Prove $H \cap K$ is itself a subgroup of $G$ if $H$, $K$ are subgroups of $G$

Let $G$ be a group. If $H$, $K$ are subgroups of $G$, then prove that $H \cap K$ is itself a subgroup of $G$.

What I need: I am hoping you wonderful people could read this over and make sure it is solid. I appreciate any advice/corrections etc.

My Work:

Associativity: Since we assume $H$, $K$ are subgroups of $G$, then $H$, $K$ inherit associativity from $G$.

Identity-As subgroups of $G$, we may conclude that $e \in H$, $e \in K$ where $e$ is the identity of $G$.

Closure- $\forall$ $a,b \in H, ab \in H$ and $\forall$ $a,b \in K, ab \in K$. This must be true since $H$, $K$ are assumed to be subgroups. Then we can conclude: $$\forall a,b \in H \cap K, ab \in H \cap K$$

which shows that $H \cap K$ has closure under the group operator for any pairing of arbitrarily selected elements that are in $H$, $K$ simultaneously.

Inverse - Because $H$, $K$ are subgroups we know: $\forall c \in H and K$ there exists $c^{-1}$ in both $H$, $K$ s.t. $cc^{-1}=c^{-1}c=e$. Then this is true for all $c \in H \cap K$.

We may also say that every element in $H$ is also in $G$ and every element in $K$ is also in $G$. This must be true since we assume $H$, $K$ are subgroups of $G$. This implies:

$$\forall x \in H \cap K, x \in G$$

Since $H \cap K$ is associative, closed under the group operation, contains an identity and has inverses, and because all elements in $H \cap K$ are also in $G$, I can conclude that $H \cap K$ is itself a subgroup of $G$.

The computer I am using just now randomly decided to not display LaTex in a rendered form. I apologize if the formatting is wrong...I did my best to be accurate after I lost the ability to check it halfway through typing this.

• This is fine. Good job. Commented Feb 16, 2015 at 4:31
• Looks good to me
– user142198
Commented Feb 16, 2015 at 4:32
• " We may also say that every element in H is also in G and every element in K is also in G. This must be true since we assume H, K are subgroups of G . This implies: ∀x∈H∩K,x∈G" I don't think showing this is necessary. That is obvious since $H,K \lt G$. Commented Jun 16, 2019 at 3:11