If I write G/H, I say G mod H and that means all the left (or right) cosets of H and G? Are quotient groups by definition, normal?

Printable View

- Nov 23rd 2012, 06:43 AMamywquotient groups
If I write G/H, I say G mod H and that means all the left (or right) cosets of H and G? Are quotient groups by definition, normal?

- Nov 23rd 2012, 08:14 AMDevenoRe: quotient groups
it means all the left (or right) cosets of H IN G, for this to even make sense, H must be a normal subgroup of G (you can form cosets of a non-normal subgroup, but the left cosets and the right cosets will be different, so the set of cosets will not form a group).

normal is a term that goes along with "subgroup", not "group". and it very much matters WHICH group the subgroup is a subgroup OF.

for example, we can have subgroups like so:

H < K < G.

H might be normal in K, but not normal in G. if however, H IS normal in G, it is also normal in any subgroup of G containing H (like K in this example).

the normal subgroups of a given group are "special". these are the groups we can make quotient groups out of. they are also the same subgroups that can be kernels of homomorphisms. abelian groups are "nice" partly because EVERY subgroup of an abelian group is normal (all the subgroups are "special" ones). this makes abelian groups very easy to classify.

what happens with a quotient group is we basically treat all of a subgroup H as "one element" (think of it as "a subgroup-chunk"). because we can chop any group G into "H-sized pieces" (cosets, one of which is H itself), we might ask when these pieces themselves (the cosets) form a group.

that is, we would like for:

Ha*Hb = Hab

in terms of elements of G, this is:

(ha)(h'b) = h"(ab). we can play with this, a bit:

multiply on the right, by b^{-1}:

hah' = h"a

multiply on the left, by h^{-1}:

ah' = (h^{-1}h")a

from this, we see that if this is going to work ah' (which is in aH), is going to have to be in Ha (since h^{-1}h" is in H, so (h^{-1}h")a is an element of Ha).

this is going to have to work out for every element h'b of Hb. so we need every element of aH to be in Ha. but aH and Ha are exactly the same size.

on the other hand, suppose aH = Ha for every element a in G.

then (Ha)(Hb) = a(HH)b = aHb (since HH = H, subgroups are closed under multiplication) = (aH)b = (Ha)b = H(ab).

the whole mess comes about primarily because in groups, we don't know that ab = ba (usually not). this makes groups "difficult" sometimes. normal subgroups are the ones that "partially" overcome this difficulty, we may not have ha = ah, but at least we have Ha = aH, so as long as we stay in "H-sized chunks", things work out.