Groups and subgroups

Question

if $G$ is a finite group. $H,K$ are subgroups of $G$ and $|H|=18$, $|K|=25$ than why the intersection of $H$ and $K$ is only the unit element and can't include more elements?

Answer 1

The intersection of any two subgroups of $G$ is again a subgroup of $G$. By lagrange's theorem $|H\cap K|| |H|,|K|$ . What is $\gcd(|H|,|K)$

Answer 2

If $G$ is a group and $H,K$ be two subgroups of it, as @Amr noted, you can prove that $H\cap K$ is a subgroup of $G$ containing in both $H$ and $K$. So $$H\cap K\subseteq H~~\text{and}~~H\cap K\subseteq K$$ so according to Lagrange theorem since $H$ and $K$ are both finite so the order of $H\cap K$,say $t$, is finite and divides $|H|$ and $|K|$. Now find the number that divides $25$ and $18$ simultaneously. It is just $t=1$.

Answer 3

Hint: Suppose $x\in H\cap K$. Use Lagrange's theorem for $\langle x\rangle$.

Answer 4

Since you did not choose any answer yet, here is more "physical" approach. The purpose of writing this is not to have an efficient answer.

If we have a nontrivial $x \in H \cap K$, then $\langle x \rangle$ has order $5$ or $25$ by Lagrange. This contradicts the fact that this cyclic subgroup is contained in $H$.