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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.

  • 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 08 '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. –  Aug 08 '12 at 22:31
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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. – Code-Guru Aug 08 '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 08 '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 08 '12 at 23:15
  • @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. – anon Aug 08 '12 at 23:23
  • @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. – M Turgeon Aug 09 '12 at 00:06
  • How to prove HaHb = Hc then HaHb = Hab – Flintoff Sep 19 '18 at 19:38

4 Answers4

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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.)

Community
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anon
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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 09 '12 at 00:05
  • 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 09 '12 at 00:07
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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. – Jack Schmidt Aug 09 '12 at 00:17
  • 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 09 '12 at 02:07
  • @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 09 '12 at 02:23
  • 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"? ). – David Wheeler Aug 09 '12 at 02:32
  • @anon, I know all that. Read my last comment above. – DonAntonio Aug 09 '12 at 02:32
  • 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). –  Aug 09 '12 at 04:05
  • 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? – Mahathi Vempati Mar 16 '17 at 17:14
  • @Tinkidinki I believe so, beacuse we just use that $1 \in HaHa^{-1}$. – Danilo Gregorin Afonso Sep 06 '17 at 14:08
  • How to prove HaHb = Hc then HaHb = Hab – Flintoff Sep 19 '18 at 19:38
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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 $X=Hg$.) – anon Sep 20 '18 at 02:06
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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 $=$.

Jack Schmidt
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Suppose g ∈ G and consider the product of two right cosets HgHg−1 .
Observe egeg−1 = e ∈ HgHg−1 and since by hypothesis HgHg−1 is a right coset, it would have to be the right coset H = He as right cosets are either equal or disjoint and in this case e ∈ H = He so we must have equality HgHg−1 = H.
Therefore, for every g ∈ G we have HgHg−1 = H so clearly gHg−1⊆ HgHg−1 = H. Hence, gHg−1 ⊆ H for all g ∈ G so H is a normal subgroup of G.

harsh kumar Chourasia
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Suppose, if possible, for all $a,b\in G$, $(Ha)(Hb)$ is a right coset of $H$. We intend to show that $H\trianglelefteq G$.

Notice that $ab$ is an element of $HaHb$. $\because ab = eaeb\in HaHb$.

As per our supposition, $HaHb$ should be a right coset of $H$. However, $Hab$ is the only right coset containing $ab$ because of the fact that cosets are either identical or disjoint. Since $ab\in HaHb\cap Hab$, we can conclude that the only possibility is $HaHb=Hab$.

For all $h\in H$, $eahb\in HaHb$ so, we need $(aha^{-1})ab\in Hab$ i.e., possible only if $aha^{-1}\in H$.

$\therefore H\trianglelefteq G\quad\blacksquare$

Alternate Proof:

Suppose, if possible, $HaHb$ is a right coset of $H$. We intend to show that $H\trianglelefteq G$.

For $h_1, \ h_2, \ h_3, \ h_4\in H$, we have two arbitrary elements $h_1ah_2b,$ $h_3ah_4b\in HaHb$. Since right cosets are equivalence classes under the relation $x\equiv y\pmod{H}$ iff $xy^{-1}\in H$ and $HaHb$ is a right coset, any two elements of $HaHb$ are related, i.e., $(h_1ah_2b)(h_3ah_4b)^{-1}\in H\tag*{}$

Using shoe-lace formula and associativity of the group operation, we have: $(h_1ah_2b)(h_3ah_4b)^{-1}\\=h_1a\left(h_2h_4^{-1}\right)a^{-1}h_3^{-1}\in H\tag*{}$ i.e., possible iff $a\left(h_2h_4^{-1}\right)a^{-1}\in H$ (by closure).

Let $h_2h_4^{-1}=h\in H$ (by closure property). Notice that $h$ is an arbitrary element of $H$ because $h_2$ and $h_4$ were picked arbitrarily. $a$ is also an arbitrary element of $G$.

Thus, we need $aha^{-1}\in H$ for all $a\in G$ and $h\in H$ i.e., $H\trianglelefteq G$ . $\blacksquare$

Proof of the converse:

Suppose $H\trianglelefteq G$. We intend to show that for all $a,\ b\in G$, $\exists$ $c\in G$ such that $(Ha)(Hb)=Hc$.

For all $h_1, \ h_2\in H$, clearly $h_1ah_2b\in HaHb$.

By normality and closure of $H$, $h_1ah_2b=\underbrace{h_1(ah_2a^{-1})}_{\in H}ab\in Hab\tag*{}$ $\therefore \ HaHb\subseteq Hab$.

For all $h\in H$ and $hab\in Hab$, $hab=(ha)(eb)\in HaHb\tag*{}$

$\therefore \ Hab\subseteq HaHb$.

Hence, $(Ha)(Hb)=H(ab)=Hc$ for all $c\in G$ such that $abc^{-1}\in H$. $\blacksquare$

Conclusion:

Product of two cosets being a coset is a necessary and sufficient condition for normality of a subgroup, so it's indeed an equivalent definition!

For $H\trianglelefteq G$, it makes perfect sense to define the direct product as $(Ha)(Hb)=H(ab)$

  1. As a consequence of the result we have proven above: If you take the set of all cosets of $H$ in $G$ i.e., $\{gH\ \forall\ g\in G\}$, it will be closed under this direct product operation. $\because$ $H\trianglelefteq G$

  2. This operation is well defined too. In other words, the result of this operation does NOT depend on the choice of coset representatives for the operand cosets. Suppose $Ha=Hx$ and $Hb=Hy$ then show that $(Ha)(Hb)=(Hx)(Hy)$ i.e., $H(ab)=H(xy)$.

  3. Check that this operation is associative.

  4. $He$ i.e., $H$ itself will serve as the identity element. $\because$ $(Ha)(He)=H(ae)=Ha$.

  5. $Ha^{-1}$ will serve as inverse of $Ha$. $\because (Ha)(Ha^{-1})=H(aa^{-1})=He$

This set of cosets is, of course, a group i.e., the quotient group $G/H$.

Nothing special
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