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(Note: This question emerged out of mere curiosity in the definition of a normal subgroup.)

Let $(G, \cdot )$ be a group, and let $H$ be a subgroup. Now, for each coset in the set of (left ) cosets, $G/H,$ pick a representative element$^{1}$. Define a binary operation $*: G/H\times G/H \to G/H$ by defining the product of two cosets to be the coset containing the product of their representatives.

Is $(G/H, *)$ a group if and only if $H$ is a normal subgroup of $G$?

I can show the $\leftarrow$ direction quite easily since the inverse and identity conditions for $*$ are trivial and associativity follows from normality. The $\rightarrow$ direction, however, evades me, and I think it might be false.

$1$: So for each coset we are choosing one element inside it. In the infinite case, we can use the Axiom of Choice to do this.

Arturo Magidin
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Vivaan Daga
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  • @Ben Why is it a group homomorphism? I don’t think it’s that simple. – Vivaan Daga Jun 17 '24 at 09:02
  • Yeah, I just realised, sorry. – Ben Jun 17 '24 at 09:10
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    I think the dihedral group of order $6$ should be a counterexample. With the usual "$r$, $s$" presentation, think about $H = {e, s}$ and the set of representatives ${e, r, r^2}$.. – Izaak van Dongen Jun 17 '24 at 10:08
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    You did not explicitly say whether the hypothesis is for all choices of representative of left coset, $(G/H,)$ is a group or exists a choice of left representative making $(G/H,)$ a group. I remark that in the former case, this implies $H$ is normal. This is a partial answer, and thus I write it as a comment instead of an answer. Proof: I claim $H$ is the identity of $G/H$. In fact, if you choose $hH$ to represent $H$, then $xH*hH=xhH=xH$, i.e. $H=hH$ is the identity on the right. (not finished) – Asigan Jun 17 '24 at 11:42
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    (continuing) Since $G/H$ is a group, we conclude $H=hH$ must be the identity. Now select any $h\in H$ and $x\in G$. If $x\in H$ then $x^{-1}hx\in H$ is clear. If $x\notin H$, then choose a representative set containing $h,x$. Since $hH$ is the identity, therefore $xH=hH*xH=hxH$ $\implies$ $x^{-1}hx\in H$. – Asigan Jun 17 '24 at 11:43
  • The proof that multiplication of cosets via representatives is well defined if and only if $H$ is normal is the first Theorem in this answer, as well as other results. – Arturo Magidin Jun 17 '24 at 15:35
  • So, to clarify: first you pick a representative from each coset, then you define the multiplication exclusively relative to those representatives, rather than define it via any representative, right? – Arturo Magidin Jun 17 '24 at 15:36
  • @ArturoMagidin Yes, which is why that answer does not apply in my case. – Vivaan Daga Jun 17 '24 at 15:39
  • As the answers and the comment shows, this can occur when $H$ has a normal complement as well. I wonder if it is the case that this happens if and only if $H$ is either normal, or has a normal complement (i.e., is a retract). Clearly those two conditions suffice. – Arturo Magidin Jun 17 '24 at 16:41

3 Answers3

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Let $G=N\rtimes H$ be a semidirect product of $H$ and $N$. While $N$ is always a normal subgroup of $G$, in general $H$ is not normal.

Note that $G=\{gh: g\in N, h\in H\}$, therefore the left cosets of $H$ are $\{ghH=gH:g\in N\}$. Now we pick $g\in gH\cap N$ to be the representative of $gH$, then clearly $(G/H, *)\simeq N$ is a group, but $H$ is not necessarily normal. This examples has the stronger property that the representatives form a group.

Just a user
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    I specified in the question that I am talking about left cosets, but again I think this answers suffers from the same issues raised earlier. The issue is that I don’t think it’s trivial that the “obvious” homomorphism is actually a homomorphism, if you see the definition of * in the question carefully. – Vivaan Daga Jun 17 '24 at 09:51
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    @VivaanDaga Sorry for the misunderstanding. Now fixed. – Just a user Jun 17 '24 at 11:41
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Let $G$ be the group of invertible $2\times 2$ real matrices and let $H$ be the subgroup of diagonal matrices with non-zero diagonal entries.

Note that $H$ is not normal in $G$ since, for example, $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 1\end{pmatrix}$ is not diagonal.

Now we look at cosets $AH$ where $A\in G$. Note that $BH=AH$ if and only if $B=AD$ for some invertible diagonal matrix $D$. This means that a left coset $AH$ just corresponds to different scalings of the individual columns of $A$.

So let us pick as the representative for each class the one that has normalized columns. We can represent such a matrix as $\begin{pmatrix} \cos a & \sin b \\ \sin a & \cos b\end{pmatrix}$ with $a,b\in [0,2\pi)$. So we will denote the representative of the coset $AH$ as $A_{a,b}$ with this convention.

Clearly $A_{0,0}$ will be the identity and we can take as inverse for $A_{a,b}$ the matrix $\begin{pmatrix} \cos b & -\sin b \\ -\sin a & \cos a\end{pmatrix}$ after normalizing its columns. The product will be the in coset of the identity. So I think this will make $(G/H,*)$ a group.

AnCar
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  • What is a bit unusual here and quite confusing is that when one defines $*$ on $(G/H)\times(G/H)$, one really means that the function acts on the specifically chosen representatives of the chosen cosets. This circumvents the usual need for the function being well defined w.r.t. choosing different representatives. – AnCar Jun 17 '24 at 10:46
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The key to this is showing that your binary operation on the set of cosets is well defined. (It won't be if the result can depend on the choice of representative elements for the two cosets in the product.)

It is well defined if and only if the subgroup is normal. The Wikipedia page on Quotient Groups goes through the detail.

Michael
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    The binary operation in my question is perfectly well defined in that such a binary operation exists. This is because we start by picking representatives from the offset. I think you have misread my question. – Vivaan Daga Jun 17 '24 at 09:34
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    It is often convenient to think about cosets using representative elements of the parent group, but cosets are objects in their own right. If you define an operation using representatives, it is an operation from an ordered pair of representatives to a coset. You can't simply decree that it is an operation from an ordered pair of cosets to a coset. – Michael Jun 17 '24 at 09:59
  • Try an order 2 subgroup of the non-abelian subgroup of order 6, and play about with different representative of the same two (distinct) cosets. You can get different results for different representatives of the same two (distinct) cosets. – Michael Jun 17 '24 at 10:04
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    Hi Michael. I think OP's operation is well-defined. The key point is that we fix a choice of representatives before defining $\ast$, and we don't harbour any pretense that $\ast$ does not depend on this choice. – Izaak van Dongen Jun 17 '24 at 10:12
  • That's my point: there is such a well defined binary operation from ordered pairs of representative elements (elements of the parent group set) to the set of cosets. It gives a well defined binary operation from ordered pairs of cosets to the set of cosets, when combined with a procedure choosing representative elements of cosets, if and only if the subgroup is normal. – Michael Jun 17 '24 at 13:00
  • This is incorrect, as shown in Just a User's answer. Take $G=S_3$, $H={e,(12)}$. Pick coset representatives for the three cosets to be $e$, $(123)$, and $(132)$. The representatives form a group, and we can define an operation on the cosets by $(gH)\odot (g'H) = gg'H$, where $g,g'\in {e, (123), (132)}$ (so $gg'$ also lies in ${e,(123),(132)}$. This makes the cosets into a group under this operation, defined on representatives, but $H$ is not normal. – Arturo Magidin Jun 17 '24 at 16:26