Question: Let $G$ be a finite group and $H$ and $K$ be subgroups of G. Prove the following:
$1)$ The order of $H \cap K$ is a common divisor of the order of $H$ and $K$.
$2)$ Suppose H and K are not equal and have the same prime order, $p$. Then $H \cap K$ = $ \{ e \} $
Here are my proofs:
$1)$ Let $ord(G) = n$. Then by Lagrange's Theorem, $ord(G)$ is multiples of $ord(H)$ and $ord(K)$. The set $H \cap K$ is a subgroup of $H$ and $K$ since $H$ and $K$ are both subgroups of $G$. Thus, by Lagrange's Theorem, $ord(H)$ and $ord(K)$ are multiples of $ord(H \cap K)$. Therefore, $ord(H \cap K)$ divides $ord(H)$ and $ord(K)$.
$2) $ Let $H \neq K$. Suppose $p$ is a prime integer such that $ord(H) = p$, and $ord(K) = p$. Obviously, $ \{e \} \subset H \cap K$ since $H \leq G $ and $K \leq G$.
Let $x \in H \cap K$. Since $ord(H) = ord(K) = p$, by Lagrange's Theorem, $H$ and $K$ are both cyclic groups. In other words, there exists $h \in H$ and $k \in K$, such that, $<h> = H$, and $<k> = K$. So $x = h^m$ and $x = k^q$, for some integers $m, q$. Therefore, this states that $h^m = k^q$. However, $H$ and $K$ have prime order. This means that no power of $h$ equals to any power of $k$ except for the identity element, $e$. This forces $x \in \{e \}$. Thus , $H \cap K = \{ e \}$.
Am I proving these correctly? Thank you very much!
You are correct.
Part 2 can be simplified though: Using part 1, $H\cap K$ has either order 1 or $p$, since a prime number has only two divisors.
It cannot be $p$, since $H$ and $K$ are not equal. Thus $H\cap K$ has order 1, and as you observed that one element has to be $e$.