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I cannot seem to find a way to prove that if $H$ is a subgroup of $G$ such that the product of two right cosets of $H$ is also a right coset of $H,$ then $H$ is normal in $G.$

(This is from Herstein by the way.)

Thank you.

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There might be something missing from your question. What is your definition of the product of two cosets? Isn't it always a coset? – M Turgeon Aug 8 '12 at 22:27
For $a,b \in G,$ $HaHb = \{(h_1a)(h_2b) | h_1,h_2 \in H\}.$ i.e. we want to show that if for any $a,b \in G,$ we have that for some $c \in G, HaHb=Hc,$ then $H$ is normal. – limac246 Aug 8 '12 at 22:31
@MTurgeon $G/H$ is a group only when $H$ is normal. If $H$ is not normal, multiplication of cosets is not well-defined. – Code-Guru Aug 8 '12 at 22:58
@limac246 What have you tried? I suggest starting with the definitions of a normal subgroup and coset multiplication. – Code-Guru Aug 8 '12 at 22:59
@Code-Guru I know this. My point is that the usual way to define multiplication shows that the product of two cosets is a "coset", e.g. $(aH)(bH)=(ab)H$. The point is the well-definedness; I was wondering if this is what the OP was trying to show. – M Turgeon Aug 8 '12 at 23:15
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2 Answers

up vote 7 down vote accepted

Hint. $~$ If $Ha^{-1}Ha$ is a right coset it must be $H$ because it contains the identity.

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I'm not sure how this, as it is written, answers the OP's question: we know $\,H\,$ is a subgroup and we want to show that if a product of right cosets (the pointwise product, I guess, stemming from the group operation) is again a right coset then this sbgp. is in fact normal. Now, this follows exactly from the above: $$\forall a\in G\,\forall x,y\in H\,\exists h\in H\,\,s.t.\,\,xa^{-1}ya=h\Longrightarrow a^{-1}ya=x^{-1}h\in H$$and we're done. But how the identity in $\,H\,$ helps here? – DonAntonio Aug 9 '12 at 0:05
Hmmm...and still the above isn't complete (im my mind, of course) as the rightmost rightcoset doesn't have to be $\,H\,$, it could be $\,Hb\,$, say.. – DonAntonio Aug 9 '12 at 0:07
@DonAntonio: this is a complete proof. If $Ha^{-1}Ha=H$, then $a^{-1}Ha=H$ and $H$ is normal. In particular, it is not necessary to assume that $HaHb=Hc$ for all pairs $(a,b)$, but only for pairs where $b=a^{-1}$. The reason we need the identity for $c$ is so that you get your $xa^{-1}ya = hc$ to actually be in $H$. If $c$ were not in $H$, then we would not conclude $H$ is normal, but rather reach a contradiction. – Jack Schmidt Aug 9 '12 at 0:17
Exactly my point, @JackSchmidt ! How can we know a priori that $\,Ha^{-1}Ha=H\,$ ? This is the whole point. Of course, we can argue that since $\,Ha^{-1}Ha=Hc\,$ then for all $\,x,y\in H\,$ we have that $\,xa^{-1}ya=hc\,$ , in particular if we choose $\,y=1\in H\Longrightarrow xa^{-1}a=x=hc\in G\Longleftrightarrow c=1\,$ as we know right cosets are a partition of $\,G\,$...hmm, perhaps this is what anon meant...yes, I think it is and I didn't see clearly his hint though I knew that taking $\,Ha\,,\,Ha^{-1}\,$ is the way to prove the claim...damn, what a nice though aethereal hint! +1 – DonAntonio Aug 9 '12 at 2:07
@DonAntonio It is a fact that the only coset of a given subgroup containing the identity is the subgroup itself. Another fact is that for any subset $S\subseteq G$ and subgroup $H\le G$, we have $SH\subseteq H\iff S\subseteq H$. (Same for $HS$.) In my opinion these should be standard exercises. – anon Aug 9 '12 at 2:23
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up vote 3 down vote

A (not quite as) short alternate proof:

If $HaHb=Hc$ then $HaHb=Hab$. @anon's short proof chooses $b=a^{-1}$, but you can also choose $b=1$, since $$HaH = Ha \iff 1aH \subseteq Ha$$

Of course to get equality, we also have to use $$Ha^{-1}H =Ha^{-1} \iff a^{-1} H \subseteq Ha^{-1} \iff Ha \subseteq aH $$

In general, $HaHb=Hab \iff aHb \subseteq Hab$, so if we want $aH=Ha$ we choose $b=1$ and if we want $aHa^{-1}= H$ we choose $b=a^{-1}$. If groups are finite, we don't even have to pay attention to $\subseteq$ versus $=$.

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