I have an abstract algebra proof I can not complete. The proposition goes as follows: Consider a group $G$ of order $m$. If $n$ divides $m$, prove that $G$ has a subgroup of order $n$. This question is based off the fundamental theorem of finitely generated abelian groups.   Proof: My idea was to let $m = P_{1}^{\alpha_{1}}...P_{k}^{\alpha_{k}}$ and $n = P_{1}^{\beta_{1}}...P_{l}^{\beta_{l}}$ where $\beta_{i} \leq \alpha_{i}$. Now after this, I am a little unsure what to do. Applying the theorem can let me write this as a direct product, but I am not sure how that leads to show $G$ has a subgroup of order $n$.   I am open to answers since this is something I am independently studying in my spare time.  Thanks!  NOTE: I am not looking to prove this by induction. I want to use the fundamental theorem of finitely generated abelian groups.
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)?
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