Re: Finite Group

From: Zbigniew Fiedorowicz (fiedorow_at_hotmail.com)
Date: 03/18/05


Date: Fri, 18 Mar 2005 07:40:38 -0500

William Elliot wrote:
> Let G be a group for which every proper subgroup is finite.
> Is G finite?
>

Let p be a prime. Consider the multiplicative group of p-th
power roots of unity.

> One observes all elements of G have finite order.
> Pick g /= e and let H = a maximal subgroup not containing g.
> H is finite. <g,H> the group generated by g and G, equals G.
>
No.

> When G is Abelian, then G = <g,H> = { ng + h | n in Z, h in H }
> Now as order of g is finite as also H, <g,H> is finite.
> In fact |G| = |<g,H>| = o(g) * |H|. So far, so good.
>
> What however, if G isn't Abelian?



Relevant Pages

  • Re: Finite Group
    ... On Fri, 18 Mar 2005, Zbigniew Fiedorowicz wrote: ... >> Let G be a group for which every proper subgroup is finite. ... > power roots of unity. ... >> What however, if G isn't Abelian? ...
    (sci.math)
  • Re: abelian..
    ... >If every proper subgroup of group G is cyclic group ... >but S_3 is not abelian. ... Derek Holt. ...
    (sci.math)
  • Re: abelian..
    ... >If every proper subgroup of group G is cyclic group ... >but S_3 is not abelian. ... >thank you for your advice. ...
    (sci.math)
  • Re: Finite Group
    ... >William Elliot wrote: ... >> Let G be a group for which every proper subgroup is finite. ... >> One observes all elements of G have finite order. ... infinite nonabelian groups in which all proper subgroups have order ...
    (sci.math)
  • abelian..
    ... If every proper subgroup of group G is cyclic group ... but S_3 is not abelian. ... thank you for your advice. ...
    (sci.math)