Prove $H$ is a proper normal subgroup of $G$ if $H$ is generated by $\{[x,y] \mid x,y \in G\} \cup \{x^p \mid x \in G\}$.

Question

I am trying to solve the following problem:

Let $G$ be a nontrivial finite $p$-group, where $p$ is a prime, and let $H$ be the subgroup of $G$ generated by the set $\{[x,y] \mid x,y \in G\} \cup \{x^p \mid x \in G\}$.

Prove that $H$ is a proper normal subgroup of $G$.

It is not hard to see that $H$ is a normal subgroup of $G$. Indeed, any element of $g \in G$ commutes with any element of the commutator. Moreover, $gx^pg^{-1} = (gxg^{-1})^p$. Since any generator of $H$ commutes with any element of $G$, the same is true for $H$. However, I do not know how to prove $H \neq G$.

Any thoughts?

Answer 1

Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.

Answer 2

It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.

As for why $H<G$, note that $G'\le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.