Intersection of subgroups of a finite group 
Let $H$ and $K$ be subgroups of a group $G$, and assume that $|H|$ is a prime number. Prove that either $H\subseteq K$ or$ H\cap K =\left \{ 1 \right \}$.

We have $H\cap K \leq H$ and hence by Lagrange’s theorem $|H\cap K|$ divides $|H|$. 
Similarly, $H \cap K ≤ K$ and hence by Lagrange’s theorem $|H \cap K|$ divides $|K|$. 
Thus $|H \cap  K|$ is a common divisor of $|H|$ and $|K|$. Since $\gcd(|H|, |K|) = 1$, it follows that $|H \cap  K| = 1$ and hence $|H \cap  K| = \left \{  1\right \}$.
Is there a way to prove that both $H$ and $K$ are relatively prime, or is it obvious?
Would this be true for the second assertion, and how would I go about proving the first assertion false?
Any hints would be greatly appreciated!
 A: We don't need to dwell on $|H|$ and $|K|$ being relatively prime. All we need is that $|H\cap K|$ divides $|H|=p$. This leads to $|H\cap K|=1$ or $|H\cap K|=p$. The last condition is equivalent to $H\cap K=H$ which means $H\subset K$.
A: In any case, $|H\cap K|$ divides $|H|$, so either $|H\cap K|=1$ or $p$. Hence $H\cap K = {1} \space or\space H$ ,and you 're done.
A: Just note that $H \cap K$ is itself a subgroup and the results follow with Lagrange's Theorem since $|H \cap K|$ divides $p$, which is a prime, but then primes only have two factors. Then you do the case work get the desired results. 
It is not immediately obvious (to me at least) why $(|H|, |K|) = 1$ and it doesn't help with your second result anyways. 
"Never underestimate the power of a counting theorem" - Unknown
A: If $|H|$ is prime then the only subgroups of $H$ are $\{1\}$ and $H:$
For $x\in H$ let $Ord(x)$ be the least $n\in \mathbb N$ such that $x^n=1,$ and let $g(x)= \{x^j: 1\leq j\leq Ord(x)\}.$ Then $g(x)$ is a subgroup of $H$. By Lagrange's theorem, $Ord(x)=|g(x)|$ divides $|H|.$ So if $x\ne 1$ then $|g(x)|>1$  and $|H|=|g(x)|=Ord(x).$
So if $J$ is a subgroup of $H$ and $1\ne x\in J$ then $J\supset g(x)=H,$ so $J=H.$
For your Q: $H\cap K$ is a subgroup of $H.$ So  $H\cap K=\{1\}$ or $H\cap K=H$.
