Exercise 3.5.9 ($\langle (1 \ 2)(3\ 4), (1 \ 3)(2 \ 4) \rangle \unlhd A_4$)

Proof. By Exercise 8, the unique subgroup of order $4$ is

The table given in Exercise 8 shows that $H$ is the group of elements in $A_4$ of order $1$ or $2$, so

If $g\in G$ and $h\in H$, then $h^2=1$, therefore

so $\mathit{𝑔h}{g}^{-1}\in H$. This shows that $H⊴A_4$.

Moreover $H$ has no element of order $4$, so $H$ is not cyclic. Since every group of order $4$ is isomorphic to $Z_4$ of $V_4=Z_2\times Z_2$, we obtain

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