Exercise 14.4.18

Proof. (a) Let $g\ne e$ be in $A$, and consider

the subgroup of $M$ generated by the elements $\mathit{hg}{h}^{-1}$ as $h$ varies over all elements of $M$. Since $g\in A$, where $A$ is normal in $M$, then $\mathit{hg}{h}^{-1}\in A$ and $h{g}^{-1}{h}^{-1}\in A$ for all $h\in M$, thus $B\subset A$.

Moreover $B$ is normal in $M$: if $m\in M$, for all generators $\mathit{hg}{h}^{-1}$ of $B$, $m(\mathit{hg}{h}^{-1})m^{-1}=(\mathit{mh})g{(\mathit{mh})}^{-1}\in B$, and similarly $m{(\mathit{hg}{h}^{-1})}^{-1}{m}^{-1}=(\mathit{mh})g^{-1}{(\mathit{mh})}^{-1}\in B$.

If $g_1=h_{1}g{h}_1^{-1},\dots ,g_l=h_{l}g{h}_l^{-1}$ are the generators of $B$, every element $b\in B$ is of the form

Therefore

Here $B\subset A\subset M$, where $B$ is normal in $M$, and $A$ is a minimal normal subgroup of $M$, therefore $B=\{e\}$ or $B=A$. But $B\ne \{e\}$, since $e\ne g\in B$. Thus $B=A$:

(b) Let $g\in M,b\in Z(A)$. $A$ is normal in $G$, thus for all $a\in A$, $a^{\prime }=g^{-1}\mathit{ag}\in A$. Since $b\in Z(A)$, $a^{\prime }b=b{a}^{\prime }$, that is $(g^{-1}\mathit{ag})b=b(g^{-1}\mathit{ag})$, which implies, by left-multiplication by $g$ and right-multiplication by $g^{-1}$,

This relation is true for every $a\in A$, thus $\mathit{gb}{g}^{-1}\in Z(A)$, for all $g\in M$ and for all $b\in Z(A)$. This proves that $Z(A)$ is a normal subgroup of $M$.

Note: More generally, if $K\subset A\subset M$, where $K$ is a characteristic subgroup of $A$, and $A$ a normal subgroup of $G$, then $K$ is normal in $G$ (see [Issacs] Finite Group Theory, Lemma 1.10 p. 11): “conjugation by $g$ maps $A$ onto itself, and it follows that the restriction of this conjugation map to $A$ is an automorphism of $A$ (but not necessarily an inner automorphism of $A$). Since $K$ is characteristic in $A$, it is mapped onto itself by this automorphism of $N$”. Here $Z(A)$ is a characteristic subgroup of $A$, so this Lemma gives a solution to part (b). □