Prove that A4 has no subgroup of order 6 in this way (and this way only):(adsbygoogle = window.adsbygoogle || []).push({});

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)

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# Permutation 3 (brace yourself)

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