Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The group is generated by a,b,c,d with relations





I checked using GAP that it is $A_4$ mod its center. But what is the center? Torsion free of rank=?

share|improve this question
The center has torsion. In fact it is an abelian group, isomorphic to $\mathbb{Z}\times\mathbb{Z}_2$. –  Steve D Aug 21 '12 at 0:34

1 Answer 1

up vote 4 down vote accepted

Try the following GAP code:

> g:=FreeGroup("a","b","c","d");
<free group on the generators [ a, b, c, d ]>

> h:=g/ParseRelators(GeneratorsOfGroup(g),"ab=bc=ca,ac=cd=da,ad=db=ba,bd=dc=cb");
<fp group on the generators [ a, b, c, d ]>   

> z:=Centre(h);
Group(<37 generators>)

> IsAbelian(z);

> AbelianInvariants(z);
[ 0, 2 ]
share|improve this answer
Thanks. What are generators for the cyclic summands of the center? There must be a lot of collapsing of elements that is not apparent to me. –  Menton Aug 22 '12 at 15:15
Yes, $a$ and $d$ are superfluous generators. The infinite part of $Z(G)$ is generated by $b^3$ (which is equal to $c^3$), and the torsion part is generated by $(b^{-1}c)^2$. –  Steve D Aug 22 '12 at 19:01
The group of this question is just a specific example of the groups in your last question, which I analyzed as well. The number of generators here is even, so things are slightly different. The group in this question is the semidirect product $Q_8\rtimes \mathbb{Z}$, where $\mathbb{Z}$ is generated by $b$, and the usual generators of $Q_8$ are $i=cb^{-1}$ and $j=b^{-1}c$. $b$ acts as the permutation $(i,j,k)$; that is why $b^3$ is central. And of course $i^2=j^2=(b^{-1}c)^2$ is central. –  Steve D Aug 23 '12 at 6:04
But how do you see that $b^3=c^3$ and $(b^{-1}c)^4=1$? –  Menton Aug 23 '12 at 22:05
@Menton: Simply manipulate the relations you've given! You can work out that $a=bcb^{-1}$ and $d=cbc^{-1}$. The first relation then gives $bc=cbcb^{-1}$, or $bcb=cbc$. Continue to work from there. –  Steve D Aug 23 '12 at 23:51

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.