Prove that $H_n=\{\sigma \in S_n \mid \sigma (i) \equiv i \pmod 3 \}$ with $n≥2$ is a subgroup of $S_n$. Prove that $H_n=\{\sigma \in S_n \mid \sigma (i) \equiv i \pmod 3 \}$ with $n≥2$ is a subgroup of $S_n$.

I'm doing this problem and I don't know if my approach is correctly done.

First of all, we see that $id\in H_n$, because $id(i)=i\equiv i\! \pmod 3$, and $H_n$ is not empty.
We have to see now that if $\sigma$, $\tau \in H_n$, then $\sigma\tau^{-1}\in H_n$.
First, we see that if $\tau \in H_n$, then $\tau (i) \equiv i$ (mod $3$). Furthermore, $\tau^{-1} (i) \equiv i \pmod 3$. So  $$\sigma\tau^{-1}(i)=\sigma(\tau^{-1}(i))=\sigma(i+3k)\equiv i+3k \!\!\!\pmod 3 \equiv i\!\!\! \pmod 3 \in H_n.$$
So $H_n$ is a subgroup of $S_n$.

I know that this is an easy exercise, but I'm having some troubles with the permutation groups, so I would like someone to say if I made any mistakes
Thank you.
There seems to be absolutely no problem with your proof. You don't seem to have any issues with symmetric groups at all. What I sometimes find useful is just to draw a picture. Something like this is what I'm talking about. It helps you comprehend how elements are "mapped" to each other which is very helpful in giving you intuition and insight before coming up with a proof. For an example if you drew the picture with respect to this problem you will see that any element must be mapped to itself or can be displaced by three elements at a time either way. You  can also discover that you can draw out all the elements that are congruent to each other modulo $3$ and see that elements in this set must be mapped to itself.  

Another advice would be to delve deeper into exercises you find easy and look for other hidden patterns. For an example, given this problem it would be a fabulous exercise to think of how many elements actually are in $H_n$. That is given $S_n $ find out the number of mappings $\sigma \in S_n$ such that $\sigma(i) \equiv i \pmod 3$. I think for $n = 3(k + 1)$ there are $ 3 k! $, for $n = 3k + 2$ there are $ 2k! + (k- 1)! $ and for $n = 3k + 1$ there are $k! + 2(k-1)!$ number of such mappings. You wanna get better at manipulating symmetric groups? Then check if my claim is right.
Complete proof without any mistake