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.

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}
\begin{equation} gxg'=gyg' \end{equation}
\begin{equation} x=y \end{equation}

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

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

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

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


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

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

\begin{equation} =(g*x*g')*(g*y*g') \end{equation}
\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.

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