Is left multiplication well defined on a quotient set of a group?

Question

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.

Answer 1

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 $G$.

Answer 2

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.

Answer 3

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.

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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.

Answer 4

Just note that by associativity of multiplication in $G$, we also have associativity of complex-multiplication (=multiplication of subsets of arbitrary $G$) and hence $$(gx)H=g(xH)=g(yH)=(gy)H$$