Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?


share|cite|improve this question
up vote 7 down vote accepted

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":

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!

share|cite|improve this answer

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.

share|cite|improve this answer
I just wonder whether it can be expressed in some way i don't know since it is infinite. – Yuan Oct 18 '10 at 3:22
@0592: But what does "express" mean? We can "describe" every element of the infinite group $F$ easily enough (reduced group words in $a$ and $b$). Is that a way of "expressing $F$"? We can simply say $N$ consists of products of conjugates (by elements of $F$) of the generators and their inverses; is that a way of "expressing $N$"? – Arturo Magidin Oct 18 '10 at 3:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.