Group theory doubt

Question

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

Comments

Q.2 answer is C

Answer 1

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..

Comments

nice explaination.. but can u plzeplain this a bit...

 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.

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how (a*b)(a*b) = (a*a)(b*b) ??

@habib sir

a = a^-1 exists for all element is to prove it is a group right??

for abelian group (a*b) =(b*a) na?

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for finding the order of a^8 simply put = 10/gcd(8,10) = 5 thats the shortest way to find it out.

Answer 2

5 should be ans

Comments

please discuss details.it will be helpful.