Question

If G is a simple group that has a subgroup K of index n, prove that $\left|\mathit{G}\right|$ divides n! . [Hint : Let T be the set of distinct right cosets of K and consider the homomorphism $\mathit{\phi }\mathbf{}\mathbf{:}\mathbf{}\mathit{G}\mathbf{\to }\mathit{A}\left(T\right)$of Exercise 41 in Section 8.4. Show that $\mathit{\phi }$is injective and note that $\mathit{A}\left(T\right)\mathbf{\cong }{\mathit{S}}_{\mathbf{n}}$(Why?) .

Short answer

It is proved that,$\left|G\right|$divides n! .

Step-by-step solution

Referring to Exercise 41 of Section 8.5

Considering $\phi :G\to A\left(T\right)$ given by $\phi \left(a\right)=f_{a^{-1}}$.

And $f_{a^{-1}}:T\to T$, is given by $f_{a^{-1}}\left(kb\right)=kb{a}^{-1}$.

Proving that G divides n!  

According to the hint, consider T to be the set of a distinct right cosetsss of K.

Let $\phi :G\to A\left(T\right).$given by $\phi \left(a\right)=f_{a^{-1}}$where $f_{a^{-1}}:T\to T$is given by $f_{a^{-1}}\left(kb\right)=kb{a}^{-1}$.

As we know, G is a simple group.

So the kernel $\phi$must be trivial and injective.

Therefore, the image of $\phi$is a subgroup of $A\left(T\right)$ with the same ordering of elements as G.

This implies that, $\left|G\right|$divides $\left|A\left(T\right)\right|$.

Since the order of $\left|A\left(T\right)\right|$is n! , hence $\left|G\right|$divides n! .

Thus, $\left|G\right|$divides n! .