Order of the intersection of groups

Suppose H and K are subgroups of a group G. If |H| = 12, |K| = 35. Find $\displaystyle |H \cap K|$

My solution:

By a theorem, I know that |H| and |K| divides |G|, and that the order of every element h in H and k in K divides the order of H and K, respectively. i.e. |h| divides |H| and |k| divides |K|.

Then the element in $\displaystyle |H \cap K|$ is an element that is commonly shared by both H and K, meaning the elements must have an order that divides both 12 and 35.

Well, the only number that does that is 1, and the only element with order 1 is the identity. So $\displaystyle |H \cap K|$ = 1.

Is that right?