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July 23, 2020 1:54 pmpts 30 pts
IEXAMPLE 4 We determine the number of elements of order 5 in Z2s Z By Theorem 8.1, we may count the number of elements (a, b) in Z,s7, with the property that 5 = I(a, b)l = lcm(lal, Ib). Clearly this requires that either lal = 5 and lbl = I or 5, or Ibl = 5 and lal= 1 or 5. We consider two mutually exclusive cases. Case 1 lal = 5 and lbl = I or 5. Here there are four choices for a (namely, 5, 10, 15, and 20) and five choices for b. This gives 20 ele- ments of order 5. Order of an element m in Za is n/(n.m) Case 2 lal =I and lbl 5. This time there is one choice for a and four choices for b, so we obtain four more elements of order 5. Case 1: An element a in Zas Will have order 5 if satisties the Thus, Z,, 2, has 24 elements of order S. equation: 5-25/(25,a). So possible values O a are 5.1015. 20. Case ' An element b in Zg will have order itf satisties the equation: 1-5/(5.5). So possible value of b is only 1 i.e. 5. Similarly. An element b in Zs will have order 5 if satisfies the equation: 5-5/(5.b). So possible value of b are 1.2.3.4. Therefore. there are five Lchoicesfor.b -
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  • Deepak singh thankyou
    order of a is 1 or 5 , in first case order of a is 5 and in second case order of a is 1
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  • Piyush
    No, O(a)=5, first case me consider kar liya hai. second case me keval 1 order hai.
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  • Deepak singh
    see attached
    • cropped-1922389788.jpg
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