Proving that a subgroup is normal if two equivalence relations on $A$ coincide

Here is the setup for the proposition I am trying to prove.

Let $A$ be a group and $A' \subset A$ a subgroup. We define the left relation $\sim^{l}$ on $A$ with respect to $A'$ by declaring that $a_1 \sim^{l} a_2$ if and only if there exists $a' \in A'$ such that $a_2 = a' a_1$. We define the right relation $\sim^{r}$ on $A$ wth respect to $A'$ by declaring that $a_1 \sim^{r} a_2$ if and only if there exists $a' \in A'$ such that $a_1 = a_2 a'$.

I did the verification that $\sim^{l}$ and $\sim^{r}$ are equivalence relations and was able to prove that if $A'$ is normal in $A$, then $\sim^{l}$ and $\sim^{r}$ coincide. Now I'm trying to prove the converse, namely that if $\sim^{l}$ and $\sim^{r}$ coincide, then $A'$ is normal.

The definition of normal I'm working with is:

A subgroup $A' \subset A$ is normal if for any $a \in A$ and $a' \in A'$, $aa' a^{-1} \in A'$.

I'm very unsure on how to proceed. I know that need to prove that given any $a \in A$ and $a' \in A'$, the element $a a' a^{-1}$ lives in $A'$. $A$ and $A'$ are groups, so I can fix $a \in A$ and $a' \in A$. I can't form a relation amongst them because I don't know walking into the problem that these are actually related under $\sim^{l}$ or $\sim^{r}$, though I know I need to use the hypothesis in some way.

I also tried to reverse the proof that if $A'$ is normal in $A$, then $\sim^{l}$ and $\sim^{r}$ coincide, but without any success.

I'd greatly appreciate some hints on how to proceed. I'll update this post as I work on this problem further.

Solution 1:

Hint: Let $x\in A'$ and $a\in A$. We want to show $axa^{-1}\in A'$.

Note that $a\sim^r ax$, hence $a\sim^{l}ax$, and therefore (since it is an equivalence relation), $ax\sim^l a$. Thus, there exists $y\in A'$ such that $ya=ax$.

Solution 2:

Yes, you need to prove that $aa'a^{-1}\in A'$ if $a'\in A'$. Note that $a\sim^r aa'$. Since $\sim^r=\sim^l$, $a\sim^l aa'$. So $aa'=a''a$ with $a''\in A'$. Then $aa'a^{-1}=a''aa^{-1}=a''\in A'$ as desired.