# The Unapologetic Mathematician

## Cosets and quotients

I know I’ve been doing a lot of theory here for a while, but just hold on a little longer, it’s about to pay off.

Now we know what a group is and what a subgroup is. Today I want to talk about the cosets of a subgroup $H$ in a group $G$.

A subgroup is sort of like a slice through a group. The cosets of a subgroup are like slices “parallel” to the group. We know that $H$ passes through the identity of $H$ already, so we want to find a “parallel copy” through any other element we choose.

Now, cosets come in two flavors: left and right. We get a left coset $gH$ by composing every element of $H$ on the left with $g$. On the other hand, we get a right coset $Hg$ by composing every element on the right.

Let’s let $G$ be the group $S_3$, and $H$ be the subgroup $\{e, (1 2)\}$, where $e$ is the identity element. For each element of $G$, let’s consider its left and right cosets.

 $g$ $gH$ $Hg$ $e$ $\{e, (1 2)\}$ $\{e, (1 2)\}$ $(2 3)$ $\{(2 3), (1 3 2)\}$ $\{(2 3), (1 2 3)\}$ $(1 2)$ $\{(1 2), e\}$ $\{(1 2), e\}$ $(1 2 3)$ $\{(1 2 3), (1 3)\}$ $\{(1 2 3), (2 3)\}$ $(1 3)$ $\{(1 3), (1 2 3)\}$ $\{(1 3), (1 3 2)\}$ $(1 3 2)$ $\{(1 3 2), (2 3)\}$ $\{(1 3 2), (1 3)\}$

Notice that the element $g$ is in both $gH$ and $Hg$, since the identity $e$ is in $H$. Also notice that if $g'$ is in $gH$ then $gH$ is the same as $g'H$, and similarly for right cosets. In fact, you should be able to verify that

• $g$ is in $gH$
• If $g$ is in $g'H$, then $g'$ is in $gH$
• If $g$ is in $g'H$ and $g'$ is in $g''H$, then $g$ is in $g''H$

and three other similar statements for right cosets. Together, these say that we can separate the elements of $G$ into “equivalence classes”. If $g$ and $g'$ are not in the same class then $gH$ and $g'H$ share no elements at all. But if $g$ and $g'$ are in the same class, $gH$ and $g'H$ are the same set — the equivalence class itself.

Okay, so we’ve sliced up the group $G$ into (left) cosets of the subgroup $H$. If we consider two of them, $gH$ and $g'H$, we can multiply everything in the first by $g'g^{-1}$ to get things in the second, and everything in the second by $gg'^{-1}$ to get things in the first. These two transformations undo each other, so each coset of $H$ is “the same size”. If $G$ is a finite group this means that each coset has the same number of elements — the number of elements in $H$. We’ve sliced $G$ into a bunch of pieces that never overlap, and all having the same number of elements. This merits some emphasis.

If $G$ is a finite group and $H$ is a subgroup, then the number of elements in $H$ must divide the number of elements of $G$!

Now we’ve got a set of cosets of $H$, which we write $G/H$. What’s really nice is that sometimes this set is a group too! The natural idea to multiply two cosets is to take an element of each and multiply them and see what its coset is. Unfortunately this doesn’t always work. The answer might depend on which elements we choose

To see what goes wrong, let’s pick two elements in the first coset. We know that $g$ is in $gH$, and we can see that so is $gh$ if $h$ is an element of $H$. We just pick $g'$ again from $g'H$. Now we multiply to get $gg'$ for one choice and $ghg'$ for the other. These are only in the same coset if there is some $h'$ in $H$ so that $gg'h'=ghg'$. That is, so that $h'=g'^{-1}hg'$. The requirement is this: for every $g'$ in $G$ and for every $h$ in $H$ the composition $g'^{-1}hg'$ must land back in $H$. We call a subgroup with this property “normal”.

So if $H$ is normal, then our naïve idea for how to multiply two cosets does work right, and doesn’t depend on how we choose the element of each coset to multiply: $gH\circ g'H=(gg')H$. You should verify that this composition on the set $G/H$ of cosets of $H$ in $G$ actually satisfies the group axioms. We call this the “quotient group” of $G$ by $H$, or “$G$ modulo $H$“.

Other exercises:

• Check to see that in the example given above (where $G$ is $S_3$) that the subgroup isn’t normal. Find one that is, and see what the quotient is.
• Show that any subgroup of an abelian group is normal.
• Consider the subgroup of $\mathbb{Z}$ (addition as the composition) consisting of all multiples of 12. Call it $12\mathbb{Z}$. What is $\mathbb{Z}/12\mathbb{Z}$? What if we change 12 to any other number $n$?

February 15, 2007

## Subgroups

A subgroup is pretty straightforward. It’s a little group living inside a bigger group. If you’ve got a group $G$ and some collection $H$ of elements of $G$ so that $H$ is a group using the same composition as $G$, then $H$ is a subgroup. To be more explicit, you need that

• If $x$ and $y$ are in $H$ then $xy$ is in $H$.
• If $x$ is in $H$ then $x^{-1}$ is in $H$.
• The identity $e$ of $G$ is in $H$. [added at the suggestion of Toby Bartels]

We say that a subgroup is “closed” under composition and inverse, meaning that if we start with elements of $H$ and take compositions and inverses we never leave the subgroup.

The collection of all even integers is a subgroup of the group ${\mathbb Z}$ of all integers (with addition as the operation).
The subset $\{e,(1\,2\,3),(1\,3\,2)\}$ is a subgroup of the group $S_3$.
Every group has two “trivial” subgroups: the whole group itself, and the subgroup consisting of just the identity element.
There are two ways of getting subgroups that I want to spend a bit more time on: “images” and “kernels”.