Commutator Subgroup in a $p$-group

Question

Let $G$ be a finite non-trivial $p$-group. Show that $G'$ (the commutator subgroup of $G$), is a proper subgroup of $G$.

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How could one show this result? I was thinking of first arguing that the center $Z(G)$ is non-trivial since $G$ is a non-trivial finite $p$-group. I'm not sure on how to proceed though.

Answer 1

Since $G$ is a $p$-group, all maximal subgroups must be normal.

If $M$ is a maximal subgroup, then $G/M$, being of order $p$, is cyclic, hence $......$ (can you complete this sentence?)

Answer 2

Proceed by induction on the order of $G$ and use the fact that for a $p$-group $G$, its center $Z(G)$ is non-trivial. If $|G|=p$, then $G$ is cyclic and hence $G'=\{1\}$. Now let $G$ be a $p$-group, then $G/Z(G)$ has definitely an order smaller than that of $G$. By induction, $(G/Z(G))'=G'Z(G)/Z(G) \subsetneq G/Z(G)$. Hence $G'Z(G) \subsetneq G$, so $G' \subsetneq G$.