For $x\in G$, define $H_x=\{g^{-1}xg \mid g\in G\}$. Under what conditions on $x$ will $H_x \leq G$. Further, if $H_x \leq G$, will $H_x \lhd G$?

Question

For $x\in G$, define $H_x=\{g^{-1}xg \mid g\in G\}$. Under what conditions on $x$ will $H_x \leq G$. Further, if $H_x \leq G$, will $H_x \lhd G$?

My attempt is as below:

$G$ is a group so $e\in G$ and $e^{-1}xe=x \in H_x$.This means that $H_x$ will be non-empty.

For $H_x$ to be a subgroup we must show that for any $g_1^{-1}x g_1$ and $g_2^{-1}xg_2$ in $H_x$ , $(g_1^{-1}xg_1)(g_2^{-1}xg_2)^{-1}\in H_x$.

Also $(g_1^{-1}xg_1)(g_2^{-1}xg_2)^{-1}=(g_1^{-1}xg_1g_2^{-1}x^{-1}g_2)$.

Now if $x=e$ then $x \in H_x \Rightarrow e\in H_x$ and we have $(g_1^{-1}xg_1g_2^{-1}x^{-1}g_2)=e \in H_x$. So We can say that $H_x \leq G$. In this case since $H_x=\{e\}$, we have $H_x \lhd G$.

Please let me know if my reasoning is right. Also is $x=e$ the only possible condition on $x$ which makes $H_x$ a subgroup?

Answer 1

A necessary condition for the set $H_x$ to be a subgroup is that $e\in H_x$. So you need some $g\in G$ such that $$ e=gxg^{-1} $$ This implies $$ g^{-1}eg=g^{-1}gxg^{-1}g $$ and you're done, aren't you?

Answer 2

Your reasoning is correct. To answer your final question, think that something to hold for every $g_1, g_2$ must hold in particular for $g_1=g_2$. Explicitly:

\begin{alignat}{1} H_x \le G &\iff \forall g_1,g_2 \in G, \exists g\in G \mid g_1^{-1}xg_1(g_2^{-1}xg_2)^{-1}=g^{-1}xg \\ &\Longrightarrow \forall g_1 \in G, \exists g\in G \mid g_1^{-1}xg_1(g_1^{-1}xg_1)^{-1}=e=g^{-1}xg \\ &\Longrightarrow \exists g\in G \mid xg=g \\ &\Longrightarrow x=e \end{alignat}

You have already shown that $x=e \Longrightarrow H_x \le G$, so indeed: $H_x \le G \iff x=e$.

Answer 3

So following suggestion by @Orat I am posting an answer.

If we recognize $H_x$ as conjugacy class of $x$ (or orbit of x under conjugation) then we have this result that $\forall g \in G, \vert g^{-1}xg \vert=\vert x \vert$. Now if $H_x$ is required to be a subgroup then $H_x$ must contain identity element $e$. But then $\vert g^{-1}xg \vert=\vert e \vert=1$. Now $e$ is only one element in any group which has order 1.So $ g^{-1}xg=e \Rightarrow x=geg^{-1}=gg^{-1}=e$. So we can conclude that x=e is the only condition which will make $H_x$ a subgroup of $G$.

Now for normality. We know that $x=e \Rightarrow \{g^{-1}xg|g \in G\}=\{e\}$ and $\{e\}$ is always a normal subgroup of $G$.