Equivalence classes and subgroups of bijections of $S_3$ 
Consider the group $S_3 = Sym({1,2,3})$, which we recall is the group of all bijections from $\{1,2,3\}$ to $\{1,2,3\}$ and where the group operation is composition. We have seen that this group has order 6 and is not abelian. We let
  S3 equal this set
For instance, $a_5$ is the function $a_5 : \{1,2,3\} \to \{1,2,3\}$ with $a_5(1) = 2, a_5(2) = 3,$ and $a_5(3) = 1$.
(a) Explain in one or two sentences why $H = \{1,a_1\}$ is a subgroup of $S_3$.
(b) Recall that there is an equivalence relation ~ defined in $S_3$ by $u$ ~ $v$ iff $uv^-1 \in H$. Find all the distinct equivalence classes for ~. 
(c)Which of the obtained equivalence classes are subgroups of $S_3$?

So for (a), I know that for each $x \in$ group compositioned with identity element yields $x$, so I know that 1 belongs in subset $\{1, a_1\}$, and same can be applied for $a_1$. But I do not know how to summarize that into a sentence.
For (b) and (c), I'm having trouble exactly what equivalence classes are... Are they just subsets that partition the group in 2?

Let's write out the elements of $S_3$.  They are $\{(12), (23), (13), (123), (132),e\}$.  (In the picture they are $a_1=(12), a_2=(23), a_3=(13), a_4=(132)$ and $a_5=(123)$.)
Now, it's clear that there are $3$ equivalence classes.  They are $[(12)]=\{e, (12)\}, [(13)]=\{(123),(13)\}$ and $[(23)]=\{(132), (23)\}$.  Only the first is a subgroup.
