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I need to show the following:

Let $H$ be a subgroup of $G$. $H$ is normal iff it has the following
property: for all $a,b in G$, $ab in H$ iff $ba in H$.

I have to use the following definition of a normal subgroup:

Let $H$ be a subgroup of $G$. $H$ is called a normal subgroup of $G$
if it is closed with respect to conjugates, i.e. if for any $a in H$
and $x in G$, $xax^{-1} in H$.

I tried to prove the $(Rightarrow )$ part, but I could not suceed. However, I proved the $(Leftarrow )$ part.
Here it is:

Let $h in H$ be arbitrary. Then for any $xin G$, $eh = (x^{-1}x)h=(x^{-1})(xh) in H$. Hence, it follows by the property that $(x^{-1})(xh) in H Rightarrow xhx^{-1} in H$. Thus, we are done.

For the $(Rightarrow )$ part, I know I need to pick any $a, b in G$ and assume $ab in H$ then I need to show $ba in H$ and also show the converse. Here's how I start: since $ab in H$, for any $xin G$, we have $xabx^{-1} in H$. I tried a couple of things but I didn't get anywhere. Can I get some hints?

forward implication: Let $ab in H$: Since $H$ is normal, then also $a^{-1}aba=ba in H$. (The other implication of the property follows by its symmetry in $a$ and $b$).

A different way to see it: $abin H$ if and only if $a$ and $b^{-1}$ are in the same right coset of $H$. $bain H$ if and only if $b$ and $a^{-1}$ are in the same right coset of $H$, which is equivalent to saying that $a$ and $b^{-1}$ are in the same left coset of $H$.

Thus the property under consideration means exactly that left and right cosets coincide, which you can probably easily see as equivalent to the definition you have given.