# 1) Let a be an element of order 8 in a group G. Find the order of each of the following a^2, a^3, a^4, a^5, a^6, a^7, a^8.

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### 1 Answer

The order of a is the smallest power n such that `a^n=1`

So if the order of a is 8, then `a^8=1` and for every k<8, `a^k != 1`

(where "1" is the identity element of the group)

The order of `a^2` :

`a^2 !=1`

`(a^2)^2=a^4 !=1`

`(a^2)^3=a^6 !=1`

`(a^2)^4=a^8 =1`

Thus the order of `a^2` is 4, because 4 is the smallest power we can raise `a^2` to and get 1

The order of `a^4` :

`a^4 !=1`

`(a^4)^2=a^8 =1`

Thus the order of `a^4` is 2, because 2 is the smallest power we can raise `a^4` to and get 1

The order of `a^6` :

`a^6 !=1`

`(a^6)^2=a^(12) = a^4*a^8 = a^4*1 !=1`

`(a^6)^3=a^(18) = a^2*(a^8)^2 = a^2*1^2 !=1`

`(a^6)^4=a^(24) = (a^8)^3 =1^3 =1`

Thus the order of `a^6` is 4, because 4 is the smallest power we can raise `a^6` to and get 1

The order of `a^8` :

`a^8 =1`

That means `a^8` is already the identity, and the identity always has order 1.

The order of `a^3`, `a^5`, and `a^7` is 8:

This is because 3,5, and 7 are all relatively prime to 8. (That is, the greatest common divisor of 3 and 8 is 1; the gcd of 5 and 8 is 1; the gcd of 7 and 8 is 1)

To see (intuitively) why this is true, try writing out the powers of `a^3`:

`(a^3)^1 = a^3 != 1`

`(a^3)^2 = a^6 != 1`

`(a^3)^3 = a^9 = a*a^8 = a*1 != 1`

`(a^3)^4 = a^(12) = a^4*a^8 = a^4*1 != 1`

`(a^3)^5 = a^(15) = a^7*a^8 = a^7*1 != 1`

`(a^3)^6 = a^(18) = a^2*(a^8)^2 = a^2*1^2 != 1`

`(a^3)^7 = a^(21) = a^5*(a^8)^2 = a^5*1^2 != 1`

`(a^3)^8 = a^(24) = (a^8)^3 = 1^3 = 1`

So the smallest power you can raise `a^3` to and get 1 is 8. And so the order of `a^3` is 8

Basically you need to raise a to a multiple of 8 to get 1. The smallest multiple of 8 that is also a multiple of 3 is 24, so we need:

`(a^3)^8 = a^(24)`, and the order of `a^3` is 8. This situation will happen any time you are working with a power of a that is relatively prime to the order of a.