Here is the question:

Using the hint it was very easy to show a contradiction. Briefly the result isIf p > q are primes, a group of order pq has at most one subgroup of order p.

Hint: Suppose H, K are distinct subgroups of order p. Show that $\displaystyle H \cap K = <e>$ and then use $\displaystyle [H \vee K : H ] \geq [K : H \cap K ] $ to show a contradiction.

$\displaystyle [ H \vee K : H] = \frac{p^2 - 1}{p}$ and $\displaystyle [K : H \cap K] = p$

I have two questions. The first: Is there a specific reason that $\displaystyle H \cap K = <e>$ uses <e> rather than {e}?

The second is by way of a counter-example. I note that $\displaystyle \mathbb{Z} _2 \oplus \mathbb{Z} _2$ contains 3 subgroups of order 2. Now this doesn't directly contradict the theorem since p = q = 2 and the problem statement said p > q. However the proof by contradiction never used this fact and I can find nothing in the proof to indicate that p = 2 and |G| = 4 should be a special case?

Thanks.

-Dan