Exercise 1.7.17 ($\varphi_g: x \mapsto gxg^{-1}$ is an automorphism of $G$)

Question

Let $G$ be any group and let $A = G$. Show that the maps defined by $g\cdot a = gag^{-1}$ for all $g, a \in G$ do satisfy the axioms of a (left) group action (this action of G on itself is called conjugation).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
For any $g\in G$, we define the map 
$$
\varphi_g\ 
\left\{
\begin{array}{ccl}
G & 	o & G\
 x & \mapsto & g x g^{-1}
\end{array}
ight.
$$
Then $\varphi_g$ is a homomorphism: if $x, y \in G$, then
$$\varphi_g(x) \varphi_g(y) = g x g^{-1}\  g y g^{-1} = g xy g^{-1} = \varphi_g(xy).$$

Note that for all $x \in G$, $(\varphi_g \circ \varphi_{g^{-1}})(x) = g(g^{-1} x g) g^{-1} = x$, so $\varphi_g \circ \varphi_{g^{-1}} = \mathrm{Id}_G$, and similarly, replacing $g$ by $g^{-1}$, $\varphi_g^{-1} \circ \varphi_g = \mathrm{Id}_G$. This show that $\varphi_g$ is bijective, so $\varphi_g$ is an automorphism of $G$:
$$\forall g \in G,\ \varphi_g \in \mathrm{Aut}(G).$$

Since $\varphi_g$ is a homomorphism, $\varphi_g(x^k) = (\varphi_g(x))^k$ for all $k \in \Z$, so
$$\forall k \in \Z,\ gx^k g^{-1} = (g x g^{-1})^k.$$
Hence, for all $k \in \Z$, 
$$x^k = e \iff g x^k g^{-1} = e \iff (g x g^{-1})^k = e.$$
(where $e = 1_G$ is the identity element of $G$.)

If $x$ has infinite order, $x^k = e$ is always false, so $(g x g^{-1})^k = e$ is also always false, so $gxg^{-1}$ has infinite order.

If $x$ has finite order, then 
\begin{align*}
|x| = d &\iff \forall k \in \Z, \ (x^k = e \iff d \mid k)\
& \iff \forall k \in \Z, \ ((g x g^{-1})^k = e \iff d \mid k) \
&\iff |gx g^{-1}| = d),
\end{align*}
so $x$ and $gxg^{-1}$ have the same order for all $x$ in $G$.

Let $A$ be a subset of $G$. Then $gAg^{-1} = \varphi_g(A)$. Since $\varphi_g$ is bijective, the restriction
$$
\left\{
\begin{array}{ccl}
A & 	o & \varphi_g(A)\
x & \mapsto & \varphi_g(x)
\end{array}
ight.
$$
is also bijective, hence $|\varphi_g(A)| = |A|$, thus
$$|gAg^{-1}| = |A|.$$
\end{proof}

Exercise 1.7.17 ($\varphi_g: x \mapsto gxg^{-1}$ is an automorphism of $G$)

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Exercise 1.7.17 ($\varphi_g: x \mapsto gxg^{-1}$ is an automorphism of $G$)

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