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Question → If (G,*) is a group of order 960 and there exist a in G such that a^m=e for some integer m<=960 where e is identity element of G then total number of possible value of m is___________

 

Answer==28
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${if\ the\ element\ of\ a\ group\ G\ is\ order\ n,then\ a}^{m}=e\ if\ and\ only\ if\ n\ is\ divisor\ of\ m$

this is a property of subgroup

so in ques have to find number of subgroups

$m\leqslant 960$

factors of m =  $2^{6}*3^{1}*5^{1}$

possible value of m = number of divisors of m

$=(6+1)*(1+1)*(1+1)$

$=28$

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Correction number of divisors should be $(6+1)(1+1)(1+1) $

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