# Proof of Conjugate Subgroup Isomorphism

Let $G$ be a group, and let $H$ be a subgroup of $G$. Prove that if $a$ is an element of $G$, then the subset $aHa^{-1} = \{g ∈ G | g = aha^{-1} \text{ for some } h \in H\}$ is a subgroup of $G$ that is isomorphic to $H$.

Proof:
Let $G$ be a group and $H$ is a subgroup of $G$. Suppose $a$ is any element of $G$. Now $ϕ: H \rightarrow aHa^{-1}$ be defined by $ϕ(h) = aha^{-1}.$ So $$ϕ(h_1 h_2) = ah_1h_2a^{-1} = ah_1a^{-1} ah_2a^{-1} = ϕ(h_1) ϕ(h_2)$$ Thus, ϕ is a homomorphism.

(one-to-one)
Now $$\begin{eqnarray*}h \in \operatorname{ker}(ϕ) &\text{if and only if}&ϕ(h) = 0\\ &\text{if and only if}&aha^{-1} = 0\\ &\text{if and only if}&h = 0\\ \end{eqnarray*}$$

Hence, $\operatorname{ker}(ϕ) = \{0\}$. Thus $ϕ$ is one-to-one.

(onto)
Let $y \in aHa^{-1}$. Then $y = aha^{-1}$ for some $h \in H$. As $h \in H$, $ϕ(h) = aha^{-1} = y$. Thus $ϕ$ is onto.

Hence, $ϕ$ is an isomorphism.

How do I show that $aHa^{-1}$ is a subgroup of $G$?

• Use the fact that $H$ was a subgroup to begin with to show that if $aha^{-1},aga^{-1}\in aHa^{-1}$ then their product lies in $aHa^{-1}$, and that if $aha^{-1}\in aHa^{-1}$ then it's inverse, $(aha^{-1})^{-1}=ah^{-1}a^{-1}$ is in $aHa^{-1}$. – Eric Naslund Apr 7 '14 at 1:47

Let me show you how I would write each step for the subgroup verification.

(identity)
$aa^{-1}=1\in aHa^{-1}$, so $aHa^{-1}$ has identity.

(inverses)
If $x\in H$ then $x^{-1}\in H$, so $$\left(axa^{-1}\right)^{-1}=ax^{-1}a^{-1}\in aHa^{-1}.$$

(associativity)
Associativity is trivially inherited from the supergroup.

(closure)
Take $axa^{-1}\in aHa^{-1}$ and $aya^{-1}\in aHa^{-1}$ with $a,y\in H$. $$axa^{-1}aya^{-1}=axya^{-1}$$ By closure of $H$, $xy\in H$, so $axya^{-1}\in aHa^{-1}$.

Alternatively, you could compose your homomorphism $\phi$ with the zero homomorphism $x\mapsto 1$. The image of $\phi$ is the kernel of this homomorphism, and you have proven that $\phi$ is onto.

First, we need to establish the existence of inverses:

Consider any element $aha^{-1} \in aHa^{-1}$. Then $(aha^{-1})^{-1} = ah^{-1}a^{-1}$. Certainly, this is in $aHa^{-1}$ since $h^{-1} \in H$.

Next, we check that it is closed under multiplication:

Consider a second element $ah'a^{-1} \in aHa^{-1}$. Then $ah'a^{-1}aha^{-1} = ah'ha^{-1}$. Certainly, this is in $aHa^{-1}$ since $h'h \in H$.

I'll let you check that the identity is also contained in our subgroup.

It suffices to check: $e \in aHa^{-1}$ and given arbitrary elements $x, y \in aHa^{-1}$, we have $xy^{-1} \in aHa^{-1}$.

The first condition follows from the fact that $H$ is a subgroup and so is the the second: let $x = ala^{-1}$ and $y = aka^{-1}$, then $xy^{-1} = a(lk^{-1})a^{-1} = aha^{-1} \in aHa^{-1}$.