Showing that the intersection of 2 subgroups is a subgroup Suppose $H_{1}$ and $H_{2}$ are both subgroups of the group G.
We want to show $$H_{1}\bigcap  H_{2}\leq G$$
This means(by the sub-group test):
$$b^{-1} \in H_{1}\bigcap H_{2}$$
and
$$ab \in H_{1}\bigcap H_{2}$$ for all $a,b$
The intersection of the 2 subgroup is non-empty since the identity element of is contained in both subgroups and thus their intersection must necessarily contain e.
I proceed next to show that if $$ab$$ is in the intersection then it must necessarily be in both subgroups. This implies from the inverse property of subgroup that $$b^{-1}$$ is in both subgroup. 
But does this necessarily means that $$b^{-1}$$ is in the intersection of the 2 subgroups? I'm not inclined to say yes because while it is necessarily true that anything that is in the intersection must be in the 2 subgroup, the converse is certainly not true.
 A: There are some issues with your reasoning. You want to show that $H_1\cap H_2$ is a subgroup of $G$.
For this you need to show that $b^{-1}\in H_1\cap H_2$ and $ab\in H_1\cap H_2$ for all $a$ and $b$ in $\mathbf H_1\cap \mathbf H_2$
(You didn't write the part in bold above, which is crucial.)
Then you write:
I proceed next to show that if $ab$ is in the intersection then it must necessarily be in both subgroups.
This is a tautology and will lead to nothing. If $ab$ is in the intersection $H_1\cap H_2$ then by definition it is in both $H_1$ and $H_2$. This has nothing to do with group theory.
What you ought to so is show that if $a, b\in H_1\cap H_2$ then $ab\in H_1\cap H_2$. This is true. For this gives $a, b\in H_1, H_2$, and by the subgroup property, we have $ab\in H_1, H_2$, which is same as saying that $ab\in H_1\cap H_2$.
Similarly for the inverse part.
A: This is how it's usually done: Take $a, b \in H_1\cap H_2$. That means that we have $a, b \in H_1$, and we have $a, b \in H_2$. Since both $H_1$ and $H_2$ are subgroups of $G$, we know that $ab \in H_1$ and $ab \in H_2$. Therefore, $ab \in H_1\cap H_2$.
Showing that $a^{-1} \in H_1\cap H_2$ is very similar: Since $a \in H_1$, we know that $a^{-1} \in H_1$. In the same way we have $a^{-1}\in H_2$. This means exactly that $a^{-1} \in H_1\cap H_2$.
A: First you need to show that $H_1 \cap H_2$ is non empty , this is true because $e \in H_1$ and $e \in H_2$ because they are both subgroups of $G$ and so $e \in H_1 \cap H_2$
Now you need to show that if $a \in H_1 \cap H_2$ and $b \in H_1 \cap H_2$ then $ab \in H_1 \cap H_2$, this is also true because  $a \in H_1 \cap H_2 \implies  a \in H_1$ and $a \in H_2$ and the same thing for $b$ and so $ab \in H_1$ and $ab \in H_2$ and so $ab \in H_1 \cap H_2$
Finally you need to show that if $a \in H_1 \cap H_2$ then $a^{-1} \in H_1 \cap H_2$ , this is also true because  $a \in H_1 \cap H_2 \implies a \in H_1 \implies a^{-1} \in H_1$ and also  $a \in H_1 \cap H_2 \implies a \in H_2 \implies a^{-1} \in H_2$ and so $a^{-1} \in H_1 \cap H_2$ and so the intersection is a subgroup
