Let $H\le G$ as groups. Show $g^{-1}Hg$ is a subgroup of $G$. Let $G$ an aribitrary group and $H$ a subgroup of $G$. Then, for all $g\in G$ we have that $g^{-1}Hg$ is another subgroup of $G$.
I did not have problem proving that the identity and the inverse of any $h$ are in $g^{-1}Hg$, but had problems proving the closure property for subgroups.
Thanks in advance.
 A: I will use the one-step subgroup test.
Fix $g\in G$.
Since $e\in H$, we have $g^{-1}eg=g^{-1}g=e\in g^{-1}Hg$, so that $g^{-1}Hg\neq\varnothing$.
Since $g^{-1}Hg=\{g^{-1}xg\mid x\in H\}$ and $H\le G$, we have $g^{-1}Hg\subseteq G$.
Let $a,b\in g^{-1}Hg$. Then there exist $h,k\in H$ such that $a=g^{-1}hg, b=g^{-1}kg$. Now
$$\begin{align}
ab^{-1}&=(g^{-1}hg)(g^{-1}kg)^{-1}\\
&=(g^{-1}hg)(g^{-1}k^{-1}(g^{-1})^{-1})\\
&=(g^{-1}h)(gg^{-1})k^{-1}g\\
&=g^{-1}(hk^{-1})g,
\end{align}$$
which is in $g^{-1}Hg$ since $hk^{-1}\in H\le G$.
Hence $g^{-1}Hg\le G$.
A: That is easy, let $x,y\in g^{-1}Hg$, then by definition of the subgroup we have $x=g^{-1}h_{1}g$ for some unique $h_1\in H$, and similary let $y=g^{-1}h_{2}g$ for some unique $h_{1}\in H$, and now consider $xy$, which is by definition
$xy= (g^{-1}h_{1}g)(g^{-1}h_{2}g)=g^{-1}h_{1}h_{2}g$ since $h_{1},h_{2}\in H$ then with $h_{3}=h_{1}h_{2}\in H$, a unique element, since $xy=g^{-1}h_{3}g$, we have $xy\in g^{-1}Hg$.
The others three things you can verify and are easy.
A: We write
$$ g H g^{-1} = \{ ghg^{-1} : h \in H\} \subseteq G.$$
Edit: Thanks to Shaun in the comments for pointing out this isn't necessarily obvious. Let $ghg^{-1} \in gHg^{-1}$. We have $h \in H \subseteq G$, $g \in G$, and we use the fact that $G$ is closed under multiplication, which means that $gh \in G$. Since $g^{-1} \in G$ as well, we have $ghg^{-1} \in G$. So for any $ghg^{-1} \in gHg^{-1}$, we have $ghg^{-1} \in G$.
The goal now is to show that this is a subgroup of $G$ given that $H$ is a subgroup. First, we see that $e \in gHg^{-1}$, since $e \in H$ (by property of being a subgroup) and $geg^{-1} = gg^{-1} = e \in gHg^{-1}$. Next, recall that $(ab)^{-1} = b^{-1}a^{-1}$. Using this fact, we have that if $ghg^{-1} \in gHg^{-1}$, then $(ghg^{-1})^{-1} = (g^{-1})^{-1} h^{-1} g^{-1} = gh^{-1}g^{-1}$. Since $H$ is a subgroup, $h^{-1} \in H$, so $gh^{-1}g^{-1} \in gHg^{-1}$. Finally, if $ghg^{-1}, gh'g^{-1} \in gHg^{-1}$, then $ghg^{-1} \cdot gh'g^{-1} = gh(g^{-1}g)h'g^{-1} = gh(e)h'g^{-1} = ghh'g^{-1}$. Since $H$ is a subgroup, $hh' \in H$, so $ghh'g^{-1} \in H$. This shows $gHg^{-1}$ is a subgroup.
