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Let $G$ be a group and $H$ a subgroup, now for every $g \in G$ we define $\sigma_{g}:G/H \rightarrow G/H: xH \mapsto gxH$. Note that we know nothing about the subgroup $H$.

My question is whether or not the function $\sigma_{g}$ is even well-defined in all cases. After all if $xH=yH$, why should $\sigma_{g}(xH) = gxH = \sigma_{g}(yH) = gyH$? It seems to me like the only way this works for every $g \in G$, is if $H$ is a normal subgroup. Am I missing something?

Context: I had this question after reading option 2 in https://www.math3ma.com/blog/4-ways-to-show-a-group-is-not-simple.

Kan't
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H. de Gracht
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4 Answers4

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Yes, this operation is well defined, and no, it does not require $H$ to be normal.

Recall that $xH=yH$ if and only if $x^{-1}y\in H$. So you to show it is well defined we need to show that if $xH=yH$, then $gxH = gyH$. And indeed, because $$\begin{align*} gxH = gyH &\iff (gx)^{-1}(gy)\in H\\ &\iff (x^{-1}g^{-1})(gy)\in H\\ &\iff x^{-1}(g^{-1}g)y\in H\\ &\iff x^{-1}y\in H\\ &\iff xH=yH. \end{align*}$$ So the map is not only well-defined, it is also one-to-one. Note that normality is not used in any way.

This shows that left multiplication is a left group action on the set of left cosets of a subgroup $H$. This fact has important consequences: for example, it is used to show that if $[G:H]=n$, then there exists a normal subgroup $N\triangleleft G$ with $N\leq H$ and $[G:N]\leq n!$. This in turn is used to show that if $[G:H]$ is the smallesst prime that divides $|G|$, then $H$ must be normal in $G$.

Arturo Magidin
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It is well defined. Why not just check it by definition? $xH=yH$ means that $y^{-1}x\in H$. It follows that:

$(gy)^{-1}(gx)=y^{-1}g^{-1}gx=y^{-1}x\in H$.

And so $gxH=gyH$, as desired.

Mark
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  • Thanks, I forgot that the inverse of multiplication swaps the elements, I guess that explains why I had such a hard time showing it. – H. de Gracht Oct 30 '23 at 18:24
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Actually there is always a way to get a well-defined operation without using representatives.

Let $σ(g)$ be the function on $G/H$ such that $σ(g)(C) = gC = \{ \ gc : c{∈}C \ \}$ for each $C{∈}G/H$. Then $σ(g)(C) = gxH$ for every $x{∈}G$ such that $xH = C$. Thus $σ : G/H→G/H$ and you get the desired property too.

~ ~ ~

That's it. In case you are wondering why this looks so much simpler, it's because it is really that simple, once you understand cosets as just sets. Arturo's way is just convoluted; don't think of "$xH = yH$" as meaning "$y^{-1}x∈H$"; it literally means $\{ \ xh : h{∈}H \ \} = \{ \ yh : h{∈}H \ \}$ and nothing else.

You will find that a lot of facts become much simpler and clearer when you do things this way, as you will lessen unnecessary inverses. Algebraically, there is also a fundamental advantage this way, as you will be able to see better when inverses are actually necessary. In this case, obviously not.

user21820
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  • If you hide the representative behind the symbol "$C$", I think that the very membership "$C\in G/H$" is questionable, by definition of $G/H$. – Kan't Oct 31 '23 at 07:11
  • @citadel: Your objection does not seem to make sense. I defined σ(g) to be the function with domain G/H such that σ(g)(C) = gC for each C∈G/H. This is 100% well-defined, trivially. The only thing we nee to prove is that the range of σ(g) is within G/H, and that is clear from the next sentence since g(xH) = (gx)H. Please think again.. – user21820 Oct 31 '23 at 07:18
  • Can you write explicitly down $G/H$? – Kan't Oct 31 '23 at 07:25
  • @citadel: G/H here is the set of left-cosets. Their definition is standard and you can find it on wikipedia too: G/H = { gH : g∈G } = { { gh : h∈H } : g∈G }. – user21820 Oct 31 '23 at 07:56
  • I'm quite aware of that, and just because of that definition, I don't see how you can get rid of that "$g$" when dealing with your "$C$". Btw, when you quote another user's contribution, it'd be fair to inform them with the @. – Kan't Oct 31 '23 at 08:00
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    @user21820 I agree with the spirit of your answer, and that this is the way to think about it. Anything else is overcomplicating it. But I would word it a bit differently: the map $2^G \to 2^G: A \mapsto gA$ is clearly a well-defined bijection. It maps $xH \mapsto g(xH) = (gx)H$, so the map $xH \mapsto (gx)H$ is well-defined as a restriction of something well-defined. – J. De Ro Oct 31 '23 at 08:15
  • @citadel: Please specify exactly which line you do not understand. For each g∈G, we define the function σ(g) by specifying its domain to be G/H and that for each C in its domain we have σ(g)(C) = gC. I even clearly stated the standard definition of gC, namely { gc : c∈C }. This is a trivially well-defined definition, so I've absolutely no clue what you mean by "get rid of that g". – user21820 Oct 31 '23 at 08:18
  • @J.DeRo: That's an interesting way to see it (i.e. as a restriction of a function)! I am not sure about whether it's pedagogically better or not, though, because students need to learn to grasp simple definitions first before looking at defining a function via a restriction... XD – user21820 Oct 31 '23 at 08:22
  • @J.DeRo: By the way, we don't need your (bigger) map to be a bijection, for the properties we are looking at here. Naturally, it's useful for later things, but not at the moment. It's part of my remark that inverses are not necessary here. =) – user21820 Oct 31 '23 at 08:26
  • My point is: how can you state "$C\in G/H$" else than assuming $C=xH$ for some $x\in G$? Without that, your map is a legitimate one just from the powerset of $G$ to itself, I think. Let me think more about that, though. – Kan't Oct 31 '23 at 12:34
  • @citadel: I think you are completely misreading. It is a completely standard thing to define a function exactly like I did. For example, "Let f be the function on ℝ such that f(x) = x^2+1 for each x∈ℝ." is a perfectly legitimate definition of a function f! I'm doing exactly the same here, defining a function on G/H instead of ℝ! – user21820 Oct 31 '23 at 13:17
  • This certainly works; though here the "well-definedness" is the issue of verifying that the defined function from subsets of $G$ to subsets of $G$ will actually map subsets-that-are-cosets to subsets-that-are-cosets (that is, whether this is a well-defined function from $G/H$ to $G/H$ depends on the output objects actually being of the desired kind rather than the well-definedness being an issue of verifying that different representations of the input lead to the same output). It might be worth noting that we have traded what "well-defined"ness needs to be checked, even if this is easier. – Arturo Magidin Oct 31 '23 at 19:27
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Just note that by associativity of multiplication in $G$, we also have associativity of complex-multiplication (=multiplication of subsets of arbitrary $G$) and hence $$(gx)H=g(xH)=g(yH)=(gy)H$$

Hagen von Eitzen
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