Proving that $H\cap K = \{e_G\}$ given $|H|=3$ and $|K|=5$

Question

I found the following question in an old test paper:

Suppose $H$ and $K$ are two subgroups of $G$ such that $|H|=3$ and $|K|=5$. Prove that $H\cap K = \{e_G\}$ where $e_G$ is the identity of $G$.

My attempt:

Every group of prime number order is isomorphic to $\Bbb Z_{p}$ ($p$ being the prime number) and thus they are necessarily cyclic. Thus, $H$ is isomorphic to $\Bbb Z_3$ and $G$ is isomorphic to $\Bbb Z_5$. After this I'm not sure how to proceed. I thought $\Bbb{Z_3}$ and $\Bbb{Z_5}$ have no element in common since the elements of $\Bbb{Z_3}$ are $\{\bar{0}_3,\bar{1}_3,\bar{2}_3\}$ while the elements of $\Bbb{Z_5}$ are $\{\bar{0}_5,\bar{1}_5,\bar{2}_5,\bar{3}_5,\bar{4}_5\}$. But again, I'm confused because since $H$ and $K$ are subgroups they must be sharing the identity element of $G$.

Questions:

So, could someone please clarify this problem of mine? Clearly $\Bbb Z_3$ and $\Bbb Z_5$ have no element in common, but then how do $H$ and $K$ have an element (the identity in common)?

Moreoever, what would the correct way to approach the quoted problem?

Answer 1

$H\cap{}K$ is a subgroup of both $H$ and $K$ and thus by Lagrange’s theorem its order divides both 3 and 5 thus it has to be just the identity element. The identity element in both groups is the same as in G because they are subgroups and the identity element of a group is unique.

Answer 2

Any element of $H\cap K$ distinct from $e_G$ will have both order $3$ and $5$, but no such element of $g$ exists. To be more precise, if $g\in G$ is such that $g^3=g^5=e_G$, then $g=e_G$.