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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $G$.

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