Recently, my professor wrote down the following theorem without a proof:
Theorem. Let $G$ be a finite abelian group, and $H<G$ be a subgroup. If $|H|=m$ and $|G/H|=n$ with $\operatorname{gcd}(m,n)$, then $G\simeq H\times G/H$.
This is equivalent to the statement that the short exact sequence of groups
$$0\longrightarrow H\longrightarrow G\longrightarrow G/H\longrightarrow0$$ splits.
After some searching on the internet, I found out that this is a general theorem that goes by the name of Schur-Zassenhaus Theorem. The proofs I found use complicated machinery in group theory, group cohomology, which I haven't seen yet. Is there any proof known of the special case where $G$ is finite abelian?
Suppose that $G$ is a finite abelian group and $|G| = mn$, where $\gcd(m,n) = 1$. Let $H < G$ be a subgroup of order $m$. Here is a proof which only works in this special case.
First note that $H = \{ x \in G : x^m = 1\}$. Indeed if $x \in H$, then $x^m = 1$ since $H$ has order $m$. Conversely if $x^m = 1$, then $(xH)^m = H$ in $G/H$. On the other hand $(xH)^n = H$ since $G/H$ has order $n$, so $xH = H$ since $\gcd(m,n) = 1$. Thus $x \in H$.
Let $K = \{ x \in G : x^n = 1\}$. For any $x \in G$, we can write $x = yz$, where $y^m = 1$ and $z^n = 1$. (Proof: write $am + cn = 1$ for $a,c \in \mathbb{Z}$, and take $y = x^{cn}$ and $z = x^{am}$).
Thus $G = HK$. Since $\gcd(m,n) = 1$ we have $H \cap K = 1$, so $K \cong G/H$ and $G \cong H \times K$.