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I was watching the following online lecture on normal subgroups, and came across the following proof: Normal and quotient groups

If we have some group, G, and some subgroup of G, let's say N, we can construct left cosets like so:

Each left coset: $gN$ for some $g \in G$.

Now we pick two cosets: $xN, yN$. Now since $e \in N$,

$x.e = x \in xN$ and $y.e = y \in yN$.

Now here is the part that I don't understand:

For cosets to act like a group:

$x.y \in (xN)(yN)$

Why must this be true? And what is $(xN)(yN)$.Is it the element-wise group operation of each element in $xN$ and $yN$?

Then the proof said: $i.e. (xN)(yN) = xyN$.

This does not make sense to me, what is so obvious about this statement?

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    If you want to be able to multiply cosets as if they were a group, you need (i) the element-wise product of two cosets to be a coset; and (ii) since $xy$ lies in the element-wise product of $xN$ and $yN$, you need the coset that results from doing $(xN)(yN)$ to be the coset $xyN$. This holds for normal subgroups and only for normal subgroups. By the way: it is impossible to help you understand their justificaiton unless we know what lecture you are talking about. There are multiple ways to establish these things. By not saying, you prevent help. – Arturo Magidin Oct 10 '21 at 02:38
  • @ArturoMagidin I added the lecture, sorry for the confusion. –  Oct 10 '21 at 02:51
  • @ArturoMagidin Could you also explain why $(xN)(yN) \in xyN$? –  Oct 10 '21 at 03:18
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    It is not true that $(xN)(yN)\in xyN$; the left side is a set, not an element. For normal subgroups, $gN=Ng$ for all $g\in G$. So $(xN)(yN) = x(Ny)N = x(yN)N = xyNN = xyN$, because for any subgroup $H$, $HH=H$. This is all done in the post I linked to. – Arturo Magidin Oct 10 '21 at 03:37

1 Answers1

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The condition $$xy\in(xN)(yN)\tag{1}$$ is not necessary for the set of cosets to be a group. Instead, (1) is the condition for multiplication on cosets to be deducible from multiplication in your original group.

(In fact, if there are $n$ cosets, then we can always make them a group by choosing an arbitrary bijection to the cyclic group on $n$ elements. But this usually isn't a very interesting group structure, because it is entirely unrelated to the original $G$ we started out with.)

Recall that each coset is the equivalence class of elements of $g$ that differ by an element of $N$; that is, $$xN=\{nx:n\in N\}$$ This set has a representation as $xN$, but this representation is not unique: if $n\in N$, then $(xn)N=xN$. (For example, consider $D_3/C_2$. This means: take the group of symmetries of a triangle; any reflection $r$ generates a (non-normal) subgroup. That subgroup has three cosets, which I will call $C_2$, $\rho C_2$, and $\rho^2C_2$. You can verify that $rC_2=C_2$, $\rho rC_2=\rho C_2$, etc.)

We want a multiplication on cosets that arises from multiplication in $G$, so that we can deduce $(xN)(yN)$ just by computing $(xy)N$. But we have no guarantee that the coset $(xy)N$ doesn't depend on which $x$ and $y$ we choose to represent $xN$ and $yN$.

One way to fix this is to say: given two cosets $K_1$ and $K_2$, the natural multiplication is $$K_1K_2=\{k_1k_2:k_1\in K_1, k_2\in K_2\}$$ This allows us to define $(xN)(yN)$, and one can check that the result is a coset precisely when $N$ is normal. We want this multiplication to line up with multiplication in $G$: $$(xN)(yN)=xyN$$ To ensure this, we follow Arturo Magidin's comment: if $xy\in(xN)(yN)$, then $$xyN\subseteq(xN)(yN)N=(xN)(yN)$$ Since both sides are cosets, they cannot contain each other strictly. Thus $xyN=(xN)(yN)$, as desired.

Jacob Manaker
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