Question

A group G is said to be metabelian if it has a subgroup N such that N is abelian, N is normal in G, and $\mathit{G}\mathbf{/}\mathit{N}$is abelian.

Show that${\mathit{S}}_{\mathbf{3}}$ is metabelian.

Short answer

It is proved that$S_3$ is metabelian.

Step-by-step solution

Step by Step Solution Step 1: Lagrange’s Theorem

If K is a subgroup of a finite group G, then the order of K divides the order of G. In particular, $\mathbf{\left|}\mathbf{G}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{K}\mathbf{\right|}\mathbf{[}\mathbf{G}\mathbf{:}\mathbf{K}\mathbf{]}$.

Proving that S3 is metabelian

We Know that, $S_3$is a symmetric group of degree 3. The group of even permutations $A_3$is the normal group of order 3. Since it has only 3 elements, it is abelian.

Consider the quotient group $S_3/A_3$.

From Lagrange’s theorem order of $S_3/A_3$as:

$\begin{array}{c}O(S_3/A_3)=\left|S_3\right|/\left|A_3\right|\\ =\frac{6}{2}\\ =2\end{array}$

Since the only group of order 2 is ${\mathbb{Z}}_2$, ${\mathbb{Z}}_2$is abelian.

Hence, it is proved that$S_3$ is metabelian.