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This is a question from Artin's algebra textbook.

The tetrahedral group of rotations has 1 element of order 1, 8 elements of order 3 (rotations of $120^°$ around a vertex), and 3 elements of order 2 (rotations of $180^o$ around the axis through the midpoints of opposite edges).

Is there a way to find the class equation without using brute force on $A_4$?

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It has 12 elements. Any force you use won't be brutish.

The conjugacy classes in $S_n$ are easy to describe. And you can prove that each conjugacy class in $S_n$ which consists of even permutations (i.e., which is contained in $A_n$) will either stay a conjugacy class in $A_n$ or split in half (see here). For $A_4$, there's only one conjugacy class (of size 8) that even has the potential to split in half. You can just check if that happens.

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    $\begingroup$ Your wording is confusing, because $S_n$ has even and odd permutations $\endgroup$ – J. W. Tanner Feb 11 at 21:25
  • $\begingroup$ Thanks -- I guess the parentheses don't do what I wanted them to do. I mean the ones which consist of even permutations, i.e., the only ones which are actually in $A_n$. I'll make an edit. $\endgroup$ – csprun Feb 11 at 22:23
  • $\begingroup$ Yes, your phrase "which consists of even permutations" is restrictive, not parenthetical. The wording is better now. Thank you $\endgroup$ – J. W. Tanner Feb 11 at 23:03
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The class equation of $A_4$ is well known to be $12=1+4+4+3$, since the cycle structure is preserved by conjugation. In this case, the class of $3$-cycles of size $8$ splits.

The subgroup of the tetrahedral group you refer to is also well known to be $A_4$.

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