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suppose that G IS cycliic group of order 10 with generator "a" belongs to G , order of $a^{8}$ will be ?

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As a is the generator so order of a =n i.e o(a)=10 in our case.

We know a^10=e. we know that a^8 is not an identity.Let us say a^8=b.

Now we need to find order of b.

o(b) must divide order of group.It can be 1,2,5(+ve divisors of 10). Order 1 is not possible as b is not identity element.

2,5,10 are left options.

let o(b) as x.We need value of x in b^x=e.

put b as a^8.

Now we need (a^8)^x=e

.Try with x=2.it becomes a^16.

a^16= a^10*a^6. ,which means e*a^6.Hence this cannot give identity

Try x=5

(a^8)^x=e.

means a^40 which is e. Hence 5 is our answer.

so this mean b^5=e and 5 is answer .As 5 is satisfying so no need to check for 10 as we need smallest +ve  integer.

Please let me know if anything is wrong in the solution
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