A subgroup $H$ of $G$ is normal iff for all $a,b \in G$, $ab \in H \iff ba \in H$.

Question

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 $x\in G$, $eh = (x^{-1}x)h=(x^{-1})(xh) \in H$. 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 $x\in G$, we have $xabx^{-1} \in H$. I tried a couple of things but I didn't get anywhere. Can I get some hints?

Answer 1

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 $b$).

Answer 2

A different way to see it: $ab\in H$ if and only if $a$ and $b^{-1}$ are in the same right coset of $H$. $ba\in H$ 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.