The normal subgroup generated by $a^2, b^2$.

Question

This is Exercise 2.1.4(c) 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 $F$:

(c) $a^2, b^2$.

My Attempt:

We have that $F/N\cong \langle a', b'\mid a'^2, b'^2\rangle$ by the mapping $aN\mapsto a', bN\mapsto b'$. I don't know the order of $F/N$. I'm guessing it's infinite.

The elements of $N$ are of the form $$a^{2n_1}b^{2n_2}a^{2n_3}b^{2n_4}\dots a^{2n_{m-1}}b^{2n_m}$$ for $n_i\in\Bbb Z, m\in\Bbb N\cup\{0\}$.

Please help :)

Answer 1

Generally, an element of $F$ can be described as a product of integer powers of $a$ and $b$, but in the quotient any integer power can be simplified to an exponent of either $0$ or $1$, so the resulting elements all look like alternating products of $a$s and $b$s. Indeed, $F/N$ is generated by $a$ and $ab$, evidently $a$ has order $2$ and $ab$ has infinite order. Notice $a(ab)a^{-1}=ba=(ab)^{-1}$ so in fact we have that $F/N$ is an internal semidirect product $\langle ab\rangle\rtimes\langle a\rangle\cong\mathbb{Z}\rtimes\mathbb{Z}_2$. (I am abusing notation in referring to elements of $F/N$ by their representatives in $F$.)

Answer 2

Just some things regarding what you wrote (at first before editing). Hope they help to get a better grasp. Firstly, your description of $F/N$ as the presentation $\langle a',b'|a^2,b^2\rangle$ is not correct; you don't need the accents $'$, just write $\langle a,b|a^2,b^2\rangle$ (the relations in a presentation must be among generators, so taking $a',b'$ to be generators and $a^2,b^2$ to be the relations forces the question: What are the $a$ and $b$?) Also the element you wrote $a'^{2n_1}\cdots$ in the quotient is the identity element, (as anon pointed) since $a^2=b^2=1$ in the quotient.

Now to your question, what you have is the presentation $$\langle a,b|a^2,b^2\rangle:=\frac{F(a,b)}{N(a^2,b^2)}$$ where $F(a,b)$ is the free group on two generators and $N(a^2,b^2)$ is the normal subgroup generated by $a^2,b^2$. So you have a group generated by two elements, each of order $2$. Someone may identify this group with the free product of $\Bbb{Z}_2*\Bbb{Z}_2$ (but you don't really need this).