Proving that a function is surjective and the equality holds

Question

This is exercise 2.76 (ii) from Rotman's "Introduction to the theory of Groups".

I was able to do (i) but I have no idea how to proceed to prove (ii).

Answer 1

You know that: $$[G:H_1\cap\dots \cap H_n]=[G:H_i][H_i:H_1\cap\dots \cap H_n]$$

Hence $[G:H_i] \;\; | \;\; [G:H] \; \forall i$, but they are coprime so:

$$\prod_{i=1}^{n} [G:H_i] \;\; | \;\; |G/H_1\cap\dots \cap H_n| $$

But $G/(H_1\cap\dots \cap H_n)\sim Im(\varphi) \leq G/H_1\times\dots \times G/H_n \;$ by first isomorphism theorem. So $|G/H_1\cap\dots \cap H_n|=|Im(\varphi)|\;\; | \;G/H_1\times\dots \times G/H_n;|= \prod_{i=1}^{n} [G:H_i] \implies \prod_{i=1}^{n} [G:H_i] = |G/H_1\cap\dots \cap H_n|$.