# 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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