# The alternating group on five elements is simple: Simpler 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 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 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.

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