Converse of Lagrange's theorem for abelian groups

Solution 1:

This proof is correct, and it is the natural way to argue, given the inputs that you have. In some sense the indices are also natural: they encode all the relevant data.

If you want to remove some of them, though, here is one standard approach:

• First assume that $G$ has $p$-power order, and prove the result in this case.
  (I.e. prove your Claim first.) This eliminates your index $i$ in this part of the argument. (Note by the way that your indices $\beta$ should actually be decorated with $i$ as well as $j$, but in this approach they don't need to be.)

• Now explain how to deduce the general case from the $p$-power order case. (This amounts to joining together more-or-less the first and last paragraphs of your proof. Now you need the index $i$, but you don't need the $\beta$s or $\delta$s, because they were only used in the proof of the claim.)

I call this a "standard appraoch" because reorganizing steps of a proof so that various claims, etc., get proved first is a standard method for avoiding an overgrowth of notation. Ultimately, this is often why steps of the proofs of theorems are broken up into lemmas.