# Re: Group Theory

mareg@xxxxxxxxxxxxxxxxxxxxxxxx wrote:
In article <1160046334.164177.156020@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Tonico" <Tonicopm@xxxxxxxxx> writes:

mareg@xxxxxxxxxxxxxxxxxxxxxxxx wrote:
In article <19790289.1159493051233.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Cezanne123 <Cezanne123@xxxxxxx> 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.
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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.
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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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