Intersection of 2 subgroups must be a subgroup proof.

Question

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.

Answer 1

To avoid subscripts, let $H_1 = P,\; H_2 = Q$. Closure is addressed in Hint 2.

$P$ and $Q$ are subgroups of a group $G$. Prove that $P \cap Q$ is a subgroup.

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 $G$?

Answer 2

You've done almost all of it already:

Let $G$ be a group, $H_1, H_2$ subgroups of $G$, $H=H_1\cap H_2$. Then

• $H$ is not empty: If $e$ is the neutral element of $G$, then $e\in H_1$ and $e\in H_2$, because these are subgroups. Hence also $e\in H$ and $H\ne \emptyset$.
• If $a,b\in H$, then $ab^{-1}\in H$: Indeed, $a,b\in H$ implies $a,b\in H_1$ as $H\subseteq H_1$, hence $ab^{-1}\in H_1$ because $H_1$ is a subgroup. Similarly, $ab^{-1}\in H_2$ and hence $ab^{-1}\in H$.

You should know the subgroup criterion: A subset $H$ of a group is a subgroup iff $H\ne\emptyset$ and $a,b\in H$ implies $ab^{-1}\in H$. Hence we have just shown that $H$ is a subgroup of $G$.