Question: > 
> 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](https://i.stack.imgur.com/AzDqG.png)
> 
> 
> 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 $S\_3$?
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> 
> 

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?

Answer: 
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)).
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.