# The Center of a p-Group is Not Trivial

## Problem 10

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 $G\neq Z(G)$.
Then by the class equation, we have
$|G|=|Z(G)|+ \sum_{i=1}^r |G:C_G(g_i)|,$ where $g_i$ are representatives of the distinct conjugacy class not contained in the center $Z(G)$, and $C_G(g_i)$ is the centralizer of $g_i$.
(Since we are assuming that $G \neq Z(G)$ such $g_i$ exist.)

Since $g_i \not \in Z(G)$, the groups $C_G(g_i)$ are proper subgroups of $G$ and hence $p$ divides $|G: C_G(g_i)|$. Of course $p$ divides $|G|$, thus $p$ should divide $|Z(G)|$ as well.
Therefore the center $Z(G)$ cannot be trivial.

## Comment.

This problems is a simple/nice application of the class equation of  group theory. The only information about the group is that its order is a prime power. From this we can conclude that the center of the group is not trivial.

### More from my site

• The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ Let $p$ be a prime number. Let $G$ be a non-abelian $p$-group. Show that the index of the center of $G$ is divisible by $p^2$. Proof. Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$. Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
• Normalizer and Centralizer of a Subgroup of Order 2 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 […]
• All 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 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 […]
• 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 […]
• Subgroup Containing All $p$-Sylow Subgroups of a Group Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$. Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$. Then show that $N$ contains all $p$-Sylow subgroups of […]
• Equivalent Definitions of Characteristic Subgroups. Center is Characteristic. Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$. (a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$. (b) Prove that the center […]
• Group Homomorphism, Conjugate, Center, and Abelian group Let $G$ be a group. We fix an element $x$ of $G$ and define a map $\Psi_x: G\to G$ by mapping $g\in G$ to $xgx^{-1} \in G$. Then prove the followings. (a) The map $\Psi_x$ is a group homomorphism. (b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the […]

### 1 Response

1. 07/26/2016

[…] The center of a p-group is not trivial (post 1) […]

##### A Group of Linear Functions

Define the functions $f_{a,b}(x)=ax+b$, where $a, b \in \R$ and $a>0$. Show that \$G:=\{ f_{a,b} \mid a, b \in \R,...

Close