Show $S_3$ is not cyclic and find the subgroups of order $2$

abstract-algebra

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).

  1. Show that $S_3$ is not cyclic.
  2. Write down a complete list of all subgroups of order $2$ in $S_3$.
  3. Show that none of the above subgroups is normal in $S_3$.

Solution

  1. 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.

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

  2. I have no idea how to do this, any hints are appreciated.

Best Answer

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 $(ac)(ab) = (abc)$.

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