G is a group and contained within it are two subgroups H and K. Prove/Disprove that H intersect K is a subgroup. G is a group and contained within it are two subgroups H and K. Prove/Disprove that H intersect K is  a subgroup.
What I said for H intersect K:
PROVING:
Assume H and K are subgroups of G. Since H and K are subgroups of G, they satisfy the properties of closure, unique identity and unique inverse. With H intersect K, we will only have the shared elements of our two subgroups contained with our set, which implies closure and so we at least (if no other elements are shared) have our identity and inverse contained. It is clear to see that H intersect K is a subgroup of G.
Showing closure: If a,b belong to H, then a*b belongs to H. If a,b belong to K, then a*b belong to K. Thus, a,b belong to H intersect K which implies that a*b belongs to H intersect K.
So my question is: was this stated correctly or am I just plain wrong? I've been trying to comprehend the material and THINK I get it, but just would like some outside feedback.
 A: Your "Proving" header seems to be more of a sketch than an actual proof (maybe this was intended). In that case, to show that $H \cap K$ is a subgroup, you want to show three things: closure under multiplication, closure under inverses, and nonemptiness. The existence of a unique identity will follow from the first two (and is often a good way of showing the third). Arguments of the form in your "Showing closure" heading are what you want to be using to prove these properties.
For those arguments, rather than stating your arguments "if $a,b \in H$ and $a,b \in K$ satisfy given properties then $a,b \in H \cap K$ satisfy given properties," it would be more clear to state them in the following form:
Let $a,b \in H \cap K$. Then $a,b \in H$ and $a,b \in K$. Since $H$ is a group, we have that $a\cdot b \in H$ and since $K$ is a group, we have that $a \cdot b \in K$. Thus
$a \cdot b \in H \cap K$.
Closure under inverses can be proved similarly. For nonemptiness, maybe my hint earlier was enough, but if it wasn't:

Show that the identity is in $H \cap K$.

