Conjugate Subgroups of a Group

# Conjugate Subgroups of a Group

Recall from the Conjugate Elements in a Group page that if $G$ is a group and $g_1, g_2 \in G$ then $g_1$ is said to be a conjugate of $g_2$ if there exists an $a \in G$ such that:

(1)\begin{align} \quad g_1 = ag_2a^{-1} \end{align}

We can similarly define conjugacy of subgroups $H_1$ and $H_2$ of a group $G$.

Definition: Let $G$ be a group and let $H_1$ and $H_2$ be subgroups of $G$. Then $H_1$ is said to be a Conjugate of $H_2$ if there exists an $a \in G$ such that $H_1 = aH_2a^{-1}$. |

As with conjugacy of elements - conjugacy of subgroups of a group $G$ is not that interesting when $G$ is an abelian group.

Proposition 1: Let $G$ be a group. If $G$ is abelian and $H_1$, $H_2$ are subgroups of $G$ then $H_1$ is a conjugate of $H_2$ if and only if $H_1 = H_2$. |

**Proof:**$\Rightarrow$ Suppose that $H_1$ is a conjugate of $H_2$. Then there exists an $a \in G$ such that $H_1 = aH_2a^{-1}$, i.e., $H_1a = aH_2$. Since $G$ is abelian, we have that $H_1a = H_2a$. Multiplying on the right by $a^{-1} 4]] gives us [[$ H_1 = H_2$.

- $\Leftarrow$ Suppose that $H_1 = H_2$. Let $e \in G$ be the identity element. Then $H_1 = eH_2e^{-1}$. So $H_1$ is a conjugate of $H_2$. $\blacksquare$

Proposition 2: Let $G$ be a group and let $H$ be a subgroup of $G$. Then $H$ is a normal subgroup of $G$ if and only if the only subgroup of $G$ that is conjugate to $H$ is $H$ itself. |

**Proof:**$\Rightarrow$ Suppose that $H$ is a normal subgroup subgroup of $G$. Then $aHa^{-1} = H$ for all $a \in G$. Thus $aHa^{-1} \neq K$ for any other subgroup $K$ of $G$. Thus the only subgroup of $G$ that is conjugate to $H$ is $H$.

- $\Leftarrow$ Suppose that the only subgroup of $G$ that is conjugate to $H$ is $H$ itself. Let $a \in G$. Then $aHa^{-1}$ is a subgroup of $G$, and thus, it must equal $H$. So $H$ is a normals subgroup of $G$. $\blacksquare$