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 $G$.

  • 1
    $\begingroup$ 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? $\endgroup$ – Noble Mushtak Feb 13 '16 at 23:17
  • 3
    $\begingroup$ $\;H\;$ 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). $\endgroup$ – DonAntonio Feb 13 '16 at 23:17
  • $\begingroup$ noble- i wrote word for word the prompt. it said '"if .....then H is a normal subgroup. $\endgroup$ – mac5 Feb 13 '16 at 23:21
  • $\begingroup$ Does my statement from ii) and iii) work? Did i actually multiply the same representative elements? $\endgroup$ – mac5 Feb 13 '16 at 23:24
  • 1
    $\begingroup$ 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 $\;aH=Ha'\;$ . $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.