If $H$ is a group with a subset $X$ such that $H$ is generated by $\left<X, Z(H) \cap H'\right>$, show that $H=\left<X\right>.$

Question

If $H$ is a group with a subset $X$ such that $H$ is generated by $\left<X, Z(H) \cap H'\right>$, show that $H=\left<X\right>.$

I've been having difficulty with this for quite a while.

I've been given the hint to evaluate $[zx, y]$ if $z$ commutes with both $x$ and $y$ and worked out that it would equal $[x, y]$ but I can't begin to work out how this is helpful.

I'm also a bit confused as to if this is even possible. Since $Z(H)$ is normal, so is $Z(H) \cap H'$ and so couldn't you just define $X$ as $H/(Z(H) \cap H')$ as a counterexample?

Answer 1

Hint: show that $H’ \cap Z(H) \subseteq \Phi(H)$, the Frattini-subgroup of $H$ (the intersection of all maximal subgroups of $H$). The elements of this subgroup are non-generators.

Here is a proof: if $M$ is maximal in $H$, then either $H=MZ(H)$ or $M=MZ(H)$ (note $MZ(H)$ is a subgroup between $M$ and $H$). In the first case: $H'=M'$, whence $H' \cap Z(H) \subseteq H'=M' \subseteq M$. In the second case, $Z(H) \subseteq M$, so $H' \cap Z(H) \subseteq Z(H) \subseteq M$. So either way, $H' \cap Z(H) \subseteq M$. It follows that $H' \cap Z(H) \subseteq \Phi(H)$.