# The index of the normal subgroup generated by $a^3, b^2, aba^{-1}b^{-1}$ in $\langle a, b\rangle$.

This is Exercise 2.1.4(d) of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by W. Magnus et al.

## The Question:

Let $F=\langle a, b\rangle$. If $N$ is the normal subgroup of $F$ generated by each of the following sets of words, find the index of $N$ in $F$:

(d) $a^3, b^2, aba^{-1}b^{-1}$.

## My Attempt:

The quotient $F/N$ is isomorphic to $\langle a, b\mid a^3, b^2, ab=ba\rangle\cong C_3\times C_2$, so the index of $N$ in $F$ is $6$.

Is this correct?

(NB: I've asked a lot of similar questions lately. I'm due to start my PhD (which is fully funded by a scholarship!) this October and I'll be working closely with group presentations, so I'm brushing up on them as much as I can.)

• It is correct indeed. – user128787 Aug 10 '17 at 10:43
• @user128787 Would you elaborate on that, please, with some feedback in an answer, so I can accept it? :) – Shaun Aug 10 '17 at 10:48

Since the answer is completely yours, I can write one of my favorite theorems from group theory, which essentially proves why $\langle a,b|a^3,b^2,ab=ba\rangle$ and $C_3\times C_2$ are isomorphic. That would be von Dyck's theorem.

So let $G$ be a group with presentation $\langle a_1,\dots, a_n|r_1,\dots, r_m\rangle$ and $H$ some other group. The $r_i$'s are elements of $G$ so they can be written in terms of the generators, i.e. $r_i=a_{i_1}^{\epsilon_1}\cdots a_{i_k}^{\epsilon_k}$ where $\epsilon_j=\pm 1$ for every $i$.

Suppose that there is map $f:\{a_1,\dots,a_n\}\to H$ ($f$ is just a map defined on the generators, not a homomorphism) with the property $$f(a_{i_1})^{\epsilon_1}\cdots f(a_{i_k})^{\epsilon_k}=1_H$$ Note: If $f$ was a homomorphism that would be just $f(r_i)$, so, what we are assuming is just that $f$ vanishes on each relation of $G$.

Then, $f$ can be extended to a homomorphism $\bar{f}:G\to H$ such that $\bar{f}(a_i)=f(a_i)$ for each $i$.

The proof of that is quite obvious, set $\bar{f}(a_i)=f(a_i)\,\forall i$, extend $\bar{f}$ holomorphicaly and voilà.

Now that is important for your answer since it proves why $G:=\langle a,b|a^3,b^2,ab=ba\rangle$ and $C_3\times C_2$ are isomorphic. Write $C_3\times C_2$ as $C_6$ and define $f(a)=2$, $f(b)=3$. Then $f$ vanishes on each relation and by von Dyck's theorem you have a homomorphism $\bar{f}:G\to C_6$ which is surjective, since $2$ and $3$ generate $C_6$. Now define $g:C_6\to G$ by setting $g(1)=b\cdots a^{-1}$. If I am not mistaken, $g\circ f=id_G$ and $f\circ g=id_{C_6}$, so they are indeed isomorphic.

The rest of the proof is exactly as in your comment.