When you have an implication, "if $P$, then $Q$", the converse is the implication "if $Q$, then $P$."
Lagrange's Theorem says:
Let $G$ be finite with $|G|=n$. If $d$ is the order of a subgroup of $G$, then $d$ divides $n$.
The converse would therefore say:
Let $G$ be a finite group with $|G|=n$. If $d$ divides $n$, then $d$ is the order of a subgroup of $G$.
This is an implication. In order to show that an implication is false, you need to show that it is possible for the antecedent ("$d$ divides $n$") to be true and at the same time, the consequence ("$d$ is the order of a subgroup of $H$") to be false.
So in order to show that the converse of Lagrange's Theorem is not true, what you need to do is:
- Exhibit a specific finite group $G$;
- Exhibit a specific number $d$ that divides $G$; and
- Prove that $G$ does not have any subgroups of order $d$.
In other words, you need to give an example that shows that the converse statement does not have to be true. (A "counterexample".)
What you did makes no sense, and does not prove anything. The "converse" has nothing to do with the reciprocal $|H|/|G|$, and is certainly not an assertion that the order of $G$ divides the order of $H$.
Now, just to help in your search: if $G$ is abelian, then it will satisfy the converse of Lagrange's Theorem, so try looking for nonabelian groups; if $|G|$ is a power of a prime, or the product of two distinct primes, then $G$ will satisfy the converse of Lagrange's Theorem as well, so avoid those. But try the very smallest size that is left once you throw out all of those orders.
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.
Best Answer
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.
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:
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:
Again $A_4$ shows that this result is best possible.
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 $k(G)$.
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.
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.