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# LS11-3 - S UMMARY OF 11/3 L ECTURE We reviewed a couple of...

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S UMMARY OF 11/3 L ECTURE We reviewed a couple of notions. Most importantly, recall that the order of an element g of a group G is defined to be the order of the cyclic subgroup generated by that element: | g | = |h g i| . Another way to think about | g | is that it is the smallest positive integer n such that g n = 1 (i.e. g 6 = 1 , g 2 6 = 1 , g 3 6 = 1 , . . . , g n - 1 6 = 1 , but g n = 1 ). It’s a good exercise to figure out why the order satisfies this property. (Also, note that g, g 2 , . . . , g n are all distinct!) Next, we discussed direct products. Given two sets A and B we define a new set A × B , called the direct product of A and B , to be the collection of all ordered pairs of elements, the first from A , the second from B . In mathish, A × B = { ( a, b ) : a A, b B } . For example, R × Z is the set of all ordered pairs of the form ( x, n ) where x is any real number and n is any integer. Thus, (3 . 14159 , - 7) R × Z , but (3 , 2 . 1) 6∈ R × Z . We then concluded class by going over Cauchy’s theorem once more, this time completing the proof. Since the bulk of the proof is in the previous lecture summary, I’ll pick up where that

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