Let H be a subgroup of a multiplicative group G such that the product of any two of left cosets of H is a left coset of H. Is H normal in G? Prove or disprove.

I got a hint to solve the problem as follows: Let $a, b, c\in G$ then by the given condition, we have

$$(aH)(bH)=cH \implies a(Hb)H=cH$$

$$\implies Hb=a^{-1}cH\implies b\in a^{-1}cH \implies bH=a^{-1}cH=Hb$$

I am unable to understand the entire second line.

Problem 1: How to get $a(Hb)H=cH\implies Hb=a^{-1}cH$

Problem 2: $Hb=a^{-1}cH\implies b\in a^{-1}cH$ (Is it due to the fact that as $e\in H$ then $eb\in Hb$ and consequently $eb=b\in a^{-1}cH$)

If it is correct, please help me to understand the solution.

Another hint is given as $(aH)(bH)=cH \implies cH=abH$. Using it how to solve the problem.

In the case of 2nd Hint, clearly, $e\in H \implies ae\in aH ~~\& ~~be\in bH \implies aebe\in aHbH ~~ i.e. ~~ab \in cH$ but $abe\in abH \implies ab\in abH$ i.e $ab$ is common in $aHbH$ and $abH$ then $aHbH=abH$. (As we know that any two left cosets are either equal of disjoint.)

But how to conclude the proof.

  • $\begingroup$ The main idea in the second line seems to be that the (left) cosets are disjoint, meaning that if they share any one element (for instance, $bH$ and $a^{-1}cH$ both contain $b$), then they are in fact equal. I have to think a little bit about the last $=Hb$. $\endgroup$ – Arthur Aug 25 '16 at 5:26
  • $\begingroup$ @Arthur the last $=Hb$ is just from the first equality on that line. $\endgroup$ – stewbasic Aug 25 '16 at 5:33

Edited to make the argument a bit more efficient.

I think this is more or less what the first hint is suggesting:

$aHbH = cH$, so $ab = aebe \in aHbH = cH$, where $e$ is the identity element. Multiplying on the left by $a^{-1}$ gives us $b \in a^{-1}cH$. Thus the left coset containing $b$ is $a^{-1}cH$, and so $bH = a^{-1}cH$.

Now, $aHb = aHbe \subseteq aHbH = cH$, so $Hb \subseteq a^{-1}cH = bH$ (the last equality was proved in the previous paragraph).

We have established that $Hb \subseteq bH$. This holds for any $b$, so in particular it holds for $b^{-1}$, and so $Hb^{-1} \subseteq b^{-1}H$. Multiplying on the left and right by $b$ gives us $bH \subseteq Hb$.

We have shown both containments, so $bH = Hb$, hence $H$ is normal.

| cite | improve this answer | |
  • $\begingroup$ I have edited the first equality of the 2nd line. Please see. $\endgroup$ – user1942348 Aug 25 '16 at 5:40
  • $\begingroup$ $Hb=a^{-1}cH$ not $HbH = a^{-1}cH$ as written in the Hint. $\endgroup$ – user1942348 Aug 25 '16 at 6:01

An alternative approach is to observe that since $ab\in(aH)(bH)$, we must have $(aH)(bH)=abH$. We can now define an operation $*$ on the set $\mathscr{L}$ of left cosets of $H$ by $aH*bH=abH$. It's straightforward to check that this operation is a well-defined group operations on $\mathscr{L}$ with identity $H=1_GH$ and inverses given by $(aH)^{-1}=a^{-1}H$. The map $h:G\to\mathscr{L}:a\mapsto aH$ is therefore a homomorphism, and since $\ker h=H$, $H$ must be normal in $G$.

| cite | improve this answer | |

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.