Let $G$ be an abelian group of order $m$. If $n$ divides $m$, prove that $G$ has a subgroup of order $n$. [duplicate]

So I know $|G|=p_1p_2p_3\cdots p_n$ but I don't know where to go from there. I'm having trouble figuring out how they both relate.

Hint: Step 1: By the fundamental theorem of finite abelian groups, you know that $G\cong\Bbb Z/(p_1^{k_1})\times\cdots \Bbb Z/(p_r^{k_r})$ (note that $p_i$ and $p_j$ need not be distinct for $i\neq j$). Then if $\left|G\right| = m$, what is $m$ in terms of the $p_i^{k_i}$'s? (What is $\left|\Bbb Z/(p^r)\right|$? If $H$ and $K$ are finite abelian groups, what is $\left|H\times K\right|$?)

Step 2: Writing $m = \prod_{p\textrm{ prime}}p^{r_p}$, $n = \prod_{p\textrm{ prime}}p^{s_p}$, what can you say about the $s_p$'s in terms of the $r_p$'s? (Is one bounded by the other for all $p$, perhaps?) How does the factorization of $m$ into primes relate to the expression you found for the order of $G$ in step 1?

Step 3: Next, you want to use your knowledge of the subgroups of a cyclic group: in particular, what are the subgroups of $\Bbb Z/(N)$ for arbitrary $N\in\Bbb Z$ (and what are their orders)?

Step 4: Using this information, you'll be able to use the decomposition of $G$ given by the fundamental theorem of finite abelian groups to find subgroups $H_i$ of each factor $\Bbb Z/(p_i^{k_i})$ such that $\left|H_1\times\cdots\times H_r\right| = n$. Here you'll need to take a bit of care, as even though $\left|G\right| = p_1^{k_1}\cdots p_r^{k_r}$, $G$ need not be isomorphic to $\Bbb Z/(p_1^{k_1})\times\cdots\times\Bbb Z/(p_r^{k_r})$ - so you may find that you have a prime factor $q$ of $n$ for which there is no factor $\Bbb Z/(p_i^{k_i})$ of $G$ with $q$ dividing $p_i^{k_i}$ (Why is this relevant? Think about step 3.). However, this can be remedied using the observations you made in step $2$.

If $|G|={p_1}^{k_1}{p_2}^{k_2}....{p_n}^{k_n}$ and $ n={p_1}^{r_1}{p_2}^{r_2}....{p_n}^{r_n}$ where $n|m$, and $p_i's$ may be equal or distinct and as $G \cong \mathbb{Z_{p_1^{k_1}}} \times \mathbb{Z_{p_2^{k_2}}}......\mathbb{Z_{p_n^{k_n}}} $ pick, $a_1 \in \mathbb{Z_{p_1^{k_1}}} $ such that $|a_1|=p_1^{r_1}$, and similarly $a_2,a_3,,,,a_n$ etc i.e. $|a_i|=p_i^{r_i}$. Then $a_1.a_2.a_3...a_n$ is an element of order $n$ as $G$ is abelian.

Apply induction on order of $G$. It is true for $m=1$.

Let $p$ be a prime dividing $n$ ,then $p$ divides $m$. So $G$ has an element $a$ of order p. Consider $K=\langle a \rangle$, thus $O(G/K)=m/p$, which is less than $m$.

Let $n=m_1p$. So by the induction hypothesis $G/K$ has a subgroup of order $ m_1 $ (say, $H/K$) where $H$ is a subgroup of $G$ containing $K$. Hence, $O(H)=m_1p=n$.