In Chapter 11 of Armstrong: Groups and Symmetry, Armstrong gives an argument (see bottom of page 59 and top of page 60 of Armstrong - attached)

The argument runs as follows: (see attached)

does not contain a subgroup of order 6. For suppose that H is a subgroup of which has six elements. If a 3-cycle belongs to H, its inverse must also belong to H, so the number of 3-cycles in H is even. There cannot be six as we need room for the identity element. Suppose there are four, say , , , . Then , , , , , , are all distinct and belong to H, contradicting our assumption that |H| = 6.

Finally, if only two 3-cycles lie in H, then H must contain the subgoup { , (12)(34), (13)(24), (14)(23)}. But 4 is not a factor of 6 so Lagrange's Theorem rules out this case too. We conclude that no subgroup of order 6 exists.

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Firstly, why does Armstrong focus on 3-cycles in his reasoning regarding a subgroup of order 6?

Secondly, I cannot follow the statement:

"Finally, if only two 3-cycles lie in H, then H must contain the subgoup { , (12)(34), (13)(24), (14)(23)}. "

Why does this if-then statement follow?

Can someone please help?

Peter