# Normalizer and Centralizer of a Subgroup of Order 2 ## Problem 94

Let $H$ be a subgroup of order $2$. Let $N_G(H)$ be the normalizer of $H$ in $G$ and $C_G(H)$ be the centralizer of $H$ in $G$.

(a) Show that $N_G(H)=C_G(H)$.

(b) If $H$ is a normal subgroup of $G$, then show that $H$ is a subgroup of the center $Z(G)$ of $G$. Add to solve later

## Definitions.

Recall that the centralizer of $H$ in $G$ is
$C_G(H)=\{g \in G \mid gh=hg \text{ for any } h\in H\}.$

The normalizer of $H$ in $G$ is
$N_G(H)=\{g\in G \mid gH=Hg \}.$

## Proof.

### (a) Prove $N_G(H)=C_G(H)$

In general, we have $C_G(H) \subset N_G(H)$. We show that $N_G(H) \subset C_G(H)$.

Take any $g \in N_G(H)$. We have $gH=Hg$. Since $|H|=2$, let $H=\{1,h\}$.
Then $gH=\{g,gh\}$ and $Hg=\{g, hg\}$. Since $gH=Hg$, we have $gh=hg$. Namely $g\in C_G(H)$.

This proves that $N_G(H) \subset C_G(H)$, hence $N_G(H)= C_G(H)$.

### (b) If $H$ is normal, then $H$ is a subgroup of $Z(G)$

Suppose that $H$ is a normal subgroup of $G$, that is $G=N_G(H)$.
By part (a), this implies that $G=C_G(H)$. Hence $H < Z(G)$. 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 ##### A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$

Let $G$ be a group of order $|G|=pqr$, where $p,q,r$ are prime numbers such that $p<q<r$. Show that $G$ has...

Close