Is it possible that pair of isomorphic subgroups of finite group is conjugate in some larger group

Let $G$ be a finite group. Let $H$ and $I$ be a pair of isomorphic but non-conjugate subgroups of $G$. Does there always exists a finite group $L$ containing $G$ such that $H$ and $I$ is conjugate in $L$?

This is how far I've got: by Cayley's theorem, it suffices to assume that $G$ is a finite symmetric group. I have a simple example but fail to prove the general case. The example is: let $$G=S_4=\langle a,b,c \mid a^2 = b^3 = c^4 = abc = 1\rangle$$ $$H = \langle a \rangle; \quad I = \langle c^2 \rangle.$$ It is easy that $a$ is not conjugate to $c^2$ in $S_4$. Now we construct group embedding

$$\begin{align} \varphi: S_4 &\to S_6,\\ a &\mapsto (23)(56),\\ b &\mapsto(134),\\ c&\mapsto(56)(1432) \end{align}$$

It is easy that $\varphi(a)$ conjugates to $\varphi(c^2)$ in $S_6$.

Solution 1:

Thanks to Derek Holt's comment.

This is the rigorous proof: Suppose that the underlying set $G$ can be written as disjoint union of right cosets:

$$G = \bigsqcup_i Hg_i = \bigsqcup_iIs_i$$

where $g_i \in G$, $s_i \in G$ and $i = 1,2,\ldots, \frac{|G|}{|H|}$. We embed $G$ into $S_{|G|}$ via:

$$L: G \to S_{|G|}, \; g \mapsto (L_g:g' \mapsto gg').$$

Let $\varphi: H \to I$ be an isomorphism. We construct bijection

$$b: G \to G, \; hg_i \mapsto \varphi(h)s_i$$

for $h \in H$ and $i = 1,2,\ldots, \frac{|G|}{|H|}$. One checks that $b \circ L_h = L_{\varphi(h)} \circ b$ for all $h \in H$. Therefore $L(H)$ is conjugate to $L(I)$ in $S_{|G|}$.

• Remark that the bijection $b$ relates to three things: a set of representatives $g_i$ of right cosets of $H$, a set of representatives $s_i$ of right cosets of $I$, and a group isomorphism $\varphi: H \to I$. In this regard, any pair of subgroups of $G$ is conjugate in $S_{|G|}$.