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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