# 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.

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 $$eb=b\in a^{-1}cH$$)

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.

• 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 $=Hb$. – Arthur Aug 25 '16 at 5:26
• @Arthur the last $=Hb$ is just from the first equality on that line. – 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.

• I have edited the first equality of the 2nd line. Please see. – user1942348 Aug 25 '16 at 5:40
• $Hb=a^{-1}cH$ not $HbH = a^{-1}cH$ as written in the Hint. – 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 $G$.