Question

Prove that ${\mathit{A}}_{\mathbf{5}}$has no subgroup of order 30.

Short answer

It is proved that,$A_5$ has no subgroup of order 30.

Step-by-step solution

Obtain A5 has no subgroup of order 30

Consider that any subgroup of index 2 is normal.

Now, order of $A_5$is:

$\begin{array}{c}\frac{5!}{2}=\frac{120}{2}\\ =60\end{array}$

Obtain result

Therefore, if there is a subgroup $K$of $A_5$of order is 30 then, it has index 2.

By Lagranges theorem, the index of subgroup of $H$of finite group $G$is, $\frac{\left|G\right|}{\left|H\right|}$, and $\left|A_5\right|=60$and $\left|K\right|=30$.

Hence,$K$ is normal subgroup of $A_5$of order 30 and this is not possible as$A_5$ is simple group.

Thus, it is proved that has$A_5$ no subgroup of order 30.