Finite group generated by two finite subgroups

Question

Question

Let $G=\langle H,K\rangle$ where $H,K$ are two finite subgroups of $G$. If $H$ is subnormal in $G$, then show that $G$ is finite.

Attempt

I know that $D_{\infty}=\mathbb{Z}_2\ast\mathbb{Z}_2$ is generated by two elements of order $2$, but it is not finite. I can not see how to put the subnormal condition into play.

Definition: $H\leq G$ is a subnormal subgroup of $G$ if there there is a finite chain of subgroups $H\lhd H_1\lhd\ldots\lhd H_k=G$

Answer 1

Use induction on the subnormal depth $k$ of $H$ in $G$. If $k=1$, then $K \le N_G(H)$ and the result holds.

Let $H^K := \langle H^k \mid k \in K \rangle$ be the subgroup generated by the finitely many conjugates of $H$ under the elements of $K$. Since $K \le N_G(H^K)$, it is enough to show that $H^K$ is finite.

The subgroups $H^k$ are all subnormal of depth (at most) $k-1$ in $H_{k-1}$, so we can use the inductive hypothesis to prove that $H^K$ is finite.

More precisely, putting $K = \{ k_1,\ldots,k_n\}$, we can use the inductive hypothesis to show successively that $\langle H^{k_1} \rangle$, $\langle H^{k_1},H^{k_2}\rangle, \ldots, \langle H^{k_1},H^{k_2},\ldots,H^{k_{n-1}}\rangle$, $H^K$ are all finite.