# Every two finite solvable groups of the same order have the same composition factors

I proved the following statement:

The composition factors of every finite solvable group are isomorphic to cylic groups of prime order.

I want to use this result to prove that every two finite solvable groups of the same order have the same composition factors.

If I knew that for every prime number in the decomposition of the order of the groups there is a composition factor that is cyclic of that order I would obviously be finished. However, I'm not even sure if this statement is true and I don't have a good idea on how to proceed in case the statement isn't true.

Help will be appreciated! :)

• I wonder whether both you and Babak are misunderstanding what it means for two groups to have the same composition factors. If you know that all composition factors are cyclic of prime order and that the product of the orders of the composition factors is equal to the order of the group, then what you are trying to prove follows immediately. – Derek Holt Jan 1 '13 at 16:52

If $1=G_0\lhd G_1\lhd \ldots \lhd G_n=G$, then $|G|=\prod_{i=1}^n [G_{i}:G_{i-1}]$.