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I cannot see how 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?2012-08-08
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    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.2012-08-08
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    @MTurgeon $G/H$ is a group only when $H$ is normal. If $H$ is not normal, multiplication of cosets is not well-defined.2012-08-08
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    @limac246 What have you tried? I suggest starting with the definitions of a normal subgroup and coset multiplication.2012-08-08
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    @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.2012-08-08
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    @MTurgeon $(aH)(bH):=(ab)H$ is not the "usual way to define multiplication" that I am aware of; it just happens to be true when $H$ is normal (indeed this equation is ill-defined otherwise because the output depends on choice of representatives). Usually the multiplication of two subsets of a group is defined just the way limac246 intimated, $$AB:=\{ab:a\in A,b\in B\}.$$ On this it is not immediate (nor true) that products of cosets are necessarily cosets.2012-08-08
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    @anon I just asked for the definition of multiplication the OP was using in order to understand what she was asking. I don't know why this spanned so many comments.... This is all besides the point.2012-08-09
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    How to prove HaHb = Hc then HaHb = Hab2018-09-19

2 Answers 2

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Hint: if $HaHa^{-1}$ and $Ha^{-1}Ha$ are right cosets they must be $H$ because they contain the identity.

(I have updated my hint to involve both $HaHa^{-1}$ and $Ha^{-1}Ha$ because $aHa^{-1}\subseteq H$ is not by itself equivalent to $aHa^{-1}=H$ when $H$ is infinite; see counterexamples here, here, here.)

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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?2012-08-09
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    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..2012-08-09
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    @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.2012-08-09
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    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! +12012-08-09
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    @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.2012-08-09
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    Well, duh, $He = H$ and if $S \subseteq H$ surely $SH \subseteq HH = H$. OTOH, if $sh \in H$ then $s \in HH^{-1} = H$ (or should I have said "derp"? ).2012-08-09
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    @anon, I know all that. Read my last comment above.2012-08-09
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    Thank you very much (to all) for the help. Strange that I haven't seen this question anywhere but Herstein. Most sources usually just show that if H is normal then the product of two (right) cosets is a (right) coset, and not consider the other direction (i.e. the converse).2012-08-09
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    Why do we need the information that product of two right cosets is a right coset for this to hold true? If I just have the information that the product of two right cosets is a coset (not necessarily right) , then I can solve this right?2017-03-16
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    @Tinkidinki I believe so, beacuse we just use that $1 \in HaHa^{-1}$.2017-09-06
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    How to prove HaHb = Hc then HaHb = Hab2018-09-19
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    @Hansie $HaHb$ contains $eaeb$, which is $ab$, and thus $HaHb$ is $Hab$. (If $g$ is any element of a right coset $X$ then $X=Hg$.)2018-09-20
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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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    How to prove HaHb = Hc then HaHb = Hab2018-09-19