group with a composition series of length two

[SimonLee]

Here is the question:

 

If a group G has a composition series of length two, then for any two distinct normal subgroups M and N, G= M x N..How to see that?

 

what I am thinking is whether we can get G>M>1 is a composition series.(every composition series should have length two from Jordan Holder)

Thanks a lot!

 

Simon

[matt grime]

Given that G has two distinct normal subgroups, there are two decomp series

1<N<G

1<M<G

 

by jordan holder it follows that {N,G/N} = {M,G/M}

 

implying that G/M=N, and G/N=M

 

that is G= M(semi-direct product)N and G =N(semidirect product)M

 

i've not thought if the conclusion thus semidirect product must be direct, follows easily or not (i'm not really that much into my groups more their representations)

 

actually it does follow easily, doesn't it.

[SimonLee]

Thx a lot!

 

I am a little confused about the reason {N,G/N} = {M,G/M} implying that G/M=N, and G/N=M , why we cannot get G/N isomorphic to G/M, M isomorphic to N? And I have a stupid question, the intersection of N and G/N would be always trivial or not?

 

If we can get G= M(semi-direct product)N and G =N(semidirect product)M,then since both M and N are normal subgroups, so the semi-direct products are direct,that's due to the definition.

 

And I am also thinking of using the facts: M N are bothe maximal normal subgroups, (We can gurantee here that M and N are both maximal, but whether the intersection is trivial? ) then G = MN

[matt grime]

And I have a stupid question, the intersection of N and G/N would be always trivial or not?

 

 

Like I say, these aren't questions I've considered for a long time but for this bit, G/N is n't a subgroup of G so you can't talk about it's intersection with N, really.

 

Your other points are good, but I can't figure out a good answer right now.