Convincing normal subgroup proof? I wrote the following proof on an exam, I was wondering if it makes sense.
Question: let $H$ be a subgroup of $G$, written in multiplicitive notation. Prove that if the coset multiplication defined by $(aH)(bH) = (ab)H$ is well defined for all $a,b \in G$, then $H$ is a normal subgroup.
Proof: $H$ being a normal subgroup of G means that $gH = Hg$
If we define multiplication of cosets by multiplication of their representative elements, then suppose $H$ is not normal.
i) $(aH)(bH)=(ab)H$
ii) since $Ha\ne aH$, then let $Ha = a'H$
iii) $(a'H)(bH) = (a'b)H \ne (ab)H$
So we multiplied $Ha$ and $aH$ against $bH$ and got different results. Since $Ha$  and $aH$ have the same representative elements, this means that multiplication would not be well-defined. Therefore, if $H$ was not normal, then multiplication by representative elements would not be well defined.  Therefore $H$ must be a normal subgroup of $G$.
 A: Here is a theorem you may not be aware of:

Theorem: Let $H$ be a subgroup of a group $G$. Then the following are equivalent 
  
  
*
  
*$H$ is normal in G.
  
*Every left coset of $H$ is also a right coset of $H$.
  

In particular, in your second point, by saying that $aH = Ha'$, you have implicitly assumed that $H$ is normal.
To see why the theorem is true, suppose that every left coset of $H$ is also a right coset of $H$. Let $gH$ be a left coset which is equal to the right coset $Hk$.
Then $$H = gHk^{-1}\ni gek^{-1}$$from which it follows that $gk^{-1}\in H$, and hence $kg^{-1}\in H$.
So 
$$\begin{align}
gH &= Hk\\
&= Hk(g^{-1}g)\\
&= H(kg^{-1})g\\
&= Hg&&\text{ since }kg^{-1}\in H.
\end{align}$$
Since $g$ was arbitrary, it follows that $gH = Hg$ for every $g$, so $H$ is normal.

Here is a way to prove your original result. Suppose for contradiction that the multiplication is well defined, and that $H\not\lhd G$. Then there exists some $g\in G$ and some $h\in H$ such that $ghg^{-1} \notin H$, and hence 
$$\begin{align}H \ne ghg^{-1}H &= (gH)(hH)(g^{-1}H)&&\text{by the multiplication rule}\\
 &= (gH)(eH)(g^{-1}H)&&\text{since $hH = H$}\\ 
&= (gg^{-1})H \\&= H\end{align},$$
giving the required contradiction.
