Alg7.14

From Exampleproblems

proof Let $H$ and $K$ , each of prime order $p$ , be subgroups of a group $G$ . If $H\ne K$ , prove $H\cap K=<e>$ .

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 $H\ne K$ and any $k\in K$ , $k\ne e$ , generates exactly $K$ , $k\notin H$ . Likewise for any $h\in H$ , $h\ne e$ , $h\notin K$ . Since $H$ and $K$ are both groups, they both must contain the identity $e$ . Therefore, if $H\ne K$ , where $H$ and $K$ are of prime order $p$ , it must be true that $H\cap K=<e>$ .