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From: Cezanne123 on 28 Sep 2006 17:23 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?
From: on 5 Oct 2006 05:52 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.
From: Tonico on 5 Oct 2006 07:05 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. Now let G be a group as the OP proposes, and supose x is an element of infinite order in G ==> <x> is isomorphic to Z and for every n, <x^n> is a sbgp of <x>, and thus of G. But, again, the intersection of these not-trivial sbgps is trivial, so the condition isn't fulfilled by G ==> contradiction ==> there can't be an element of infinite order in such a group G. Regards Tonio
From: on 5 Oct 2006 11:37 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. >Now let G be a group as the OP proposes, and supose x is an element of >infinite order in G ==> <x> is isomorphic to Z and for every n, <x^n> >is a sbgp of ><x>, and thus of G. But, again, the intersection of these not-trivial >sbgps is trivial, so the condition isn't fulfilled by G ==> >contradiction ==> there can't be an element of infinite order in such a >group G. >Regards >Tonio >
From: Arturo Magidin on 5 Oct 2006 13:11
In article <2795489.1159493241696.JavaMail.jakarta(a)nitrogen.mathforum.org>, Cezanne123 <Cezanne123(a)aol.com> wrote: >If G has no nontrivial subgroups, show that G must be finite of prime order. > >How do you show this is true? I did it by paying attention in class and doing my own assignments. By the way: your proposition is false. -- ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org |