Quotienting by generators in free groups

Question

I feel like this is a simple result but have not touched algebra in a while and can't find the right combination of words to search for.

Suppose we have a free group on 2 generators $G = \langle a, b\rangle$ and we quotient by one of the generators, say $a$. The subgroup $\langle a \rangle$ is not normal so the quotient is not a group, but we still have well-defined left and right cosets.

Given a word (say $aba^{-1}b^{-1}$), is there any significance to "erasing" $a^{\pm 1}$, the generator by which we are quotienting? For instance, in the quotient, I would like to say $$aba^{-1}b^{-1} \rightarrow \not a b\not a^{-1} b^{-1} \equiv bb^{-1} \equiv e.$$

Is such an identification legitimate in any sense for free groups? Can anything be said about finite groups?

Answer 1

A (right) transversal $T$ a quotient $G/H$ is a set of (right) coset representatives for $G/H$ such that if $s, t\in T$ then $Hs\neq Ht$ (so $st^{-1}\not\in H$).

The (right) cosets of $F(a, b)/\langle a\rangle$ have as a transversal the set $T$ of freely reduced words which begin with $b$ or $b^{-1}$, $T=\{b^{\epsilon}W\mid \epsilon=\pm1, b^{\epsilon}W \text{ is freely reduced}\}$. This is because if $Hu=Hv$ then $a^iu=a^jv$ for some $i, j\in\mathbb{Z}$.

Therefore, $Haba^{-1}b^{-1}\neq He$. In fact, this is clearly false as $aba^{-1}b^{-1}\not\in\langle a\rangle$! The issue is that in the working from your question, when you are cancelling the $a$-terms you are implicitly assuming that $H$ is normal, which it isn't. What you are really doing is: $$ \begin{align*} Haba^{-1}b^{-1}&=Hba^{-1}b^{-1}\\ &=HbHb^{-1}ba^{-1}b^{-1}&\text{by normality. Ooops!}\\ &=HbHa^{-1}b^{-1}\\ &=HbHb^{-1}\\ &=H&\text{again by normality} \end{align*} $$ As you can hopefully see, you are assuming normality of $H$ in a fundamental way.

So, what happens if we assume that $H$ is normal? Let us call this new, normal subgroup $N$, to distinguish between $H=\langle a\rangle$ (the smallest subgroup containing $a$) and $N=\langle\langle a\rangle\rangle$ (the smallest normal subgroup containing $a$, called the normal closure of $a$). In this setting, we can erase everything as you want to do, and the notation for this is presentations. We write $\langle a, b; a\rangle$ for $F(a, b)/\langle\langle a\rangle\rangle$. As every group is the quotient of some free group, every group has a presentation (although not necessarily finite). The classic reference for presentations is the book Combinatorial group theory by Magnus, Karrass and Solitar. Anything I say now would be said better and in more detail in their book. So I will stop typing now.

Answer 2

When you present a group $G = \langle S \mid R \rangle$ with generators $S$ and relations $R$, you are considering a quotient of the free group $F = F_S$ on the generators $S$ by the normal subgroup $N = N_R$ on the relations $R$: $$ G = F/N. $$

In your example of "canceling" a generator, you're looking a the image of the free group under a homomorphism sending $a \mapsto 1$. (Such a homomorphism necessarily exists as your group is free.) Notice that the homomorphism automatically sends, not only $a \mapsto 1$, but $bab^{-1} \mapsto b1b^{-1} = bb^{-1} = 1$, as well. In fact, it wasn't necessary that $g=b$, the element of the group that you conjugated by, was a generator: for any $g \in G$, $gag^{-1} \mapsto 1$.

This shows that when you want to "cancel" a generator, you're actually quotienting by a normal subgroup (the kernel of the homomorphism). Of course you don't have to cancel only a bare generator (what a waste); you can cancel a word in the generators, and that's what relations are!