The alternating group on five elements is simple: Simpler proof

simplicity_of_alternating_group_five_simpler_proof.json

https://arbital.com/p/simplicity_of_alternating_group_five_simpler_proof

by Patrick Stevens Jun 17 2016 updated Jun 17 2016

A proof which avoids some of the heavy machinery of the main proof.

If there is a non-trivial Normal subgroup $~$H$~$ of the Alternating group $~$A_5$~$ on five elements, then it is a union of conjugacy classes. Additionally, by Lagrange's theorem, the order of a subgroup must divide the order of the group, so the total size of $~$H$~$ must divide $~$60$~$.

We can list the [alternatinggroupfiveconjugacyclasses conjugacy classes of $~$A_5$~$]; they are of size $~$1, 20, 15, 12, 12$~$ respectively.

By a brute-force check, no sum of these containing $~$1$~$ can possibly divide $~$60$~$ (which is the size of $~$A_5$~$) unless it is $~$1$~$ or $~$60$~$.

The brute-force check

We first list the [number_theory_divisor divisors] of $~$60$~$: they are $$~$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$~$$

Since the subgroup $~$H$~$ must contain the identity, it must contain the conjugacy class of size $~$1$~$. If it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $~$13$~$ (since the smallest other class is of size $~$12$~$); so it is allowed to be one of $~$15$~$, $~$20$~$, $~$30$~$, or $~$60$~$. Since $~$H$~$ is also assumed to be a proper subgroup, it cannot be $~$A_5$~$ itself, so in fact $~$60$~$ is also banned.

The class of size $~$20$~$

If $~$H$~$ then contains the conjugacy class of size $~$20$~$, then $~$H$~$ can only be of size $~$30$~$ because we have already included $~$21$~$ elements. But there is no way to add just $~$9$~$ elements using conjugacy classes of size bigger than or equal to $~$12$~$.

So $~$H$~$ cannot contain the class of size $~$20$~$.

The class of size $~$15$~$

In this case, $~$H$~$ is allowed to be of size $~$20$~$ or $~$30$~$, and we have already found $~$16$~$ elements of it. So there are either $~$4$~$ or $~$14$~$ elements left to find; but we are only allowed to add classes of size exactly $~$12$~$, so this can't be done either.

The classes of size $~$12$~$

What remains is two classes of size $~$12$~$, from which we can make $~$1+12 = 13$~$ or $~$1+12+12 = 25$~$. Neither of these divides $~$60$~$, so these are not legal options either.

This exhausts the search, and completes the proof.