TITLE: Prove that if H is a subgroup of G, then H is a normal subgroup of G iff $\forall x, y \in G$, xy $\in$ H iff yx $\in$ H.
QUESTION [6 upvotes]: So the problem is as follows:
Prove that if H is a subgroup of G, then H is a normal subgroup of G iff the following condition holds:
$$\forall x,y \in G, xy \in H \iff yx \in H$$
It's all completed now:
We are given that H is a subgroup of G.
($\def\impl{\;\Rightarrow\;}\impl$) Assume H is a normal subgroup of G. So,  $$\forall h \in H, \forall g \in G, ghg^-1  \in H.$$ Suppose $$\forall x,y \in G, xy \in H.$$ Since H is a normal subgroup of G and $$y \in G,$$ we know $$y(xy)y^-1\in H \impl yx(yy^-1)\in H \impl yxe \in H \impl yx \in H.$$ Similarly, suppose $$\forall x,y \in G, yx \in H.$$ Since H is a normal subgroup of G and $$x \in G,$$ we know $$x(yx)x^-1\in H\impl xy(xx^-1)\in H\impl xye \in H\impl xy \in H.$$
($\;\Leftarrow\;$) Assume $$\forall x,y\in G, xy\in H \iff yx \in H.$$ Now let $$a=yx \impl xa=x(yx) \impl (xa)x^-1=(xy)xx^-1 \impl xax^-1 = xy.$$ So $$ \forall x\in G, \forall a\in H,  a \in H \impl xax^-1 \in H.$$ Therefore H is a normal subgroup of G.

REPLY [6 votes]: Note first that the elements $xy$ and $yx$ are conjugate, as
$$
x^{-1} (xy) x = y x. 
$$
Conversely, if $a$ and $b$ are conjugate, $b = z^{-1} a z$, say, then
$$
b = (z^{-1} a) z, \quad a = z (z^{-1} a)
$$
so $a = xy$ and $b = yx$, for $x = z$ and $y = z^{-1} a$.
Now use the definition of normal subgroup.