is the intersection of subgroups a subgroup of each subgroup 
Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,K\leqslant G$. I know that $H\cap K\leqslant G$ but is it the case that $H\cap K\leqslant H$ and $H\cap K\leqslant K$?

I am guessing this does not hold but why?
Also I tried with the case that $H=\langle g \rangle,K=\langle h \rangle$ where $g$ and $h$ are the elements in $G$ ($\langle h \rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $H\cap K\leqslant H$ and $H\cap K\leqslant K$ hold unless I make a mistake somewhere.
Much thanks in advance!
 A: It is indeed true that $H \cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:

Definition. Let $(G, \odot)$ be  group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
  
  
*
  
*$e \in H$,
  
*if $h_1,h_2 \in H$ then $h_1 \odot h_2 \in H$,
  
*for all $h \in H$, its inverse element $h^{-1}$ with respect to $\odot$ is also in $H$.
  

These axioms make it so that $(H, \odot)$ is a group in its own right with the very same group operation $\odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $\odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H \cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H \cap K \subseteq H, H \cap K \subseteq K$ and that $H \cap K \ni e$. Moreover, because $H \cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H \cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$. 
A: The subgroup test is:

$H$ is a subgroup of $G$ if and only if for each $x,y\in H$ we've $xy^{-1}\in H$.

Applied to a collection of subgroups $\{H_s\}$, let $x,y\in\bigcap_sH_s$ be a pair of elements. Then $x,y\in H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}\in H_s$ for every index $s$, so $xy^{-1}\in\bigcap_sH_s$, hence $\bigcap_sH_s$ is a subgroup too.
A: The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.

Remark. A subgroup $H \leqslant G$ is also a subgroup of any subgroup $K \leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H \subseteq K \leqslant G$ then:
  \begin{equation}
H \text{ is a subgroup of } G
\Leftrightarrow
H \text{ is a subgroup of } K \,.
\end{equation}

Recall that a subset $H \subseteq G$ is a subgroup of $(G,\cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But,  if $H \subseteq K \leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H \leqslant G$ is exactly the same as the condition $H \leqslant K$.
Your question just corresponds to the case: $H \cap K \subseteq H,K \leqslant G$.
