Proving A4 Has No Subgroup of Order 6

[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)

 

[fresh_42]

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.

By the first Sylow theorem, there is a subgroup of $H$ of order $3$ since $|H|=6=3 \cdot 2$, i.e. a cyclic group $\langle g \,|\,g^3=1 \rangle$ which makes $g$ 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).

... since you we the consecutive application of permutations from left to right.

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.

This contradicts the third Sylow theorem and is wrong. E.g. $S_3=\langle (1), (123), (321),(12),(13),(23) \rangle$ has three elements of order $2$ and three is odd, and $|S_3|=6$.

Maybe you meant: ... of order $3$. Every element $g$ of order $3$ pairs with a different element $g^{-1}$ also of order $3$, and no pair contains an element of another pair.

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)

Again order $3$ I suppose.