Re: Group factor spaces: am I missing something here?

From: Paul Sperry (plsperry_at_sc.rr.com)
Date: 06/23/04 

Date: Wed, 23 Jun 2004 03:21:32 GMT

In article <200406222111.i5MLBef20701@proapp.mathforum.org>, Dement
<crowdog@o2online.de> wrote:

> Theorem (see, for ex., Allenby: Rings, Fields, and Groups):
> Let G be a group and N a subgroup of N. The set of all left cosets
> of N in G forms a group with respect to the multiplication
> given by aN*bN = ab*N if and only if N is a normal subgroup of G.
>
> Apparently, this is the standard definition if a factor group of
> G with respect to N. In my opinion, as important and correct as the
> above theorem is, the formulation of it might also be seen as
> misleading. It seems to suggest that, informally stated: if we are to
> divide the elements of a group G = {g_1, g_2, ..., g_n} into
> equivilance classes in such a way as to create a new group
> (to be labeled "factor group" ), these equivilance classes are
> always of the form (g_1)N, (g_2)N, ..., (g_n)N where N is a normal
> subgroup of G.
[...]

No, it does not suggest that to me. Suppose we had, say, Z_6 and
partitioned it into subsets of size 1, 2 and 3. Clearly these are not
cosets of any subgroup and equally clearly a group operation can be
defined on them.

On the other hand, if the equivalence relation is a congruence - a~b
and c~d implies ac~bd - then the equivalence class containing the
identity is a normal subgroup whose cosets are the equivalence classes. 

-- 
Paul Sperry
Columbia, SC (USA)

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