If $H$ and $K$ are subgroups of $G$ whose orders are coprime. Is $H\cap K$ a subgroup of $H$ and $K$? If $H$ and $K$ are subgroups of $G$ whose orders are coprime. Is $H\cap K$ a subgroup of $H$ and $K$?
They both have the same identity, so we know at minimum we have $\{e\}$ so it is the trivial group if they have nothing else in common.
If they have another element that isn't the identity, they will need it's inverse as well and it will generate a cycle, like $a\in H$ and $a\in K$ has $\langle a \rangle\in H,K$. Then for them to not be the same group, they need other elements, but then these other elements will need to be closed under operations with $a$, and I am lost on where to progress.
 A: Yes, it is a subgroup. We can use one step subgroup test and let $a,b$ be any two  elements of  $M=H \cap K$ where $b$ is non zero. Then $ab^{-1} \in H$ as $a,b \in H$. Similarly for $K$.
How about you try to prove that intersection of any number (finite/ infinite) of subgroups  of a group is a subgroup, and then it just becomes obvious. I don't know why you required orders to be co-prime, it is true without the condition. Co-primeness of orders of $H$ and $K$, only forces $|H\cap K|=1$, by Lagrange theorem.
A: Since you seem troubled with things such as "what's the neutral element of a subgroup", I think you might find useful this observation:

Observation: Let $(G,\,\cdot\,)$ a group whose neutral element is $e$. Let $H\subseteq G$. The following are equivalent:

*

*$(H,\,\cdot|_{H\times H}\,)$ is a group


*$\begin{cases}e\in H\\\forall a,b\in H,\ a\cdot b^{-1}\in H\end{cases}$


*$(H,\,\cdot|_{H\times H}\,)$ is a group, its neutral element is $e$ and the inverses of its elements are their inverses in $G$.

Which provides 3 equivalent definitions of "$H<G$".
Using (2), it stands clear that the intersection of an arbitrary family $\mathfrak F$ of subgroups of $G$ is a subgroup of $G$. Both conditions hold for each subgroup in $\mathfrak F$, so they both hold for their intersection.
Using (1) it stands clear that $\begin{cases} H<G\\ S<G\\S\subseteq H\end{cases}\implies S<H$
Indeed, you only need to use $\cdot|_{S\times S}=\left(\,\cdot|_{H\times H}\right)|_{S\times S}$
A: Suppose $H$ and $K$ share an element $x$. Then the order of $x$ divides  the orders of both $H$ and $K$.  But since the orders of $H$ and $K$ are relatively prime, the only number that divides both their orders is $1$.  So $x$ must have order $1$. The only element of order $1$ is $e$. So the only element $H$ and $K$ have in common is the identity
