Lemma: If H≤G has index 2, i.e. [G:H]=2, then for any a∈G we have a^2∈H.
The 12 elements of A4 are (1),(12)(34),(13)(24),(14)(23),(123),(13… (234), and (243).
The 12 elements of A4 are (1),(12)(34),(13)(24),(14)(23),(123),(13… (234), and (243).
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A4 is the set of even permutations, which can be represented as an even number of 2-cycles. They are
(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)
Suppose H is a subgroup of A4 of order 12.
Let's look at a^2 for a in A4
(1) gives (1)
(12)(34) gives (1)
(13)(24) gives (1)
(14)(23) gives (1)
(123) gives (132)
(132) gives (123)
(124) gives (142)
(142) gives (124)
(134) gives (143)
(143) gives (134)
(234) gives (243)
(243) gives (234)
If a is in G, a^2 is in G. If a is not in G, a^2 is still in G, by the lemma. Therefore all the squares in the above list have to be in G. The trouble is, there are 9 element in the list. Since 9 > 6, there is no subgroup of order 6.
A4 is the set of even permutations, which can be represented as an even number of 2-cycles. They are
(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)
Suppose H is a subgroup of A4 of order 12.
Let's look at a^2 for a in A4
(1) gives (1)
(12)(34) gives (1)
(13)(24) gives (1)
(14)(23) gives (1)
(123) gives (132)
(132) gives (123)
(124) gives (142)
(142) gives (124)
(134) gives (143)
(143) gives (134)
(234) gives (243)
(243) gives (234)
If a is in G, a^2 is in G. If a is not in G, a^2 is still in G, by the lemma. Therefore all the squares in the above list have to be in G. The trouble is, there are 9 element in the list. Since 9 > 6, there is no subgroup of order 6.