Prove that $H \cap K = H$ where $H$ and $K$ are subgroups of a cyclic group 
Let $G$ be a cyclic group of order $n, (20<n<30)$, generated by $a$ such that $H$ and $K$ are subgroups of $G$ of order $2$ and $4$ respectively. Prove that $H \cap K = H$.

What I know so far:


*

*The order of $G$ is divisible by $ord(H)$ and $ord(K)$, so we get that $n=24$ or $n=28$.

*$G$ is cyclic, thus $G$ is an abelian group which means that the subgroups $H$ and $G$ are normal subgroups.

*$\mid H \cap K \mid \ge 1$, since every subgroup of $G$ contains the identity element of $G$.

*$\mid H \cap K \mid \le 2$, since $\mid H \cap K \mid \le \min\{ \mid H \mid,\mid K \mid  \}$.

*Every subgroup of a cyclic group is a cyclic group, and so by using Euler's totient function, we get that the number of generators in $H$ and $K$ are $1$ and $2$ respectively.
I can't see how to connect all of those conclusions together, I am probably missing something basic here.
Note: 
I have not learned Homomorphism and Isomorphism of groups yet (in class that is). 
Please refrain from giving a solution to the question but rather help me by giving me hints and criticism as it will help me improve. Thanks in advance!
 A: Hint: show that a cyclic group can only have a single element (and therefore only a single subgroup) of order $2$.
More generally, for any finite, cyclic group $G$, for any order $d\mid |G|$, there is exactly one subgroup of $G$ with order $d$.
A: Given that you already know that $n = 24$ or $n = 28$ you can just write down these two groups very explicitly (e.g. as integers modulo $n$) and then pick a red pen and a green pen (and if you need one a blue one as well) and  circle the elements of the potential subgroups that could play the roles of $H$ and $K$. Then check that the answer holds in all possible cases.
Now this is of course a boring answer that won't help you much solving the similar problem with $200 < n < 300$ or $2000 < n < 3000$ EXCEPT that when you actually do it you will notice something extraordinary: YOU WON'T NEED YOUR BLUE PEN. Now (of course) this doesn't come as a surprise after I just told you it will happen while moreover before that the other excellent answer by Arthur already told you the same thing.
The reason I am telling you this anyway is because of the 'deeper' lesson below it: 

Often it is good to just start to solve problems in the stupid, boring, non-informative brute-force way, as long as you keep your eyes open to anything remarkable that happens while doing it. Often these nice surprises will show you the path to a more elegant, general, desirable solution, but equally often you wouldn't have found the more desirable solution when just trying to 'come up' with it, instead of trying the boring version first.

Brian Hayes makes a similar point in a very nice way here: of course the other children in 8-year-old Gauss's school class did not see his clever way to add the numbers 1 to 100, but had they just start doing it they would have found SOME (slightly less) clever shortcut along the way.
