TITLE: Show that conjugate by $g$ is isomorphism
QUESTION [5 upvotes]: 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, $|A|= |im(\sigma_{g|A})|=|gAg^{-1}|$.

REPLY [3 votes]: 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?