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

ab=bc=ca

ac=cd=da

ad=db=ba

bd=dc=cb

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

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1  
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
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1 Answer

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);
true

> AbelianInvariants(z);
[ 0, 2 ]
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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
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