Hw4 Problem 6

### Problem 6

Let $G$ be a group and let $g \in G$ be fixed. Show that the map $\gamma_{g}: G \rightarrow G$ defined by $\gamma_{g}(x) = gxg'$ is an isomorphism of $G$ with itself.

Solution
One-to-One
Let $x,y\in G$. Now $\gamma_{g}(x)=gxg'$ and $\gamma_{g}(y)=gyg'$. Suppose

(1)
\begin{align} \gamma_{g}(x)=\gamma_{g}(y) \end{align}
(2)
$$gxg'=gyg'$$
(3)
$$x=y$$

through cancellation. Hence $\gamma_{g}$ is one-to-one.

Onto
Let $y\in G$. Now, we want to find $x\in G$ such that

(4)
\begin{align} \gamma_{g}(x)=y\\ gxg'=y \end{align}

By subsituting $x=g'yg$, we get

(5)
\begin{align} g(g'yg)g'=y\\ (gg')y(gg')=y\\ y=y \end{align}

Hence, $\gamma_{g}$ is onto.

Homomorphism

(6)
\begin{align} \gamma_{g}(x*y)=g(x*y)g'\\ =g*x*(g'*g)*y*g, \end{align}

where $g'*g=e=1$,

(7)
$$=(g*x*g')*(g*y*g')$$
(8)
\begin{align} =\gamma_{g}(x)*\gamma_{g}(y) \end{align}

Hence, $\gamma_{g}$ is a homomorphism.

Therefore, $\gamma_{g}:G\rightarrow G$ defined by $\gamma_{g}(x)=gxg'$ is an isomorphism of $G$ with itself.