There is no subgroup of order 6 in A₄

Proof:

First recall that A₄ consists of the even permutations of S₄. This makes it a subgroup of S₄. Since:

|S₄ |=24 , then |A₄ |=12. Let’s recall the even permutations of A₄ . These are of the following form:

(1)-1 of these, (1,2,3)-8 of these, (12)(34)-3 of these. Note that (1,2,3) (and similar) are of order 3 and (12)(34) (and similar) are of order 2.

Now suppose H ≤ A₄ is a subgroup of order 6. Since there are only four non-3-cycles, there must be a 3-cycle in H. But then the inverse of that 3-cycle must be in H. Therefore H contains either 2 or 4 3-cycles.

OTOH, since the identity is in H and either 2 or 4 3-cycles, there must be at least one element of order 2 such as (12)(34) in H.

We now show the above leads to a contradiction.

Case I : There are exactly two 3-cycles in H and hence three elements of order 2. But the three elements of order 2, together with the identity, form a subgroup of order 4. But H can not have a subgroup of order 4. Therefore this case can not occur.

Case II: There are four 3-cycles (and one element of order 2 and the identity). Lets call the three cycles: θ, θ²,ω, ω². Now consider ωθ and ωθ². These are distinct and not equal to any of: (1),θ, θ²,ω, ω². But this is a contradiction as now we have at least 7 elements in H which is of order 6.