Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set

Group Theory Problems and Solutions in Mathematics

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 $gC_G(X)g^{-1}=C_G(gXg^{-1})$.
 
LoadingAdd to solve later

Sponsored Links

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 $C_G(gXg^{-1}) \subset gC_G(X)g^{-1}$.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Two Normal Subgroups Intersecting Trivially Commute Each OtherTwo Normal Subgroups Intersecting Trivially Commute Each Other Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.   Proof. It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$. In fact, if this it true then we have […]
  • If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal SubgroupIf a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$. Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$. Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.   Hint. It follows from […]
  • Abelian Normal Subgroup, Intersection, and Product of GroupsAbelian Normal Subgroup, Intersection, and Product of Groups Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$. (That is, $A$ is a normal subgroup of $G$.) If $B$ is any subgroup of $G$, then show that \[A \cap B \triangleleft AB.\]   Proof. First of all, since $A \triangleleft G$, the […]
  • All the Conjugacy Classes of the Dihedral Group $D_8$ of Order 8All the Conjugacy Classes of the Dihedral Group $D_8$ of Order 8 Determine all the conjugacy classes of the dihedral group \[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle\] of order $8$. Hint. You may directly compute the conjugates of each element but we are going to use the following theorem to simplify the […]
  • If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itselfIf a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$. Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$. Then show that $N_G(H)=H$.   Hint. Use the conjugate part of the Sylow theorem. See the second statement of the […]
  • The Center of a p-Group is Not TrivialThe Center of a p-Group is Not Trivial Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$. (Such a group is called a $p$-group.) Show that the center $Z(G)$ of the group $G$ is not trivial.   Hint. Use the class equation. Proof. If $G=Z(G)$, then the statement is true. So suppose that $G\neq […]
  • Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$ Let $D_8$ be the dihedral group of order $8$. Using the generators and relations, we have \[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\] (a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$. Prove that the centralizer […]
  • If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its InverseIf a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element. Show that $x$ is not conjugate to $x^{-1}$.   Proof. Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$. Then we have \[xg=gx^{-1}. […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

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

More in Group Theory
Group Theory Problems and Solutions
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