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?
 A: a) Because $a_1^2=1$, so $H⊆S_3$ is closed vs. group operation (=composition) and then $H \le S_3$.
b) The equivalence class of $v \in S_3$ is:
\begin{alignat}{1}
[v] &:= \{u \in S_3\mid u \sim v\} \\
&= \{u \in S_3\mid uv^{-1}\in H\} \\
&= \{u \in S_3\mid \exists h \in H:uv^{-1}=h\} \\
&= \{u \in S_3\mid \exists h \in H:u=hv\} \\
&= \{hv, h \in H\} \\
&=: Hv 
\end{alignat}
Since $H=\{1,a_1\}$, we get $Hv=\{v,a_1v\}$: from this, firstly you can get all the equivalence classes, then you can conclude that the class $Hv$ is a subgroup if (by a)) and only if (because $1$ must be in a subgroup) $v=1$ or $v=a_1$, and then $(H1=Ha_1=)H$ is the only equivalence class which is subgroup of $S_3$ (this is c)).
A: 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.
