$S_3$ is soluable but not nilpotent

Question

Claim: $S_3$ is soluable but not nilpotent

Proof: $S_3 \cong D_6$. $D_6$ has a subgroup of order 3 generated by the rotations, $\langle r\rangle$.

$|\frac{D_6}{\langle r\rangle}|=2$ so $\langle r\rangle \triangleleft D_6$.

Furthermore, $\frac{D_6}{\langle r\rangle} \cong \mathbb{Z}_2$, and $\langle r\rangle \cong \mathbb{Z}_3$,

Therefore we have an abelian series:

$1 \triangleleft \langle r\rangle \triangleleft D_6$

And thus $S_3$ is soluable.

I would be interested in somebody showing me a different way to do this, perhaps in terms of the derived series.

Now, $S_3$ is not nilpotent because $Z(S_3) = \{1\}$

Answer 1

To use the derived series:

Consider the presentation $D_3=\langle r,\bar{r}\mid r^3=\bar{r}^2=1, \bar{r}r=r^{-1}\bar{r}\rangle$. Note that $D_3'=\langle r\rangle$: since $$ [r,\bar{r}]=r^{-1}\bar{r}^{-1}r\bar{r}=r^2\bar{r}r\bar{r}=r^2r^{-1}\bar{r}\bar{r}=r $$ and similarly other commutators all in $\langle r\rangle$. So the derive series starts $$ D_3\unrhd D_3'\cong C_3 $$ and now $C_3$ is abelian, so its derived subgroup is trivial.

Answer 2

$S_3$ has a normal solvable subgroup $C_3$ with solvable quotient $C_2$, i.e., $S_3/C_3\cong C_2$. Hence $S_3$ is solvable, too. It cannot be nilpotent, because $Z(S_3)=1$, whereas nilpotent groups have non-trivial center (this was already mentioned).

Answer 3

To show $S_3$ is solvable, note that $G$ is solvable iff $N$ and $G/N$ are solvable for any $N\trianglelefteq G$. Let $N=<(1,2,3)>$. Note that $N$ and $G/N$ are cyclic hence solvable, so, $G$ is solvable. To show $S_3$ is not nilpotent. Note that for a finite group; a group is nilpotent iff it is isomorphic to the direct product of its Sylow subgroups. Observe that the Sylow subgroups of $S_3$ are all cyclic (what are they?). So, their direct product must be abelian, but, $S_3$ is nonabelian!