This is a question from Dummit and Foote.
I am still a novice at algebra so any feedback on my work would be appreciated!
Let $G$ be a group and let $G$ act on itself by left conjugation, so each
$g \in G$ maps $G$ to $G$ by
\begin{align*}
x\mapsto gxg^{-1}
\end{align*}
For fixed $g \in G$, prove that conjugation by $g$ is an isomorphism from
$G$ onto itself (i.e. an automorphism of $G$). Deduce that $x$ and
$gxg^{-1}$ have that same order for all $x \in G$ and that for any subset
$A$ of $G$, $\vert A \vert=\vert gAg^{-1}\vert$ (here $gAg^{-1}=
\{gag^{-1}\vert a \in A\}$.)
Take $g \in G$. We show that $\sigma_{g}:G \rightarrow G$ defined by $\sigma_{g}(x) = gxg^{-1}$
is a homomorphism and is bijective.
First, we show that $\sigma_g$ is homomorphism. Take $x,y \in G$.
\begin{align*}
\sigma_g(xy) &= g(xy)g^{-1} \\
&= gxg^{-1}gyg^{-1} \\
&= \sigma_g(x) \sigma_g(y)
\end{align*}
Hence, $\sigma_g$ is homomorphism.
Second, we show that injectivity by showing $ker(\sigma_g) = \{1\}$.
Suppose, $gxg^{-1}=1$. Then,
\begin{align*}
xg^{-1} &= g^{-1} \\
x &= g^{-1}g = 1
\end{align*}
Hence, $ker(\sigma_g) = \{1\}$.
Now, we show $\sigma_g$ is surjective.
Take $g \in G$. Then, for $x = g^{-1}yg \in G$ (since $G$ is a group, $x \in G$),
$g(g^{-1}yg)g^{-1} = y$.
Therefore, $\sigma_g$ is bijective and it follows that it is a isomorphism.
Since $\sigma_g$ is isomorphism, it follows that for every $x \in G$, $|x| = |\sigma_g(x)|=|gxg^{-1}|$.
Furthermore, since $\sigma_g$ is bijective on $G$, $\sigma_g$ is bijective on $A \subset G$.
In particular, $\sigma_g$ is injective when restricted to $A$.
Hence, $|A|= |im(\sigma_{g|A})|=|gAg^{-1}|$.
Best Answer
All is fine. An alternative way to establish bijectivity might be the observation that $\sigma_g\circ\sigma_h=\sigma _{gh}$ (a useful fact on its own!) and therefore $\sigma_{g^{-1}}\circ \sigma_{g}=\sigma_{g}\circ \sigma_{g^{-1}}=\operatorname{id}_G$. - And a map with left and right inverse map is bijective. Then again, this does not reall ydiffer from what you wrote, does it?