I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamental theorem of finite abelian groups. I have come up with a proof, but it seems a bit messy, and I'm not entirely sure if it's correct. It is given below.
Let the order of $G$ be $m$ = $p_1^{\alpha_1} \ldots p_k^{\alpha_k}$. It is known that $G$ is a direct product of $p$-groups, say:
$$G = G_1 \times \ldots \times G_k$$
where each $G_i$ is a $p_i$-group. By the fundamental theorem of finite abelian groups, each $G_i$ is isomorphic to a direct product of cyclic groups of the form
$$\mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}},$$
where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$. Now if $n$ divides $m$, then we must have
$$n = p_1^{\gamma_1} \ldots p_k^{\gamma_k}$$
for some $\gamma_1, \ldots, \gamma_k$ with $0 \leq \gamma_i \leq \alpha_i$.
Claim: Each $G_i$ has a subgroup of order $p_i^{\gamma_i}$
Proof: As above, we have that
$$ G_i \cong \mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}} $$
where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$.
Now since $0 \leq \gamma_i \leq \alpha_i$, we can find
$l$ numbers $\delta_1, \ldots , \delta_l$ such that $\gamma_i = \sum_{j=1}^l \delta_j$, and $0 \leq \delta_j \leq \beta_j$. (This choice of numbers is not necessarily unique).Then $p_i^{\delta_j} | p_i^{\beta_j}$ for each $j = 1, \ldots , l$. Hence, for each factor $\mathbb{Z}_{{p_i}^{\beta_j}}$, there exists a subgroup of order $p_i^{\delta_j}$, namely $\mathbb{Z}_{{p_i}^{\delta_j}}$ (using the fact that the converse of Lagrange's theorem is true for finite cyclic groups). Taking the direct product of each of these subgroups, we get a new subgroup $G_i'$ of $G_i$:
$$G_i' \cong \mathbb{Z}_{{p_i}^{\delta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\delta_l}}$$
The order of this subgroup is $p_i^{\delta_1} \times \ldots \times p_i^{\delta_l} = p_i^{\delta_1 + \ldots + \delta_l} = p_i^{\gamma_i} $. So we have found a subgroup of $G_i$ of order $p_i^{\gamma_i}$, as required.
So each factor $G_i$ in the product $G = G_1 \times \ldots \times G_k$ has a subgroup $G_i'$ of order $p_i^{\gamma_i}$.
Therefore, $G$ has a subgroup
$$G_1' \times G_2' \times \ldots \times G_k'$$of order $p_1^{\gamma_i}…p_k^{\gamma_k} = n$, which completes the proof.
I have two questions about this: firstly, does this proof seem to work? Secondly, is there a way to make the proof more concise (e.g. a way to prove the statement without using all these indices)?
Best Answer
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.