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.


2 Answers 2


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 $G$?

  • 3
    $\begingroup$ +1, I've been struggling with this and this is the most clear explanation I've found. $\endgroup$ May 7, 2013 at 15:48

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 $G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.