If $G$ is a finite, abelian group and $m\mid \lvert G\rvert$. Then there must exist a subgroup $H$ of $G$ with order $m$. The question says:

If $G$ is an abelian group with finite order and $m$ such that $m$ divides the $G$'s order. Then there must  exist a subgroup $H$ of $G$ with order $m$.

I have an idea, but I don't dnow if it is correct. Let's try this answer:

If $m$ is a prime, so the Cauchy's Theorem guarantee the existence of an element with order $m$, so only take the subgroup generated for this element.
If $m$ isn't a prime, then we can write $m = p_1 \cdots p_n$ where $p_i$ is prime for all $i$. And therefore we have $p_i$ divides the $|G|$, $\forall i = 1, \cdots, n. $  Thus, by the remark done before, we have the existence one $d \in G,$ such that $|d_1|= p_1$. For the same way, we get $d_2, \cdots, d_n.$
Since $p_i$ is prime for all $i$, we have the $\gcd (p_i, p_j) =1 , \forall i \neq j. $ Thus, the $ | \langle d_1, \cdots, d_n \rangle | =  p_1 \cdots p_n= m.$
It's done,  $ H =   | \langle d_1, \cdots, d_n \rangle |$.

Is this correct ?
 A: First of all, if there are subgroups $A,B\subseteq G$ of coprime orders $m,n$, then $AB\subseteq G$ will have order $mn$. Thus, we can reduce to the case where $m$ is a prime power $p^k$.
Let $G$ be a minimal group such that:

*

*$p^k$ divides the order of $G$; and

*$G$ has no order $p^k$ subgroups.

We can assume $G$ has order $p^n$ with $n> k$, since otherwise its $p$-primary part $G_p$ will be a smaller group satisfying the conditions, contradicting minimality.
Let $H\subsetneq G$ be a maximal proper subgroup. If $G/H$ has order $>p$, then there exists a subgroup $U\subsetneq G/H$ of order $p$ by Cauchy's theorem. However, groups $H\subseteq I\subseteq G$ and subgroups of $G/H$ correspond 1-1, so this is a contradiction (of $H$'s maximality.)
Thus $H$ must be a subgroup of order $p^{n-1}$. If $n-1=k$ we are done, and if $n-1>k$ we are again done by $G$'s minimality.
A: You are on the right track, essentially doing (strong) induction on $m$ via its factors. However, I suggest you work only with coprime factors $m=m_1m_2$ with $\gcd(m_1,m_2)=1$. Then you can take the product of the $m_1$-group and the $m_2$-group as your $m$-group. On the other hand you need prime powers instead of primes as fundamental step; use Sylow for this.
