Exercise 3.5.10 ($A_4$ is solvable)

Question

Find a composition series for $A_4$. Deduce that $A_4$ is solvable.

Richard Ganaye · Accepted Answer

\begin{proof} 
By Exercise 9, $H = \langle  (1 \ 2)(3\ 4),  (1 \ 3)(2 \ 4) \rangle \unlhd A_4$. Moreover $\langle  (1 \ 2)(3\ 4) \rangle \unlhd \langle  (1 \ 2)(3\ 4),  (1 \ 3)(2 \ 4) \rangle$, since the index of the subgroup $\langle  (1 \ 2)(3\ 4) \rangle $ in $H$ is $2$. Therefore
\begin{align}
\{()\} = H_0  \unlhd \langle  (1 \ 2)(3\ 4) \rangle =  H_1\unlhd \langle  (1 \ 2)(3\ 4),  (1 \ 3)(2 \ 4) \rangle = H_2 \unlhd A_4 = H_3.
\end{align}
Since $|H_1 : H_0| = 2$,  $|H_2 : H_1| = 2$ and  $|H_3 : H_2| = 3$ are prime numbers, the factors are cyclic, isomorphic to $Z_2, Z_2$ and $Z_3$. Therefore (1) is a composition series, and $A_4$ is solvable.

\end{proof}

Exercise 3.5.10 ($A_4$ is solvable)

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Exercise 3.5.10 ($A_4$ is solvable)

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