Prove that $g \cdot H ∶=gHg^{−1}$ for all $g \in G$ and $H \in S_G$ is an action of $G$ on $S_G.$ Let $G$ be a group and let $S_G$ be the set of all subgroups of $G.$
Prove that $g \cdot H ∶=gHg^{−1}$ for all $g \in G$ and $H \in S_G$ is an action of $G$ on $S_G.$
I think I have no problem showing its an action, I am having trouble showing it is a function first. Any help would be appreciated.
 A: One way of defining a group action of $G$ on $X$ is as a group homomorphism from $G$ to the group $\DeclareMathOperator{\Sym}{Sym}\Sym X$ of bijections on $X$.
Here $X=S_G$, the set of all subgroups of $G$.
For starters (and I think you indicated you had proved it) you need to show that, for each $g$, $gHg^{-1}$ is again a subgroup.  So far, then, we have a function from $S_G$ to $S_G$.  Call it $i_g$.
$i_g$ is in fact an element of $\Sym S_G$.  You need to show $i_g$ does define a bijection on $S_G$, for each $g\in G$.  You need to check $H\stackrel{i_g}{\mapsto} gHg^{-1}$ is injective and surjective.
Next you need to check that $\phi:G\to\Sym S_G$ by $\phi(g)(H)=i_g(H)=gHg^{-1}$ is a homomorphism.  
So take $g_1,g_2\in G$, and check that $\phi(g_1g_2)=i_{g_1g_2}=i_{g_1}i_{g_2}=\phi(g_1)\phi(g_2)$.
So $\phi(g_1g_2)(H)=i_{g_1g_2}H=g_1g_2H(g_1g_2)^{-1}=g_1g_2Hg_2^{-1}g_1^{-1}=g_1i_{g_2}Hg_1^{-1}=g_1\phi(g_2)(H)g_1^{-1}=i_{g_1}\phi(g_2)(H)=\phi(g_1)\phi(g_2)(H)\,,\forall H\in S_G$.  
Thus $\phi$ is a homomorphism.
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