Composition Series and simple groups

1. Mar 27, 2012

lmn123

1. The problem statement, all variables and given/known data
For each m >= 2, find a group with a composition series of length 1 with a subgroup of length m.

2. Relevant equations
Simple groups iff length 1.

If G is abelian of order p1^k1....pr^kr, then length G = k1 + ... + kr

If G has a composition series and K is normal subgroup of G, then Length G = Length K + Length G/K

3. The attempt at a solution

I don't really see how this is even possible considering the fact that simple groups have length 1. So this means that G must be a simple group. I do not understand how to get a group of length m from a simple group.

I think I'm getting somewhere with this:

What I was thinking was have G = the cyclic group of order p. Then G would have length 1

Then I thought to Let H be a subgroup of G such that |G|=p^m, so if G=<a>, st |a|=p, then H=<a^p^(m+1)>

But this doesn't really work because the subgroup would have higher order than G right? Am I missing something, like modular arithmetic?

Any light you guys could shed would be appreciated
Thanks

2. Mar 28, 2012

morphism

Fix m, and let G be an abelian group of length m. Can you put G inside a simple group? (What simple groups do you know?)