$G$ a group with center $\{e\}$ and $A$ a maximal subgroup of $G$ that is also abelian and not normal. How to show that $A$ is a Frobenius complement?

Question

I have been sitting on this homework problem for days now: As the title says, I have a group (which doesn't have to be finite. Even Frobenius groups aren't defined as finite in our course) which only has the neutral element $e$ as center. $A$ is a maximal subgroup of $G$, so it means there is no subgroup $H$ of $G$ with $A\subsetneq H\subsetneq G$. I want to show that if $A$ is abelian and not normal in $G$, then $G$ has to be a Frobenius group, and $A$ is a Frobenius complement. As an hint we should consider $A\cap A^g$ and $\langle A,A^g\rangle$, where $A^g=g^{-1}Ag$ for $g\in G$.

We defined a Frobenius group as a group which has a subgroup $A$ where $|A\cap A^g|=1$, when $g\in G\setminus A$. The subgroup $A$ is then called Frobenius complement. So the hint to consider $A\cap A^g$ is pretty obvious.

Per definition we have $A\subseteq\langle A, A^g\rangle\subseteq G$, we therefore know that $A=\langle A, A^g\rangle$ or $G=\langle A, A^g\rangle$ (otherwise $A$ isn’t maximal).

This is were I sit for like two days now. I have to show, $\langle A,A^g\rangle=G$ if and only if $g\in G\setminus A$. I don’t know how. I realize I haven’t found a way to use the property of $A$ being abelian. Can someone give me a hint? I'm also not quite sure if $A\cap A^g=\{e\}$ follows straightaway from that. But I haven’t really put any thought into it.

The course is named Groups and Representations. I had no problems with the homework sheets yet. I’d say it was group theory until now (in the homework problems). Maybe the representation part factors in for this, but I don’t see how.

Thanks.

Answer 1

Let $g \in G \setminus A$.

If $\langle A,A^g \rangle = A$, then $A^g \le A$. Since $A$ is maximal in $G$, so is $A^g$, so this implies that $A = A^g$. Hence $g \in N_G(A)$, so $A < N_G(A)$ and then $N_G(A) = A$, contradicting $A$ not normal in $G$. So $\langle A,A^g \rangle = G$.

Now suppose that $B = A \cap A^g$. Then $C_G(B)$ contains both $A$ and $A^g$, so it contains $\langle A,A^g \rangle = G$ and hence $B \le Z(G)$, contradicting $Z(G) = \{e\}$.