There is this one problem in Herstein's book which recieved more comments than anything else.

"Show that is a finite abelian group has subgroups of orders $\displaystyle n$ and $\displaystyle m$ then it has a subgroup of order $\displaystyle \mbox{lcm}(n,m)$".

An unfair way to solve this problem is to use the fundamental theorem on finite abelian groups, meaning since $\displaystyle n,m$ divide $\displaystyle |G|$ then so does their least common multiple. But then there exists a subgroup of that order.

I found a simplified proof to this using the first Sylow theorem. I will prove a stronger result. Let $\displaystyle n$ be a divisor of the order of a finite abelian group then it has a subgroup of that order. This is more than what is being asked but it works "easily".*

First, let $\displaystyle H,K$ be subgroups of a finite group. So that $\displaystyle \gcd (|H|,|K|)=1$. Then $\displaystyle H\cap K = \{ e \}$, because if $\displaystyle x\in H$ and $\displaystyle x\in K$ then $\displaystyle \mbox{ord}(x)$ is a divisor of $\displaystyle |H|$ and $\displaystyle |K|$ since their orders are relatively prime it means $\displaystyle x=e$. Now if $\displaystyle H$ is such a group so that $\displaystyle khk^{-1} \in H$ for all $\displaystyle h\in H , \ k\in K$ then the group product $\displaystyle HK = \{hk|h\in H, k\in K\}$ is a subgroup. In particular if the group is abelian then this property must hold and so the group product is always a subgroup. The fundamental observation is that if $\displaystyle H,K$ are subgroups with orders $\displaystyle n,m$ which are relatively prime then $\displaystyle HK$ is a subgroup (we are assuming we are working in a abelian group of course) is a subgroup of order $\displaystyle nm$. The proof behind this rests in the identity that $\displaystyle |HK|\cdot |H\cap K| = |H|\cdot |K|$, since as we explained that $\displaystyle |H\cap K|=1$ it means $\displaystyle |HK|=nm$. Now finally we can complete the proof. Let $\displaystyle d$ be a divisor of the order of the finite abelian group. Certainly if $\displaystyle d=1$ then there is a subgroup of this order. So it is safe to assume that $\displaystyle d>1$, in that case the we can factorize $\displaystyle d=p_1^{a_1}...p_k^{a_k}$ where each $\displaystyle p_i^{a_i}$ is a divisor of $\displaystyle |G|$. By Sylow's theorem there is a subgroup $\displaystyle H_i$ having this order. If we let $\displaystyle H$ be the subgroup product of all $\displaystyle H_i$ then it has order $\displaystyle d$ because $\displaystyle \gcd (|H_i|,|H_j|)=1$ for $\displaystyle \delta_{ij}=0$. Now by above it shows that $\displaystyle H$ has order $\displaystyle d$.

*)I use this expression because maybe Sylow's first theorem is considered heavy machinery. I do not consider it to be, but I have seen others think like that.