Hw4 Problem 6
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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
\begin{align} \gamma_{g}(x)=\gamma_{g}(y) \end{align}
(2)
\begin{equation} gxg'=gyg' \end{equation}
(3)
\begin{equation} x=y \end{equation}
through cancellation. Hence $\gamma_{g}$ is one-to-one.
Onto
Let $y\in G$. Now, we want to find $x\in G$ such that
\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)\begin{equation} =(g*x*g')*(g*y*g') \end{equation}
(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.