The composition length equals the sum of the exponents on all prime divisors in a prime factorization.

Question

I read that

"The composition length equals the sum of the exponents on all prime divisors in a prime factorization."

in this link:

https://groupprops.subwiki.org/w/index.php?title=Composition_length&mobileaction=toggle_view_mobile

But why? Answer or hint? I am new in algebra, maybe it is clear but I wonder it.

Answer 1

Notice what you are citing comes from a section captioned "finite solvable group."

The "solvable" part gives you a finite normal series with abelian factors. The "finite" part allows you to further refine each of these factors until they are simple.

Now, since the factors started out abelian, they will be continue to be abelian in refinements, so you'd come to a halt at finite, simple, abelian groups, a.k.a. cyclic groups of prime order.

Since $|A/B|=|A|/|B|$ when we are talking about finite groups, you simply trace back up the composition series noting that each prime step accounts for one exponent in the factorization of the group's order.