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From: Tonico on 5 Oct 2006 15:36

mareg(a)mimosa.csv.warwick.ac.uk wrote:
> In article <1160046334.164177.156020(a)m73g2000cwd.googlegroups.com>,
> "Tonico" <Tonicopm(a)yahoo.com> writes:
> >
> >mareg(a)mimosa.csv.warwick.ac.uk wrote:
> >> In article <19790289.1159493051233.JavaMail.jakarta(a)nitrogen.mathforum.org>,
> >> Cezanne123 <Cezanne123(a)aol.com> writes:
> >> >Let G be a group such that the intersection of all its subgroups which are different from (e) is a subgroup different from (e). Prove that every element in G has finite order.
> >> >
> >> >How do you show this is true?
> >>
> >> Well the property that you mention is false in the infinite cyclic group, so
> >> it is also false in any group that contains an element of infinite order.
> >>
> >> Derek Holt.
> >*********************************************************************************
> >Perhaps I missed something here: the intersection of all the
> >non-trivial sbgps of the infinite cyclic group (Z ) is zero, otherwise
> >there'd be an integer divisible by
> >ANY prime...
> >So the OP's condition on the group isn't applicable to Z.
>
> That's exactly what I said above! I was trying to give hint rather than a
> detailed proof.
>
> Derek Holt.
*********************************************
Damn, so you did! Sorry, I completely misunderstood your post, Derek.
After you wrote the explanation I thought otherwise.
Indeed it was pretty elementary.
Regards
Tonio

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