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.
Such groups are called Lagrangian, or CLT-groups. They have been studied often in the literature. There is no complete classification, but many interesting criteria. Two (out of many) references are the following:
H. G. Bray: A note on CLT groups, Pacific Journal of Mathematics 27 (1968), no. 2., 229-231.
F. Barry, D. MacHale, A. N. She: Some Supersolvability conditions
for finite groups., Math. Proceedings of the Royal Irish Academy 167 (1996), 163--177.
Definition: A finite group $G$ is called Lagrangian if and only if for each positive divisor $d$ of $|G|$ there exists at least one subgroup $H\le G$ with $|H|=d$.
It is easy to see that every Lagrangian group is solvable, and conversely every supersolvable group is Lagrangian. The inclusions are strict. In fact, every group $G=A_4\times H$ with a group
$H$ of odd order is solvable, but not Lagrangian; and for any Lagrangian group $G$, the group $(A_4\times C_2)\times G$ is Lagrangian, but not supersolvable.
The classical counterexample to Lagrange's Theorem is $A_4$.
For example, no group $S_n$ or $A_n$ with $n\ge 5$ is Lagrangian. This follows from the fact that $A_n$ and $S_n$ are not solvable for $n\ge 5$. There are some more interesting facts, which can be easily found in the literature. For example, we have:
Proposition: If $(G:Z(G))<12$ for the index, then $G$ is supersolvable, hence Lagrangian.
The group $A_4$ shows that the above result is best possible. We have $(A_4:Z(A_4))=12$.
In the paper of Barry et al. the following result is shown:
Proposition: If $|[G,G]|<4$, then $G$ is supersolvable, hence Lagrangian.
Again $A_4$ shows that this result is best possible.
Proposition: If $|G|$ is odd and $|[G,G]|<25$, then $G$ is supersolvable, hence Lagrangian.
In fact, $[G_{75},G_{75}]\simeq C_5\times C_5$ has order $25$, so that
this result is best possible. Here $G_{75}$ denotes the unique non-abelian group of order $75$.
Denote the number of different conjugacy classes of $G$ by $k(G)$.
Proposition: If $\frac{k(G)}{|G|}>\frac{1}{3}$, then $G$ is supersolvable, hence Lagrangian.
Because of $\frac{k(A_4)}{|A_4|}=\frac{1}{3}$ the result is best possible.
It means that if the average size of a conjugacy class of $G$ is less than $3$,
then $G$ is Lagrangian.
Proposition: If $|G|$ is odd and $\frac{k(G)}{|G|}>\frac{11}{75}$, then $G$ is supersolvable, hence Lagrangian.
In fact, $\frac{k(G_{75})}{|G_{75}|}=\frac{11}{75}$, so that the result is
best possible.
Finally, let us mention a result of Pinnock ($1998$), which is related to Burnside's $p^aq^b$-theorem on the solvability of groups of such order.
Proposition: Let $G$ be a group of order $pq^b$ with primes $p,q$ satisfying $q\equiv 1 \bmod p$. Then $G$ is supersolvable, hence Lagrangian.
Best Answer
Lagrange states that the order of a subgroup is always a divisor of the original group order. So we can ask if the inverse holds: Given a group $G$ of finite order $n$ and a divisor $d$ of $n$, does there exist a subgroup $H\le G$ of order $d$?
You already pointed out that this inverse is true for finite cyclic groups. Therefore, it also holds for finite abelian groups (item 1): According to the classification of finite abelian groups, these are direct sums of cyclic groups and the product of their orders is $n$. It is possible to obtain $d$ by taking appropriate divisors of the respective subgroups.
ad 2: Let $G$ be of order $8$. Let $1\ne g\in G$. If $\langle g\rangle = G$, then $G$ is cyclic and we are done. If $\langle g\rangle$ is of order $4$, it has a subgroup of order $2$ (and of course of order $1$) and we are done. Therefore, we may assume that $\langle g\rangle$ is of order $2$ for all nontrivial $g$. But a group with $g^2=1$ for all $g\in G$ is abelian (a much-loved beginner's exercise). Therefore the converse also holds for groups of order $8$.
I see that meanwhile complete answers for 3 and 4 are already available, so I'll stop here. :)