Show that $H_{1}\cap H_{2}$ is a normal subgroup of index 4 in $G$. This was given to us a few semester ago and would like to review on it (I did not get answers way back). 
Given: group $G$ has two distinct subgroups $H_{1}$ and $H_{2}$ with $\left[G:H_{1}\right]=\left[G:H_{2}\right]=2$.
i. Show that $H_{1}\cap H_{2}$ is a normal subgroup of index 4 in $G$.
Proof. Since $\left[G:H_{1}\right]=\left[G:H_{2}\right]=2$, then $H_{1},H_{2} \trianglelefteq G$. Then, $H_{1}\cap H_{2}$ is normal in $G$. Moreover, $\left[G:H_{1}\cap H_{2}\right]\leq \left[G:H_{1}\right]\left[G:H_{2}\right]=4$. Since $H_{1}$ and $H_{2}$ are distinct subgroups, it follows that the order of $\frac{G}{H_{1}\cap H_{2}}$ cannot be 1, 2 or 3 and so $\left|\frac{G}{H_{1}\cap H_{2}}\right|=4$.
ii. To which known group is $\frac{G}{H_{1}\cap H_{2}}$ isomorphic?
iii. Explain why $G$ has a third subgroup $H_{3}$ with $\left[G:H_{3}\right]=2$.
iv. Express $H_{3}$ in terms of $G$, $H_{1}$ and $H_{2}$.
 A: ii. There are only two order-4 groups to consider. And seeing as $H_1/(H_1\cap H_2)$ and $H_2/(H_1\cap H_2)$ are both distinct, proper, non-trivial subgroups of $G/(H_1\cap H_2)$, that rules out one of the options.
iii. The group $G/(H_1\cap H_2)$ has three order-2 subgroups. We have accounted for two of them. So the third one must be $H_3/(H_1\cap H_2)$ for some subgroup $H_3\subseteq G$, by the third isomorphism theorem.
iv. The answer is
$$
H_3 = (H_1\cap H_2) \cup \big[(G\setminus H_1)\cap (G\setminus H_2)\big]
$$
It's not difficult to check that this subset contains the identity, and inverses of any elements, and that if it is a subgroup, it has index $2$. It is slightly more tricky to show that it is closed under multiplication, but using the fact that $G/H_1$ and $G/H_2$ are both cyclic groups of order $2$ should help.
How to come up with this answer? It's what you get when you use the third isomorphism theorem for calculation rather than for proof of existence. So it's a natural consequence of iii.
