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I first saw this on the Missouri State problem page and it has never been solved there.

Consider the group generated by a,b,c, and d subject to the relations

ab = c, bc = d, cd = a, and da = b

Using the first relation, the second relation becomes bab = d. Using this expression and the first relation, we obtain

abbab = a and baba = b

Taking the second relation above and multiplying both sides on the left by a^-1b^-1 and on the right by a^-1, we have b = a^-2. Now first relation above becomes aa^-2a^-2aa^-2 = a or a^-4 = a, hence i = a^5. Therefore our group consists of the five elements i, a, a^2, a^3, a^4. The other elements can be expressed in terms of a as follows: b = a^3, c = a^4, and d = a^2.

Finally, we get to his month's problems. How many elements are there in the groups given by the following generators and relations?

* Generators:a,b,c
  Relations: ab = c, bc = a, ca = b

* Generators:a,b,c,d,e
  Relations: ab = c, bc = d, cd = e, de = a, ea = b

* Generators:a,b,c,d,e,f
  Relations: ab = c, bc = d, cd = e, de = f, ef = a, fa = b 

Source: John H. Conway

I recognize the first one as the quaternions but have made no progress on the other two.

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Have you tried en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm ? –  Qiaochu Yuan Feb 18 '11 at 0:37
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2 Answers 2

up vote 8 down vote accepted

These are the so called "Fibonacci groups" F(2,n), for varying n.

In D.L. Johnsons's "Presentations of Groups", Johnson gives as exercise that

  • F(2,5) is finite and cyclic
  • F(2,6) is infinite, because there exists a homomorphism $\chi \colon F(2,6)\to \mathrm{Sym}{\mathbb Z}$ such that $(\chi(a))(n) = n+1$, $(\chi(b))(n)=-n$ for $n\in \mathbb Z$.

In a preceeding example, Johnson shows that F(2,4) is cyclic by making use of van Kampen diagrams, maybe that helps for F(2,5).

I haven't tried it myself, but they seem quite tedious exercises anyway.

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If you are familiar with the Todd-Coxeter coset enumeration algorithm (which is also covered in Johnson's book), it is not hard to show (by hand even) that the subgroup $\langle a \rangle$ of F(2,5) has index 1 in F(2,5), so F(2,5) is cyclic. The fact that it is abelian then makes it relatively easy to show from the relations that it has order 11. Incedentally F(2,7) is also finite - it is cyclic of order 29 - but that is much harder - although it's easy for a computer of course. –  Derek Holt Feb 18 '11 at 10:04
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For your second group, we can write it as: $$ \langle a,b\ |\ babab^2ab=a, ab^2aba=b\rangle. $$

The first relation gives $baba=ab^{-1}a^{-1}b^{-2}$, and plugging that into the second relation gives $ab(ab^{-1}a^{-1}b^{-2})=b$, or $aba=b^3ab$. Plugging that again into the second relation gives $ab^2(b^3ab)=b$, or $ab^5a=1$, so $b^5=a^{-2}$. Thus $a^2$ is central; combining the first and second relation gives $bab^2=a^2$, and plugging that into the first relation gives $ba^4b=a$. Since $a^2$ is central, this is the same as saying $b^2=a^{-3}$. This is enough to abelianize your group, so it was abelian to begin with; a simple check then shows it is $C_{11}$.

It is easy to see the last group is infinite: the presentation can be given as $$ \langle a,b\ |\ b^2ab^2abab^2ab, abab^2aba\rangle;$$

Quotienting out by the normal subgroup $\langle a^2,b^2\rangle$ (you can check it is normal), gives the group $C_2\ast C_2$, the infinite dihedral group.

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