Normality of subgroups $G$ is a group and $H$ is a subgroup of $G$ such that $\forall a, b$ in $G, ab\in H\implies ba\in H$. Show that $H$ is normal in $G$
 A: Let $G$ be a group, and let $H$ be a subgroup. In analogy to the definition of congruence for integers, $a\equiv b\pmod{m}$ if and only if $m|a-b$, if and only if $a-b\in\langle m\rangle$, define:


*

*$a$ is congruent on the left modulo $H$ to $b$, $a\mathrel{{}_H{\equiv}} b$, if and only if $a^{-1}b\in H$.

*$a$ is congruent on the right modulo $H$ to $b$, $a\mathrel{\equiv_H} b$, if and only if $ab^{-1}\in H$.


Proposition. Both $\mathrel{\equiv_H}$ and $\mathrel{{}_H{\equiv}}$ are equivalence relations on $G$.
Theorem. Let $G$ be a group, and let $H$ be a subgroup of $G$. The following are equivalent:


*

*For all $a\in G$ there exists $b\in G$ such that $aH=Hb$; that is, every left coset of $H$ is also a right coset of $H$.

*For all $a\in G$, $aH = Ha$.

*For all $a\in G$, $a^{-1}Ha=H$.

*For all $a\in G$, $a^{-1}Ha\subseteq H$.

*For all $a,b\in G$, $a\mathrel{\equiv_H}b$ if and only if $a\mathrel{{}_H{\equiv}} b$; that is, congruence on the left modulo $H$ and congruence on the right modulo $H$ are the same equivalence relation on $G$.

*If $aH=bH$ and $xH=yH$, then $axH=byH$ (multiplication of left cosets is well-defined).

*If $Ha = Hb$ and $Hx = Hy$, then $Hab = Hxy$ (multiplication of right cosets is well-defined).


If you prove this theorem (you may want to; it's a good exercise and it gives you other ways of checking that a group is normal) then condition 5 gives you the result: 
$$\begin{align*}
a\mathrel{{}_H{\equiv}} b &\iff a^{-1}b\in H\\
&\iff ba^{-1}\in H\\
&\iff b\mathrel{\equiv_H} a\\
&\iff a\mathrel{\equiv_H} b
\end{align*}
$$
A: Your goal is to show $xH=Hx$ for all $x\in G$ so you should show set containment both ways.
Suppose $y \in xH$. This means that there exists some $h \in H$ such that $y=xh$. Notice that $x^{-1}y=h \in H$. Now invoke your hypothesis. I'll let you finish from there.
Do be careful though. If $ab=h_1 \in H$, then you know $ba \in H$. This does not mean $ba=h_1$ but just that $ba$ is some element of $H$ (it could be a different element).
A: Choose $x\in G$ and $h\in H.$ Let $a=x^{-1}.$
Suppose $ha=b.$ 
Then $ba^{-1}=h$
$\implies a^{-1}b=h_1$ for some $h_1\in H$ (by the given condition)
$\implies b=ah_1$
$\implies ha=ah_1$
$\implies a^{-1}ha=h_1\in H$
$\implies xhx^{-1}\in H.$
Consequently $H\triangleleft G.$
