# Permutation 3 (brace yourself)

1. Dec 1, 2004

### SqrachMasda

Prove that A4 has no subgroup of order 6 in this way (and this way only):

Suppose that A4 has a subgroup H of order 6. Explain (in one sentence) why H must contain a 3-cycle. WLOG(without loss of generality) let this be (1,2,3).
Then H must have iota,(1,2,3) and (1,2,3)^-1=(3,2,1).
Now apply the result from my permutation 2 thread( Prove: a group of even order must have an even number of elements of order 2) to H.
As a result all that's left to do is take three cases, according to which element of order 2 in A4 belongs to H.
In each case compute products and take inverses in H to show H always has more than 6 elements (contradiction)