Show $S_3$ is not cyclic and find the subgroups of order $2$ 
Consider the symmetric group $S_3=\{i,\alpha,\beta,\rho,\sigma,\tau\}$ of all permutations on {1,2,3}. It's operation table is
$$\begin{matrix}
\ &i&\alpha&\beta&\rho&\sigma&\tau\\
\ &\underline{\quad}&\underline{\quad}&\underline{\quad}&\underline{\quad}&\underline{\quad}&\underline{\quad}\\
i\vert&i&\alpha&\beta&\rho&\sigma&\tau\\
\alpha\vert&\alpha&\beta&i&\sigma&\tau&\rho\\
\beta\vert&\beta&i&\alpha&\tau&\rho&\sigma\\
\rho\vert&\rho&\tau&\sigma&i&\beta&\alpha\\
\sigma\vert&\sigma&\rho&\tau&\alpha&i&\beta\\
\tau\vert&\tau&\sigma&\rho&\beta&\alpha&i
\end{matrix}$$
(I hope the table makes sense).
  
  
*
  
*Show that $S_3$ is not cyclic.
  
*Write down a complete list of all subgroups of order $2$ in $S_3$.
  
*Show that none of the above subgroups is normal in $S_3$.
  

Solution


*

*We use the table to check the cycle of each element, to see if any are a viable generator.


Starting with $\alpha$
$$\alpha\alpha=\beta,\ \alpha\beta=i,\ \alpha i=\alpha$$
$\alpha$ is not a viable generator as not all elements of $S_3$ have been found.
Doing the same for the rest
$$\beta\beta=\alpha,\ \beta\alpha=i,\ \beta i=\beta$$
$$\rho\rho=i,\ \rho i=\rho$$
$$\sigma\sigma=i,\ \sigma i=\sigma$$
$$\tau\tau=i,\ \tau i=\tau$$
we can clearly see that none of the elements are a suitable generator of $S_3$, hence the group is not cyclic.


*All subgroups of order $2$ were found above, these are $\{i,\rho\},\{i,\sigma\},\{i,\tau\}.$

*I have no idea how to do this, any hints are appreciated.
 A: Assume some of this subgroups - for example, $G = \{i, x\}$ for some transposition $x$ is normal. Then for any $h$ we have $h G h^{-1} = G$, in particular $h x h^{-1} \in G$. So either $h x h^{-1} = i$ or $h x h^{-1} = x$. The first is impossible: multiplying both parts by $h$ in right and $h^{-1}$ in left we get
$$h^{-1} h x h^{-1} h = h^{-1} h$$
$$x = i$$
So for any element $h$ we have $h x h^{-1} = x$ - so $x$ commutes with all elements of $G$. But it's impossible, as transpositon $(ab)$ doesn't commute with transposition $(ac)$: $(ab)(ac) = (acb)$ and $(ac)(ab) = (abc)$.
A: You can see that $S_3$ is non-cyclic directly from the fact it is not commutative.
For the third part:
You can use the criterion for normal subgroups, i.e observe whether the subgroups you have found satisfy
$ana^{-1}\in N , \ \  \forall a\in S_3,\ n\in N$, where $N\leq S_3$.
A: It is easy to prove that a cyclic group of order $n$ has precisely one subgroup of each order $d$ which is a factor of $n$.
In particular, a cyclic group of even order has precisely one subgroup of order $2$. Since each element of order $2$ generates a distinct subgroup of order $2$, a cyclic group of even order has exactly one element of order $2$.
Your group table shows more than one by easy observation (look a the leading diagonal). Actually exhibiting two distinct elements of order $2$ would be enough.
I mention this because it links the first and second parts of your question.

For the third part, if an element $a$ of order $2$ in a group and an element $b$ of order $3$ commute with each other, their product $c=ab$ has order $6$. Since the group has order $6$ and is not cyclic, no element of order $2$ can commute with any element of order $3$.
This means $ab\neq ba$ or $bab^{-1}\neq a$ for all pairs of elements of order $2$ and order $3$ respectively. Also $bab^{-1} \neq 1$. So if $A$  is the subgroup of order $2$ generated by $a$ you have $bA\neq Ab$ or $bAb^{-1} \neq A$ 

It is, of course, possible to do everything by calculating explicitly with the actual elements of the group - and there is much to be said for getting a feel for groups with such hands-on work. This answer is intended to indicate some of the patterns which start to appear, and to indicate the kinds of arguments which emerge once such patterns are known.
