# Write-up 4

Let $H$ be a subgroup of $G$ and $a$ be in $G$. $aHa^{-1}=\left\{ aha^{-1} \mid h\in H\right\}$. $aH=\left\{ ah\mid h\in H \right\}$.

## Part 1

Theorem. $aHa^{-1}$ is a subgroup of $G$.

Proof.
Let $x$ and $y$ be arbitrary elements of $aHa^{-1}$.
Then, $x=aha^{-1}$ and $y=aka^{-1}$ for some $h$ and $k$ in $H$ by the definition of $aHa^{-1}$.

We must show that $aHa^{-1}$ is a subset of $G$ and a group.

**Subset:**
Recall $a$ is in $G$.
$h$ is in $G$ since $H$ is a subgroup of $G$.
Since $G$ is a group and $a$ is in $G$, $a^{-1}$ is in $G$.
Since $G$ is closed, $aha^{-1}$ is in $G$.
So, $x$ is in $G$.
Therefore, $aHa^{-1}$ is a subset of $G$.

**Group:**
*Closure*

\begin{align*}
xy&=aha^{-1}aka^{-1} && \text{substitution} \\
&=ahka^{-1} && \text{definition of inverse}
\end{align*}

Since $H$ is a group, $hk \in H$.
So, $xy \in aHa^{-1}$ by the definition of $aHa^{-1}$
Therefore, $aHa^{-1}$ is closed.

*Identity*
Since $G$ is a group, it has an identity $e$, which by previous theorem it shares with all its subgroups including $H$.

\begin{align*}
e&=aa^{-1} && \text{definition of inverse} \\
&=aea^{-1} && \text{definition of identity}
\end{align*}

$e$ is obviously in $H$.
So, $e$ is in $aHa^{-1}$ by the definition of $aHa^{-1}$.
Therefore, $aHa^{-1}$ has the identity.

*Inverses*
Since $G$ is a group, $a^{-1}$ exists.
Let $t=ah^{-1}a^{-1}$.
Since $H$ is a group, $h^{-1}$ is in $H$.
So, $t$ is in $aHa^{1}$ by definition.

\begin{align*}
xt&=aha^{-1}ah^{-1}a^{-1} && \text{substitution} \\
&=ahh^{-1}a^{-1} && \text{definition of inverse} \\
&=aa^{-1} && \text{definition of inverse} \\
&=e && \text{definition of inverse}
\end{align*}

\begin{align*}
tx&=ah^{-1}a^{-1}aha^{-1} && \text{substitution} \\
&=ah^{-1}ha^{-1} && \text{definition of inverse} \\
&=aa^{-1} && \text{definition of inverse} \\
&=e && \text{definition of inverse}
\end{align*}

So, $t$ is the inverse of $x$ by the definition of inverse.
Therefore, $aHa^{-1}$ has inverses.

*Associative*
Since $G$ is a group, the operation is associative.

So, $aHa^{-1}$ is a group.

Therefore, $aHa^{-1}$ is a subgroup of $G$.

## Part 2

Theorem. If $a \in H$, then $aH = H$.

Proof (by double inclusion).
Assume $a \in H$.

First, we must show $aH \subseteq H$.
Let $x$ be an arbitrary element of $aH$.
So, $x = ah$ for some $h$ in $H$, by the definition of $aH$.
Since $a \in H$ and $h \in H$ and $H$ is a group with closure, $x \in H$.
Therefore, $aH \subseteq H$.

Second, we must show $H \subseteq aH$.
Let $x$ be an arbitrary element of $H$.
\begin{align*}
x &= ex && H \text{ has closure} \\
&= aa^{-1}x && \text { definition of inverse} \\
&= a(a^{-1}x) && \text { associativity}
\end{align*}
Since $H$ is closed, $a^{-1}x \in H$.
So, $x \in aH$, by definition.
Therefore, $H \subseteq aH$.

Therefore, by double inclusion, $aH = H$.