Prove that $x^{−1}Hx $ is a subgroup of G Let $H$ be a subgroup of a group $G$ and, for $x\in G$, let $x^{-1}ax$ denote the set $\{x^{−1}ax : a\in H\}$. Prove that $x^{−1}Hx$ is a subgroup of $G$.
 A: First of all, $1\in H^x=x^{-1}Hx$ and so $x^{-1}Hx$ is not empty. Let $x^{-1}h_1x,x^{-1}h_2x\in x^{-1}Hx$. Then $x^{-1}h_1h_2x\in x^{-1}Hx$. Furthermore, $(x^{-1}h_1x)^{-1}= x^{-1}{h_1}^{-1}x\in x^{-1}h_1x$.
A: I assume $H \ne \emptyset$.
1.)  Since $H$ is a subgroup of $G$, $e \in H$ where $e \in G$ is the identity element.  Then $e = x^{-1} e x \in x^{-1}H x$.
2.)  If $y \in x^{-1}Hx$, then $y = x^{-1} z x$ for some $z \in H$.  Then $z^{-1} \in H$, since $H$ is a subgroup.  So $x^{-1} z^{-1} x \in x^{-1} H x$.  But $(x^{-1} z^{-1} x)(x^{-1} z  x) = e$ by a very easy computation, or $(x^{-1} z^{-1} x)y = e$;  this shows $y^{-1} = x^{-1} z^{-1} x \in x^{-1} H x$, so that $H$ contains the inverse of each of its elements.  
3.)  If $k_1, k_2 \in x^{-1}Hx$, then $k_j = x^{-1} h_jx$ with $h_j \in H$ for $j = 1,2$, whence
$k_1 k_2 = (x^{-1} h_1 x)(x^{-1} h_2x) = x^{-1} h_1 
h_2 x \in x^{-1} H x \;$  since $h_1h_2 \in H$.  This shows $x^{-1}Hx$ is closed under the group operation.
(1)-(3) show that $x^{-1}Hx$ satisfies the group axioms under the associative operation inhereted from $G$, hence it is a subgroup.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: $H \leq G \implies x^{-1}1_Gx=x^{-1}1_Hx \in x^{-1}Hx \implies x^{-1}Hx \neq \emptyset$   
Now, let $x^{-1}hx,x^{-1}jx \in x^{-1}Hx$.  So:
$$x^{-1}hx(x^{-1}jx)^{-1}=x^{-1}hxx^{-1}j^{-1}x=x^{-1}hj^{-1}x \in x^{-1}Hx.$$
And the subgroup criterion is fulfilled.
