Normal subgroup condition I have the following question I cannot solve:
Lt $H$ be a subgroup of $G$. Show that $H$ is normal subgroup iff $a,b\in G, ab\in H$ implies that $ba\in H$
Can someone please give me a hint how to go about this? I could not get anywhere from definitions
Thanks!
 A: 1) Suppose $\;a,b\in G\,,\,ab\in H\implies ba\in H\;$ . Then for all $\;x\in G,\,h\in H\;$ we get
$$h\cdot1=(h\cdot x)x^{-1}\in H\implies x^{-1}hx\in H\implies H\lhd G$$
2) If $\;H\lhd G\;$ then $\;xhx^{-1}\in H\;$ for all $\;x\in G\,,\,h\in H\;$ , thus for any $\;a,b\in G\;$ such that $\;ab\in H\;$ , we have that
$$ba=a^{-1}(ab)a\in H\;\;\;\text{by normality}\;\;$$ 
A: Well, if $H$ is normal in $G$, with $a, b \in G$ and $ab \in H$, we have $ab = h \in H$, whence $b = a^{-1}h \in a^{-1}H$, whence $ba = a^{-1}ha \in a^{-1}Ha = H$ by normality of $H$.
The other way is pretty simple too, as indicated by Joanpemo in his answer.  We include a minor variant of this apparently standard argument here for the sake of completeness:
Suppose then than $h \in H$ and $g \in G$.  Then we have $hg \in G$ and also $g^{-1} \in G$; but
$(hg)g^{-1} = h(gg^{-1}) = he = h \in H, \tag{1}$
where $e \in G$ is the group identity element; thus by hypothesis we have
$g^{1}hg \in H \tag{2}$
as well.  Since (2) has been shown to hold for all $h \in H$ and $g \in G$, we conclude that $H$ is a normal subgroup of $G$. 
A: $$ab \in H$$ implies $$b \in a^{-1}H$$ impies (normal subgroup) $$b \in Ha^{-1}$$ implies $$ba \in H.$$
