Re: Algebra

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 11/02/04 

Date: Tue, 2 Nov 2004 17:14:12 +0000 (UTC)

In article <c7c988e3.0411020425.bfdbc05@posting.google.com>,
TT <sweet_sorrow30@yahoo.com> wrote:
>Let G be a group of order m*(p^k), with p prime and (p,m)=1.
>Let H be a group of order p^k and K a subgroup of order p^d, with
>0<d<=k and K is not contained in H. Show that HK is not a subgroup of
>G.

What is the number of elements in HK? |H|*|K|/|H intersect K|.

Since K is not contained in H, then |H intersect K| is a proper
subgroup of K.

And, from Lagrange's Theorem, what do you know about orders of subgroups? 

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu

Relevant Pages

  • Re: Normal subgroups
    ... Stefan Fredriksson wrote: ... We do that because of the definition of normal subgroup. ... --- Calvin ... Arturo Magidin ...
    (sci.math)
  • Re: Is my algebra proof flawed?
    ... Then G = is the trivial group under composition: ... But G is NOT a subgroup of SYM. ... identity of the group will be equal to the identity function. ... --- Calvin ...
    (sci.math)
  • Re: Normal subgroups
    ... >excercise. ... I could really use a push forward. ... If you want to be very explicit: a subgroup of order 3 must be cyclic, ... --- Calvin ...
    (sci.math)
  • Re: Group factor spaces: am I missing something here?
    ... >Let G be a group and N a subgroup of N. The set of all left cosets ... So your equivalence relation is in fact induced by a normal subgroup, ... --- Calvin ...
    (sci.math)
  • Re: Group Theory Question
    ... MimsyBoro wrote: ... >Let G be a finite group of order n. ... >for every pi^mi there exists exactly one sylow pi subgroup of G. ... --- Calvin ...
    (sci.math)