The Group Action of Conjugation of a Subgroup on a Group

Let $G$ be a group and let $H$ be a subgroup of $G$. We can define a group action of the subgroup $H$ on the set $G$ as follows. For each $h \in H$ and for each $g \in G$ let $(h, g) \to hgh^{-1} \in G$. We will now check that this is indeed a group action of the group $H$ on the set $G$.

For all $h_1, h_2 \in H$ and for all $g \in G$ we have that:

(1)

\begin{align} \quad h_1(h_2g) = h_1(h_2gh_2^{-1}) = h_1h_2gh_2^{-1}h_1^{-1} = (h_1h_2)g \end{align}

And for all $g \in G$ we have that (where $e \in H \subseteq G$ is the identity):

(2)

\begin{align} \quad ge = geg^{-1} = gg^{-1} = e \end{align}

Thus $(h, g) \to hgh^{-1}$ is indeed a group action of the group $H$ on the set $G$.