# Intersection of subgroups of orders 3 and 5 is the identity

This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment 5.

Let $H, K$ be subgroups of a group $G$ of orders 3,5 respectively. Prove that $H\cap K=\{1\}$.

The order of each element of $H$ must divide 3. Since the identity element is the only element with order 1, every other element in $H$ has order 3. Similar reasoning shows every nonidentity element of $K$ has order 5. Since $H$ and $K$ are both subgroups of $G$ they share the same identity element. Therefore, $H\cap K =\{1\}$.

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.

Thanks.

Sure, this gets the job done. You can also just observe that the intersection would be a subgroup of both $H$ and $K$. Then using the fact that the order of a subgroup divides the order of the group, this would force the order of the intersection to divide both $3$ and $5$, showing that it is trivial.