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1)Suppose that G is cyclic group of order 10 with generator a €G .Order of a^8 is...

2)(G,*) is an abelian group which of the following must hold 

A)for all g€G.     g-1   =    g

B)for all g€G.     g^2 =e

C)for all a,b €G   ( a*b)^2 =a^2* b^2

2 Answers

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Solution to part 1) : 

Here we are given that the order of the group is 10 and a is the generator..

Hence  a10  =  e  where e is the identity element but we dont know about a8..Hence we have to write it in terms of a10 as that is a known result.

Let we have   y  =  a40   which can be written as :    a40  =   (a10)4   =  e4   =   e    ...(1) 

                                                            Also ,        a40   =  (a8)5     =  e     [ Follows from (1) ]

Hence it means that  a8 is repeated 5 times in order to get the identity element e which is least number of times to do so.

Also from the corollary of the Lagranges' Theorem , 

Order of an element  divides the order of the order of the group (The actual theorem is order of a subgroup divides the order of the group)

Here 5 divides the order of group i,e, 10..

Hence order of  a8   =   5 

Solution to part 2) :

For this we have to keep in mind this theorem :

                For all elements of the group G  'a' s.t.  a  = a-1  , then the group G is abelian.

So the converse is not true necessarily and the statement 1) is converse of the given theorem actually.

Also statement 2) follows from statement 1) only..Hence not necessarily true..

Coming to statement 3) , 

For abelian group , (a * b)2     =   (a * b) * (a * b)   =   (a * a) * (b * b)  [ As it is abelian group hence a * b = b * a ]

Hence statement 3) will be true here only..

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