# Equivalent Definitions of Characteristic Subgroups. Center is Characteristic.

## Problem 246

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 $Z(G)$ of $G$ is characteristic in $G$.

## Definition

Recall that an automorphism $\sigma$ of a group $G$ is a group isomorphism from $G$ to itself.
The set of all automorphism of $G$ is denoted by $\Aut(G)$.

## Proof.

### (a) If $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$

Since $\sigma$ is an automorphism, the inverse $\sigma^{-1}$ is also an automorphism of $G$.
Hence, we have
$\sigma^{-1}(H)\subset H$ by the assumption.

Applying $\sigma$, we have
$\sigma\sigma^{-1}(H) \subset \sigma(H).$ Then we obtain
\begin{align*}
H&=\sigma \sigma^{-1}(H)\subset \sigma(H)\subset H.
\end{align*}

Since the both ends are $H$, the inclusion is in fact the equality.
Thus, we obtain
$\sigma(H)=H,$ and the subgroup $H$ is characteristic in the group $G$.

### (b) The center $Z(G)$ of $G$ is characteristic in $G$

By part (a), it suffices to prove that $\sigma(Z(G)) \subset Z(G)$ for every automorphism $\sigma \in \Aut(G)$ of $G$.

Let $x\in \sigma(Z(G))$. Then there exists $y \in Z(G)$ such that $x=\sigma(y)$.
To show that $x \in Z(G)$, consider an arbitrary $g \in G$.
Then since $\sigma$ is an automorphism, we have $G=\sigma(G)$.
Thus there exists $g’$ such that $g=\sigma(g’)$.

We have
\begin{align*}
xg &=\sigma(y)\sigma(g’)\\
&=\sigma(yg’) && \text{ (since $\sigma$ is a homomorphism)}\\
&=\sigma(g’y) && \text{ (since $y \in Z(G)$)}\\
&=\sigma(g’)\sigma(y) && \text{ (since $\sigma$ is a homomorphism)}\\
&=gx.
\end{align*}
Since this is true for all $g \in G$, it follows that $x \in Z(G)$, and thus
$\sigma(Z(G)) \subset Z(G).$ This completes the proof.

## Comment.

In some textbook, a subgroup $H$ of $G$ is said to be characteristic in $G$ if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$.
Problem (a) implies that our definition of characteristic and this alternative definition are in fact equivalent.

## Related Question.

Read the post Basic properties of characteristic groups for more problems about characteristic subgroups.

### More from my site

• Basic Properties of Characteristic Groups Definition (automorphism). An isomorphism from a group $G$ to itself is called an automorphism of $G$. The set of all automorphism is denoted by $\Aut(G)$. Definition (characteristic subgroup). A subgroup $H$ of a group $G$ is called characteristic in $G$ if for any $\phi […] • Group Homomorphism, Conjugate, Center, and Abelian group Let$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 show that (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 center of the […] • Abelian Normal subgroup, Quotient Group, and Automorphism Group Let$G$be a finite group and let$N$be a normal abelian subgroup of$G$. Let$\Aut(N)$be the group of automorphisms of$G$. Suppose that the orders of groups$G/N$and$\Aut(N)$are relatively prime. Then prove that$N$is contained in the center of […] • 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 […] • Subgroup of Finite Index Contains a Normal Subgroup of Finite Index Let$G$be a group and let$H$be a subgroup of finite index. Then show that there exists a normal subgroup$N$of$G$such that$N$is of finite index in$G$and$N\subset H$. Proof. The group$G$acts on the set of left cosets$G/H$by left multiplication. Hence […] • The Preimage of a Normal Subgroup Under a Group Homomorphism is Normal Let$G$and$G'$be groups and let$f:G \to G'$be a group homomorphism. If$H'$is a normal subgroup of the group$G'$, then show that$H=f^{-1}(H')$is a normal subgroup of the group$G$. Proof. We prove that$H$is normal in$G$. (The fact that$H$is a subgroup […] • Image of a Normal Subgroup Under a Surjective Homomorphism is a Normal Subgroup Let$f: H \to G$be a surjective group homomorphism from a group$H$to a group$G$. Let$N$be a normal subgroup of$H$. Show that the image$f(N)$is normal in$G$. Proof. To show that$f(N)$is normal, we show that$gf(N)g^{-1}=f(N)$for any$g \in […]
• A Group Homomorphism is Injective if and only if Monic Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$. Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]

### 1 Response

1. 01/06/2017

[…] out the post Equivalent definitions of characteristic subgroups. Center is characteristic. for more problems about characteristic […]

##### Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable

Let $p, q$ be prime numbers such that $p>q$. If a group $G$ has order $pq$, then show the followings....

Close