# Same set of Factor Groups implies same composition factors

Let $$G$$ and $$H$$ be finite groups.
If there are normal series of $$G$$ and of $$H$$ having the same set of factor groups, then $$G$$ and $$H$$ have the same composition factors.

Let $$G=G_0\geq\dots\geq G_m=1$$ and $$H=H_0\geq \dots\geq H_n=1$$ be normal series for $$G$$ and $$H$$ respectively.
I think that it should be related to Jordan-Holder Theorem.
So I try to "insert" them in the same group which is $$G\times H$$.

So $$G\times H\geq G=G_0\geq\dots\geq G_m=1$$ and $$G\times H \geq H=H_0\geq \dots\geq H_n=1$$ are both normal series of $$G\times H$$.
By Schreier Refinement Theorem, they have refinements that are equivalent.

I am not sure whether I prove in wrong direction. Also up to here I still haven't use the assumption that $$G,H$$ have same set of factor groups.

Remark: This problem is from exercise 5.11, An Introduction to the theory of groups by Rotman (page 101)

As I understood, we have to show that isomorphic normal series $$\mathcal S$$ and $$\mathcal S’$$ have the same composition factors. The remark at the end of the fragment from the first volume of “Algebra” by B.L. van der Waerden (Springer, 2003) shows that for any refinement $$\mathcal T$$ of $$\mathcal S$$ can be found an isomorphic refinement $$\mathcal S’$$ of $$\mathcal S’$$. Moreover, in the same fragment is remarked that the refinement $$\mathcal T$$ is a composition series iff all its factors are simple groups. Since the latter property is preserved by isomorphisms of normal series, in this case $$\mathcal T’$$ is a composition series too.