Being $|G| = 20$ and $H$ and $K$ subgroups of $G$ whose order is 5, prove that $K = H$ This time I'm having trouble with the following exercise:

Being $|G| = 20$ and $H$ and $K$ subgroups of $G$ whose order is 5, prove that $K = H$.

I'm also recommended to start by proving that $|H \cap K| = 5$.

My draft:

Let $x \in (H \cap K)\setminus{\{e\}}$. Then $|x|$ divides $|H| = |K| = 5$. Because $x \neq e$, $|x|$ can only be 5. So $|<x> = H \cap K| = 5$.

Is this part OK?
But now, on to prove that $H = K$, I've been wandering around unsuccessfully (my rubber is already half the size it was earlier today :p). Can you drop any hint on this matter?
Thanks for taking the time to read!
 A: I'm afraid your draft isn't quite complete, because you're assuming such $x$ exists. In other words, you're assuming $H\cap K$ is non-trivial. However,  if you could exlude that case then you would be done.
Here's a complete reasoning, don't read it if you want to keep looking :-)
You should note that $H\cap K\leq H$, therefore $ |K\cap H|\mid |H| = 5$, because of Lagrange's theorem. In other words $|H\cap K| = 1$ or $|H\cap K| = 5$, because $5$ is prime.
Now assume $|H\cap K| = 1$ indeed, (we will try to exclude this case, by deriving a contradiction). Then $|HK| = \frac{|H|\cdot |K|}{|H\cap K|} = \frac{5\cdot 5}{1} = 25$, which is impossible in a group of $20$ elements. (I'm assuming you are familiar with that formula?) Therefore this situation is not possible and $|H\cap K| = 5$ indeed.
A: I know you stated that you have not learned the Sylow theorems, but here is an alternative method that you can look at whenever you're ready:

$ I. $ Let $p$ denote a prime and let $G$ be a group of order $n = p^am$ with $ a \in \mathbb{N}_0$ and $ m \in \mathbb{N}$ such that $p \not \mid m.$ Every subgroup of order $ p^a$ is called a Sylow $ p$-subgroup of $G.$ Here, we denote the set of Sylow $ p $-subgroups as $\operatorname{Syl}_p(G).$
$II. $ Let $G$ have order $n = p^am $ as described above. Then the number of Sylow $ p $-subgroups $|\operatorname{Syl}_p(G)|$ (this is all equal to the normalizer $N_p (G)) $
  satisfies:
  $(i) |\operatorname{Syl}_p(G)| \equiv 1 \mod p $
$(ii) |\operatorname{Syl}_p(G)| \mid m$

(You can find plenty of proofs to $ II $ online. )
Now to your question:
We have $|G| = 20 = 5 \cdot 2^2.$ By $II. $ we have that $|\operatorname{Syl}_5(G)| \equiv 1 \mod 5$ and that $|\operatorname{Syl}_5(G)| \mid 2^2 = 4,$ but this implies that $|\operatorname{Syl}_5(G)| = 1.$ This means that there is only one such subgroup of order $5.$ Any group of order $20$ will contain only one subgroup of order $5,$ so any two subgroups $ K $ and $ H $ of $ G $ of order $5$ will be the same subgroup.
Edit: Didn't realize the date on the question, disregard any irrelevance 
