Alg7.14

proof Let $H$ and $K$, each of prime order $p$, be subgroups of a group $G$. If $H\neq K$, prove $H\cap K=$.
If $H$ and $K$ are of prime order $p$, every element $h\in H$ and $k\in K$ have order $1$ or order $p$ by the corollary to Lagrange's Theorem. However, only the identity can have order $1$ and the identity is unique. Thus both $H$ and $K$ have $p-1$ elements of order $p$. Since the orders of those elements are $p$ in a group of order $p$, both $H$ and $K$ have $p-1$ elements which generate their respective groups. Now, since $H\neq K$ and any $k\in K$, $k\neq e$, generates exactly $K$, $k\notin H$. Likewise for any $h\in H$, $h\neq e$, $h\notin K$. Since $H$ and $K$ are both groups, they both must contain the identity $e$. Therefore, if $H\neq K$, where $H$ and $K$ are of prime order $p$, it must be true that $H\cap K=$.