# The Unapologetic Mathematician

## Coset Representations

Next up is a family of interesting representations that are also applicable to any group $G$. The main ingredient is a subgroup $H$ — a subset of the elements of $G$ so that the inverse of any element in $H$ is also in $H$, and the product of any two elements of $H$ is also in $H$.

Our next step is to use $H$ to break $G$ up into cosets. We consider $g_1$ and $g_2$ to be equivalent if $g_2^{-1}g_1\in H$. It’s easy to check that this is actually and equivalence relation (reflexive, symmetric, and transitive), and so it breaks $G$ up into equivalence classes. We write the coset of all $g'\in G$ that are equivalent to $g$ as $gH$, and we write the collection of all cosets of $H$ as $G/H$.

We should note that we don’t need to worry about $H$ being a normal subgroup of $G$, since we only care about the set of cosets. We aren’t trying to make this set into a group — the quotient group — here.

Okay, now multiplication on the left by $G$ shuffles around the cosets. That is, we have a group action of $G$ on the quotient set $G/H$, and this gives us a permutation representation of $G$!

Let’s work out an example to see this a bit more explicitly. For our group, take the symmetric group $S_3$, and for our subgroup let $H=\{e,(2\,3)\}$. Indeed, $H$ is closed under both inversion and multiplication. And we can break $G$ up into cosets:

\displaystyle\begin{aligned}\{e,(1\,2),(1\,3),(2\,3),(1\,2\,3),(1\,3\,2)\}&=\{e,(2\,3)\}\uplus\{(1\,2),(1\,2\,3)\}\uplus\{(1\,3),(1\,3\,2)\}\\&=H\uplus(1\,2)H\uplus(1\,3)H\end{aligned}

where we have picked a “transversal” — one representative of each coset so that we can write them down more simply. It doesn’t matter whether we write $(1\,2)H$ or $(1\,2\,3)H$, since both are really the same set. Now we can write down the multiplication table for the group action. It takes $g_1\in G$ and $g_2H\in G/H$, and tells us which coset $g_1g_2$ falls in:

$\displaystyle\begin{tabular}{c|rrr}&H&(1\,2)H&(1\,3)H\\\hline e&H&(1\,2)H&(1\,3)H\\(1\,2)&(1\,2)H&H&(1\,3)H\\(1\,3)&(1\,3)H&(1\,2)H&H\\(2\,3)&H&(1\,3)H&(1\,2)H\\(1\,2\,3)&(1\,2)H&(1\,3)H&H\\(1\,3\,2)&(1\,3)H&H&(1\,2)H\end{tabular}$

This is our group action. Since there are three elements in the set $G/H$, the permutation representation we get will be three-dimensional. We can write down all the matrices just by working them out from this multiplication table:

\displaystyle\begin{aligned}\rho_H(e)&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\rho_H((1\,2))&=\begin{pmatrix}{0}&1&0\\1&0&0\\{0}&0&1\end{pmatrix}\\\rho_H((1\,3))&=\begin{pmatrix}{0}&0&1\\{0}&1&0\\1&0&0\end{pmatrix}\\\rho_H((2\,3))&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\\\rho_H((1\,2\,3))&=\begin{pmatrix}{0}&0&1\\1&0&0\\{0}&1&0\end{pmatrix}\\\rho_H((1\,3\,2))&=\begin{pmatrix}{0}&1&0\\{0}&0&1\\1&0&0\end{pmatrix}\\\end{aligned}

It turns out that these matrices are the same as we saw when writing down the defining representation of $S_3$. There’s a reason for this, which we will examine later.

As special cases, if $H=\{e\}$, then there is one coset for each element of $G$, and the coset representation is the same as the left regular representation. At the other extreme, if $H=G$, then there is only one coset and we get the trivial representation.

September 20, 2010