Re: G' finite ==> G/Z is finite?
mareg_at_mimosa.csv.warwick.ac.uk
Date: 02/18/05
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Date: 18 Feb 2005 12:46:57 -0500
In article <cv4o0d$40v$1@dizzy.math.ohio-state.edu>,
mareg@mimosa.csv.warwick.ac.uk () writes:
>
>In article <cv3j34$2lo$1@dizzy.math.ohio-state.edu>,
> eclark@math.usf.edu writes:
>>
>>It is a theorem of I. Schur that if G is a group,
>>Z its center, and G' its commutator subgroup, then
>>
>>(*) G/Z is finite implies that G' is finite.
>>
>>Is the converse of (*) true?
>>
>>In a talk here today Peter Hilton showed that the
>>converse is true if G is finitely generated.
>>What about the general case?
>
>No it is not true in general. An infinite extraspecial p-group is a
>counterexample.
>
>Let p be prime, and let G be generated by x_i, y_i (i > 0) and z,
>subject to the relations:
>x_i^p = y_i^p = z^p = 1, [x_i,y_i] = z, [z,x_i] = [z,y_i] = 1 for all i.
>
>Then Z(G) = G' = <z> is finite of order p, but G/G' is infinite.
Sorry, I missed out some of the relations:
[x_i,x_j] = 1 for all i,j and [x_i,y_j] = 1 for all i not equal to j.
This is the same example as Caranti posted.
Derek Holt.
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