Ok we have 2 subgroups of G defined as $H_1$ and $H_2$ the question wants us to prove $H_1$ intersect $H_2$ must also be a subgroup of G.
This seems fairly intuitive, making as math usually hard to prove 🙂
we know that for any element a in $H_1$ there exists $a^{-1}$ and that $H_1$ is closed. the same holds for $H_2$ so the intersection will only contain and element c in $H_1$ Intersect $H_2$ if c and $c^{-1}$ are in $H_1$ and $H_2$ additionally we know that $H_1$ and $H_2$ must contain $e$ the identity of G thus $H_1$ intersect $H_2$ cannot be empty. my problem lies with writing this out and proving closure on this intersection.
Best Answer
To avoid subscripts, let $H_1 = P,\; H_2 = Q$. Closure is addressed in Hint 2.
Step 1:
You know that $P$ and $Q$ are subgroups of $G$. That means they each contain the identity element, say $e$ of $G$. So what can you conclude about $P\cap Q$? If $e \in P$ and $e \in Q$? (Just unpack that means for their intersection.) In showing $e \in P\cap Q$, you also show, $P\cap Q$ is not empty.
Step 2:
You know that $P, Q$ are subgroups of $G$. So they are both closed under the group operation of $G$. If $a, b \in P\cap Q$, then $a, b \in P$ and $a, b \in Q$. So $ab \in P$ and $ab \in Q$. So what can you conclude about $ab$ with respect to $P\cap Q$? This is about proving closure under the group operation of $G$.
Step 3:
You can use similar arguments to show that for any element $c \in P\cap Q$, $c^{-1} \in P\cap Q$. If $c \in P\cap Q$, then $c \in P$ and $c\in Q$. Since $P$ and $Q$ are subgroups, each containing $c$, it follows that $c^{-1} \in P$ AND $c^{-1} \in Q$. Hence $c^{-1} \in P\cap Q$. That establishes that $P\cap Q$ is closed under inverses.
Once you've completed each step above, what can you conclude about $P\cap Q$ in $G$?