Exercise 14.3.6

Proof. (a) Consider

For all $a,b\in N$, ${\phi }_g(\mathit{ab})=\mathit{gab}{g}^{-1}=\mathit{ga}{g}^{-1}\mathit{gb}{g}^{-1}={\phi }_g(a){\phi }_g(b)$, thus ${\phi }_g$ is a group homomorphism.

Moreover, ${\phi }_g\circ {\phi }_{g^{-1}}=1_N={\phi }_{g^{-1}}\circ {\phi }_g$, so that $\phi$ is bijective: $\phi$ is a group automorphism.

Since $A$ is normal subgroup of $N$, $A_g={\phi }_g(A)$ is also a normal subgroup of $N$.

Now take $H$ a non trivial subgroup of $A_g$. Reasoning by contradiction assume that $H$ is normal in $N$. Then $K={\phi }_g^{-1}(H)=g^{-1}\mathit{Hg}$ is a non trivial subgroup of $A$. Since $H$ is normal in $N$, and ${\phi }_g$ is an automorphism of $N$, $K={\phi }_g^{-1}(H)$ is normal in $N$. This contradicts the fact that $A$ is a minimal normal subgroup of $N$. This contradiction proves that $H$ is not normal in $N$. We can conclude that $A_g$ is a minimal normal subgroup of $N$. (b) Let $g$ be any element of $G$. The hypothesis gives the inclusions $A_g=\mathit{gA}{g}^{-1}\subset A{A}_{g_1}$ and $A_{g{g}_1}=(g{g}_1)A{(g{g}_1)}^{-1}\subset A{A}_{g_1}$, therefore, since $A{A}_{g_1}$ is a subgroup,

This proves that $A{A}_{g_1}$ is a normal subgroup of $G$. (c) Let $\mathcal{}A$ be the set of all subgroups of $N$ of the form $A_{g_1}\cdots A_{g_n}$ such that the map $(a_1,\dots ,a_n)\mapsto a_1\dots a_n$ defines an isomorphism

Note that $A=A_e\in \mathcal{}A$. Then pick an element $H$ of $\mathcal{}A$ of maximal order. Then

If $H=N$, then we are done. Assume now that $H\ne N$.

If $A_g\subset H$ for all $g\in G$, then $H$ is normal in $G$:

for all $g\in G$, $A_{g{g}_i}=(g{g}_i)A{(g{g}_i)}^{-1}\subset H$, thus

Since $H\ne N$, and $H\ne \{e\}$ ( $|H|\ge |A|>1$), this is impossible by the minimality of $N$. Hence there is $g_{n+1}\in G$ such that $A_{g_{n+1}}\not\subset H$.

We don’t know if $H$ is normal in $G$, but $H=A_{g_1}\cdots A_{g_n}$ is normal in $N$, since all $A_{g_i}$ are normal in $N$ (see Exercise 7(a)). Hence $H\cap A_{g_{n+1}}$ is a normal subgroup in $N$, and lies in the minimal normal subgroup $A_{g_{n+1}}$ of $N$. Since $A_{g_{n+1}}\not\subset H$, $H\cap A_{g_{n+1}}\ne H$, therefore $H\cap A_{g_{n+1}}=\{e\}$. By Exercise 7, this proves that the map $(a_1,\dots ,a_{n+1})\mapsto a_1\dots a_n$ defines an isomorphism

But $A_{g_{n+1}}\not\subset H$, thus $H=A_{g_1}\cdots A_{g_n}{A}_{g_{n+1}}⊊A_{g_1}\cdots A_{g_n}{A}_{g_{n+1}}{A}_{g_{n+1}}$, and this contradicts the maximality of $H$. This contradiction proves that $H=N$, so that

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