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I'm trying to show that any group of order 60 in Sym(5) contains all 3-cycles and all 5-cycles. I tried to do this by showing that it must contain all elements of Sym(5) of the form s^2 for some s in Sym(5). (So then any 3-cycle/5-cycle is a square)

So, I said let A be such a subgroup. It has index 2 and is normal in Sym(5). Taking s in Sym(5), we have (sA)^2 = (s^2)(A^2) = (s^2)(A)... Then I get stuck. I read that is has something to do with the quotient group being cyclic of order 2, but I don't see how this helps...

Also, out of curiosity, could whatever argument is used for this be used to show that any subgroup of index 2 must contain all "squares" in the group?

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  • $\begingroup$ Yes, any subgroup of index $2$ will contain all squares. Recall that since the quotient has order $2$, we know that $g^2 = 1$ for any $g$ in the quotient. Now see what this says about a representative for $g^2$ in the original group. $\endgroup$ – Tobias Kildetoft Dec 2 '13 at 19:19
  • $\begingroup$ I'm just trying to relate this to Z/2Z under addition. So we have that 0^2 is 0 and 1^2 is 0. Going back to the original group, then any even number squared is still even, but any odd number squares to an even number. So, is the line of thinking to say that the second case (i.e. elements not in A), when squared, will be put into A as, in the quotient, the square changes the point it's in? Sorry for the basic terminology... Just trying to get my head around it first. $\endgroup$ – user60126 Dec 2 '13 at 19:31
  • $\begingroup$ Thinking about this, if you quotient something out and an element in that group is turned to the identity, then it means it must be a part of what you quotient out... That seems kind of simple... $\endgroup$ – user60126 Dec 4 '13 at 17:13

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