let $G$ s.t $|G| = p^3$ for some prime $p$. Prove or disprove: $|G'| \leq p$.

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I couldn't think of any counter-examples so I started proving this and I'm stuck unfortunately and was hoping to seek some help.

So we know if $G$ is a non-trivial $p$-group then $Z(G) \neq \{e\}$ therfore we have $|Z(G)| \in \{p,p^2,p^3\}$.

Consider $|Z(G)| = p^3$, then $Z(G) = G \implies G$ is abelian $\implies G' = \{e\}$.

Otherwise consider $|Z(G)| = p \implies |{G/Z(G)}| = p^2 \implies {G/Z(G)}$ is abelian $\implies G' \leq Z(G) \implies |G'| \leq p$.

Now consider $|Z(G)| = p^2$ I haven't manage to prove this and I also can't think of any group of order $p^3$ which has a center $|Z(G)| = p^2$ so I'd love some help.

EDIT

So following Mark's hint consider $|Z(G)| = p^2$. This implies $|G/Z(G)| = p \implies G/Z(G)$ is cyclic which implies $G$ is abelian. If $G$ is abelian then $G = Z(G)$ which is a contradiction to the assumption $|Z(G)| = p^2$.

Answer 1

Hint: It's an easy exercise that if $G/Z(G)$ is cyclic then $G$ is abelian. So, can a group of order $p^3$ have center of order $p^2$?

Answer 2

Tried to explain it to myself:

1. Every finite $p$ group $\ne (e)$ has a non-trivial center.

2. If the quotient of a group by its center is cyclic, then that quotient is in fact trivial ( i.e. the group is abelian).

3. If a finite $p$ group is non-abelian then its center has index $\ge p^2$.

So now consider for a finite $G$ $p$- group. Its ascending central series

$G_0 = \{e\}$, $G_{i+1}/G_i = Z (G/G_i)$ stabilizes at $G$. Let $m$ minimal such that $G_m = G$. There are two cases:

1. $m=1$. We get $G$ abelian.

2. $m\ge 2$.

Now, $G_{m-1}/G_{m-2} = Z(G/G_{m-2})$, and $G/G_{m-2}$ is not abelian ( since $G_{m-1} \ne G_m$), so $|G:G_{m-1}| = |G/G_{m-2} : G_{m-1}/G_{m-2}| \ge p^2$. Now, since $G/G_{m-1} $ is abelian, ( $G_m = G$), we get $[G/G_{m-1}, G/G_{m-1}] = (\bar e)$, that is,

$[G,G] \subset G_{m-1}$ so $|G: [G,G]| \ge |G\colon G_{m-1}|$

In the end, the conclusion: if $G$ is a finite $p$-group of order $\ge p^2$, then $|G\colon [G,G]| \ge p^2$.