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\$