Question

Prove that $\left\{\left(1\right),\left(12\right),\left(34\right),\left(13\right),\left(24\right),\left(14\right),\left(23\right)\right\}$ is a subgroup of A₄.

Short answer

It is showed that $\left\{\left(1\right),\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$is a subgroup of A₄.

Step-by-step solution

Cayley’s Theorem

Theorem 7.21states that group would be isomorphicto a group of permutations.

Show that {1,1234,1324,1423} is a subgroup of A4

It is observed that every element in $\left\{\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$ contain order 2. Therefore, they become the inverse in A₄. As a result, $\left\{\left(1\right),\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$would be a subset of A₄, which have the identity and are closed under inverses.

As a result, demonstrate that it would be closed under compositions. The following equations are based on direct calculations:

$$\begin{gathered} \left(12\right)\left(34\right)\left(13\right)\left(24\right)=\left(14\right)\left(23\right)=\left(13\right)\left(24\right)\left(12\right)\left(34\right) \\ \left(12\right)\left(34\right)\left(14\right)\left(23\right)=\left(13\right)\left(24\right)=\left(14\right)\left(23\right)\left(12\right)\left(34\right) \\ \left(13\right)\left(24\right)\left(14\right)\left(23\right)=\left(12\right)\left(34\right)=\left(14\right)\left(23\right)\left(13\right)\left(24\right) \end{gathered}$$

As a result, $\left\{\left(1\right),\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$would be a subgroup of A₄.

Hence, it is shows that $\left\{\left(1\right),\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$ is a subgroup of A₄.