Characteristic Subgroups of a Group

# Characteristic Subgroups of a Group

Definition: Let $G$ be a group. A subgroup $H$ of $G$ is said to be a Characteristic Subgroup of $G$ if for all automorphisms $\phi \in \mathrm{Aut}(G)$ we have that $\phi(H) \subseteq H$. |

*In other words, a subgroup $H$ of a group $G$ is a characteristic subgroup if every automorphism $\phi$ of $G$ maps $H$ to $H$.*

Proposition 1: Let $G$ be a group and let $H$ be a subgroup of $G$. If $H$ is a characteristic subgroup of $G$ then $H$ is a normal subgroup of $G$. |

**Proof:**Consider the subset $\mathrm{Inn}(G)$ of $\mathrm{Aut}(G)$, i.e., $\mathrm{Inn}(G)$ contains all inner automorphisms of $G$, i.e., all automorphisms of the form $i_a : G \to G$ where $a \in G$ and $i_a$ is defined for all $g \in G$ by:

\begin{align} \quad i_a(g) =aga^{-1} \end{align}

- Since $H$ is a characteristic subgroup of $G$ we have that for all $\phi \in \mathrm{Aut}(G)$ that $\phi(H) \subseteq H$. In particular, for all $a \in G$ we have that $i_a(H) \subseteq H$, i.e., $aha^{-1} \in H$ for all $h \in H$ and for all $a \in G$. So $aHa^{-1} \subseteq H$ for all $a \in G$. So $H$ is a normal subgroup of $G$. $\blacksquare$

Proposition 2: Let $G$ be a group. Then $Z(G)$ is a characteristic subgroup of $G$. |

*Recall that if $G$ is a group then $Z(G)$ denotes the center of $G$, i.e., the set of all elements in $G$ that commute with every element of $G$.*

**Proof:**Let $x \in Z(G)$. Then $xg = gx$ for all $g \in G$. Then for all $g \in G$ and for all $\phi \in \mathrm{Aut}(G)$ we have that:

\begin{align} \quad \phi(xg) &= \phi(gx) \\ \quad \phi(x)\phi(g) &= \phi(g)\phi(x) \end{align}

- Since every $\phi \in \mathrm{Aut}(G)$ is a bijection, we equivalently have that $\phi(x)g = g \phi(x)$ for all $\phi \in \mathrm{Aut}(G)$. So $\phi(x) \in Z(G)$ for each $x \in Z(G)$ and for each $\phi \in \mathrm{Aut}(G)$. That is, $\phi(Z(G)) \subseteq Z(G)$ for all $\phi \in \mathrm{Aut}(G)$. So $Z(G)$ is a characteristic subgroup of $G$. $\blacksquare$

Proposition 3: Let $G$ be a group. Then $G'$ is a characteristic subgroup of $G$. |

*Recall that if $G$ is a group then $G'$ denotes the derived subgroup of $G$, i.e., the smallest subgroup of $G$ that contains all commutators of $G$.*

**Proof:**Let $G'$ be the derived subgroup of $G$. Observe that:

\begin{align} \quad G' = \langle [g_1, g_2] : g_1, g_2 \in G \rangle = \langle g_1g_2g_1^{-1}g_2^{-1} : g_1, g_2 \in G \rangle \end{align}

- For all $g_1, g_2 \in G$ and for all $\phi \in \mathrm{Aut}(G)$ we have that:

\begin{align} \quad \phi([g_1, g_2]) &= \phi(g_1g_2g_1^{-1}g_2^{-1}) \\ &= \phi(g_1) \phi(g_2) \phi(g_1^{-1}) \phi(g_2^{-1}) \\ &= \phi(g_1) \phi(g_2) [\phi(g_1)]^{-1} [\phi(g_2)]^{-1} \in G' \end{align}

- So for each $\phi \in \mathrm{Aut}(G)$, $\phi$ maps each generator of $G'$ to $G'$. So $\phi(G') \subseteq G'$. Since this holds for all $\phi \in \mathrm{Aut}(G)$ we conclude that $G'$ is a characteristic subgroup of $G$. $\blacksquare$