# The Unapologetic Mathematician

## Generators and Relations

Now it’s time for the reason why free groups are so amazingly useful. Let $X$ be any set, $F(X)$ be the free group on $X$, and $G$ be any other group. Now, every function $f$ from $X$ into $G$ extends to a unique homomorphism $f: F(X)\rightarrow G$. Just write down any word in $F(X)$, send each letter into $G$ like the function tells you, and multiply together the result!

So what does this get us? Well, for one thing every group $G$ is (isomorphic to) a quotient of a free group. If nothing else, consider the free group $F(G)$ on the set of $G$ itself. Then send each element to itself. This extends to a homomorphism $f$ from $F(G)$ to $G$ whose image is clearly all of $G$. Then the First Isomorphism Theorem tells us that $G$ is isomorphic to $F(G)/{\rm Ker}(f)$. That’s pretty inefficient, but it shows that we can write $G$ like that if we want to. How can we do better?

February 24, 2007

## A few more facts about group actions

There’s another thing I should have mentioned before. When a group $G$ acts on a set $S$, there is a bijection between the orbit of a point $x$ and the set of cosets of $G_x$ in $G$. In fact, $gx=hx$ if and only if $h^{-1}gx=x$ if and only if $h^{-1}g$ is in $G_x$ if and only if $gG_x=hG_x$. This is the the bijection we need.

This has a few immediate corollaries. Yesterday, I mentioned the normalizer $N_G(K)$ of a subgroup $K$. When a subgroup $H$ acts on $G$ by conjugation we call the isotropy group of an element $x$ of $G$ the “centralizer” $C_H(x)$ of $x$ in $H$. This gives us the following special cases of the above theorem:

• The number of elements in the conjugacy class of $x$ in $G$ is the number of cosets of $C_G(x)$ in $G$.
• The number of subgroups conjugate to $K$ in $G$ is the number of cosets of $N_G(K)$ in $G$.

In fact, since we’re starting to use this “the number of cosets” phrase a lot it’s time to introduce a bit more notation. When $H$ is a subgroup of a group $G$, the number of cosets of $H$ in $G$ is written $\left[G:H\right]$. Note that this doesn’t have to be a finite number, but when $G$ (and thus $H$) is finite, it is equal to the number of elements in $G$ divided by the number in $H$. Also notice that if $H$ is normal, there are $\left[G:H\right]$ elements in $G/H$.

This is why we could calculate the number of permutations with a given cycle type the way we did: we picked a representative $g$ of the conjugacy class and calculated $\left[S_n:C_{S_n}(g)\right]$.

One last application: We call a group action “free” if every element other than the identity has no fixed points. In this case, $G_x$ is always the trivial group, so the number of points in the orbit of $x$ is $\left[G:G_x\right]$ is the number of elements of $G$. We saw such a free action of Rubik’s Group, which is why every orbit of the group in the set of states of the cube has the same size.

February 24, 2007