# Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set ## Problem 109

Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$.

For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$. Add to solve later

## Proof.

$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for any element $x \in X$, we have
\begin{align*}
ghg^{-1}(gxg^{-1}) &=ghxg^{-1}\\
&=gxhg^{-1} \,\,\,\, \text{ since } h\in C_G(X)\\
&=gxg^{-1}(ghg^{-1}).
\end{align*}

This computation shows that $ghg^{-1} \in C_G(gXg^{-1})$ for any $h \in C_G(X)$.
Thus this proves $g C_G(X) g^{-1} \subset C_G(gXg^{-1})$.

$(\supset)$ Conversely, take any element $k \in C_G(gXg^{-1})$.
Then for any $x \in X$, we have
\begin{align*}
g^{-1}kg(x)&=g^{-1}k(gxg^{-1})g\\
&=g^{-1}(gxg^{-1})kg \,\,\,\, \text{ since } k\in C_G(gXg^{-1})\\
&=x(g^{-1}kg).
\end{align*}
This implies that $g^{-1}kg \in C_G(X)$, hence $k\in gC_G(X)g^{-1}$ for any $k \in C_G(gXg^{-1})$. This proves $C_G(gXg^{-1}) \subset gC_G(X)g^{-1}$. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Group Theory ##### Group of Invertible Matrices Over a Finite Field and its Stabilizer

Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number. Let $G_n=\GL_n(\F_p)$ be the group...

Close