presentation of quaternion group of order 8

Question

Let G be the group defined by generators a,b and relations $a^4=e$, $a^2b^{-2}=e$, $abab^{-1}=e$. Since the quaternion group of order 8 is generated by elements $a,b$ satisfying the previous relations, there is an epimorphism from $G$ onto $Q_8$. Let $F$ be the free group on $\{a,b\}$ and $N$ the normal subgroup generated by $\{ a^4, a^2b^{-2}, abab^{-1} \}$. How to write down the normal subgroup N and express the group F/N ?

Thanks

Answer 1

Two points about this. As a subgroup of a free group, N is itself a free group. There is an algorithmic procedure for finding a free set of generators of N, known as the Schreier generators. An example of this was presented in detail in a recent item "Normal Closure in groups II":

http://math.stackexchange.com/questions/6707/normal-closure-in-groups-ii

Using a computer program (MAGMA), I got the following set of 9 generators of N:

N.1 = b * a * b * a^-1
N.2 = b^2 * a^-2
N.3 = a^-1 * b^2 * a^-1
N.4 = b^-1 * a * b^-1 * a^-1
N.5 = a^4
N.6 = a^2 * b^2
N.7 = a * b * a * b^-1
N.8 = a * b^2 * a
N.9 = a * b^-1 * a * b

The second point is that the first relation $a^4=e$ in your presentation is actually redundant. The quaternion group is defined by the presentation with two generators $a,b$ and just the two relations $a^2=b^2$ and $abab^{-1} = e$. It is a nice exercise to prove that!

Answer 2

To write down the group $F/N$, the best way in this case is to find a normal form. Note that the last relation $abab^{-1}=e$ means that $ba=a^{-1}b = a^3b$ (that is, the element $b^{-1}a^{-3}ba$ is in $N$); so any coset representative that contains the string $ba$ can be replaced with a coset representative in which the string $ba$ is replaced with the string $a^3b$. The fact that $a^2b^{-2}\in N$ means that any coset representative that in which you have a $b^2$ can be replaced with one that has an $a^2$ instead. Proceeding in this way, you can conclude that every coset can be represented by an element of the form $a^i b^j$ with $0\leq i\leq 3$ and $0\leq j\leq 1$ (you can erase any $a^4$ since $a^4\in N$). Thus, the group $F/N$ has at most $8$ elements. You already know it has at least $8$ elements, so that means that each of these elements represent distinct cosets of $N$, so that gives you a way to express $F/N$ using the representatives.

"Writing down" the group $N$ is a bit more difficult, since it is an infinite subgroup in $F$. What exactly do you mean by "writing down $N$"? A way to determine if a given element of $F$ lies in $N$? This can be achieved by finding its "coset representative" from among the special set identified above. If you get $a^0b^0$, then the element was in $N$; if not, then it was not.