# Basic Properties of Characteristic Groups ## Problem 22

### 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 $\phi \in \Aut(G)$, we have $\phi(H)=H$. In words, this means that each automorphism of $G$ maps $H$ to itself.

Prove the followings.

(a) If $H$ is characteristic in $G$, then $H$ is a normal subgroup of $G$.

(b) If $H$ is the unique subgroup of $G$ of a given order, then $H$ is characteristic in $G$.

(c) Suppose that a subgroup $K$ is characteristic in a group $H$ and $H$ is a normal subgroup of $G$. Then $K$ is a normal subgroup in $G$. Add to solve later

## Proof.

### (a) If $H$ is characteristic in $G$, then $H$ is a normal subgroup of $G$.

For each $g\in G$, define a map $\phi_g: G \to G$ defined by $\phi_g(x)=gxg^{-1}$. This is an automorphism of $G$ with the inverse $\phi_{g^{-1}}$.

Since $H$ is characteristic, we have $\phi_g(H)=H$, equivalently we have $gHg^{-1}=H$.
Therefore $H$ is a normal subgroup of $G$.

### (b) If $H$ is the unique subgroup of $G$ of a given order, then $H$ is characteristic in $G$.

For any automorphism $\phi \in \Aut(G)$, we have $\phi(H)\subset \phi(G)=G$ and $|H|=|\phi(H)|$. The uniqueness of $H$ implies that $H=\phi(H)$ and thus $H$ is characteristic.

### (c) $K$ is characteristic in $H$ and $H$ is normal$G$. Then $K$ is a normal subgroup in $G$.

For each $g \in G$, consider the automorphism $\phi_g$ of $G$ defined in the proof of (a). Since $H \triangleleft G$, we have $\phi_g(H)=H$.
Hence the restriction $\phi_{g}|_{H}$ belongs to $\Aut(H)$.
Now since $K$ is characteristic in $H$, we have $\phi_{g}|_{H}(K)=K$, or equivalently we have $gKg^{-1}=K$ and $K$ is normal in $G$.

## Comment.

Let $K$, $H$ be subgroups of $G$. Suppose that $K$ is a normal subgroup of $H$, and $H$ is a normal subgroup of $G$.
In general, we cannot conclude that $K$ is a normal subgroup of $G$.
(For example, consider the dihedral group $D_8$ of order $8$.)

Thus, normality is not transitive.
Part (c) of the problem claims that if in addition, $K$ is characteristic in $H$, then $K$ is normal in $G$.

## Related Question.

Check out the post Equivalent definitions of characteristic subgroups. Center is characteristic. for more problems about characteristic subgroups. Add to solve later

### 1 Response

1. 01/06/2017

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