# 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 $G$.

• The phrase "if the coset multiplication defined by (aH)(bH) = (ab)H for all a,b that are elements of G" is not a complete clause. It's not a complete thought or a complete logical statement. What about the "coset multiplication" are you trying to say? – Noble Mushtak Feb 13 '16 at 23:17
• $\;H\;$ being a subgroup of $\;G\;$ certainly does not mean $\;Hg=gH\;$ for all $\;g\in G\;$. In fact, being a normal subgroup is equivalent to say that any left coset is also a right one (and the other way around, of course). – DonAntonio Feb 13 '16 at 23:17
• noble- i wrote word for word the prompt. it said '"if .....then H is a normal subgroup. – mac5 Feb 13 '16 at 23:21
• Does my statement from ii) and iii) work? Did i actually multiply the same representative elements? – mac5 Feb 13 '16 at 23:24
• What you wrote in (ii) already assumes $\;H \lhd G\;$ . Read again my comment above: every left coset is a right one means that for any $\;a\in G\;$ there exists $\;a'\in G\;$ such that $\;aH=Ha'\;$ . – DonAntonio Feb 13 '16 at 23:27

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

1. $H$ is normal in G.
2. 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.

• Very beautiful. +1 – DonAntonio Feb 14 '16 at 0:04
• ah, that makes much more sense. thanks a lot man. – mac5 Feb 14 '16 at 0:08