Every element of order 2 in Alt(n) is the square of an element of order 4 in Sym(n)

Prove that every element of order 2 in A_n is the square of an element of order 4 in S_n.


We know that every element of order 2 in S_n (hence also A_n) is a product of commuting 2-cycles by a previous theorem. Write \sigma \in A_n of order 2 as \sigma = (a_{1,1}\ a_{1,2})(b_{1,1}\ b_{1,2})(a_{2,1}\ a_{2,2})(b_{2,1}\ b_{2,2}) \cdots (a_{2k,1}\ a_{2k,2})(b_{2k,1}\ b_{2k,2}). Note that the number of 2-cycles in the decomposition of \sigma is even since \sigma \in A_n.

Note that (a_{i,1}\ b_{i,1}\ a_{i,2}\ b_{i,2})^2 = (a_{i,1}\ a_{i,2})(b_{i,1}\ b_{i,2}). Thus \sigma = (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})^2 (a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2})^2 \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2})^2 = \left( (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2}) \right)^2 using a previous theorem, and the element (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2}) has order 4 in S_n by §1.3 #15.

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