Find the order of element 3 in the group Z/8Z. I constructed the cayley table an

Question

Find the order of element 3 in the group Z/8Z.

I constructed the cayley table and located the element 3.
3^2=2
3^3=2*3=4
3^4=4*3=1
so the order of element 3 would be 4 because 3^4=1 which is the identity. Or so I thought, but 3 is not the correct answer.
I thought my cayley table was correct but that is the only thing I can think of that could be wrong.
I know there are different ways to construct your cayley table but any way you construct it should yield the same result as long as it is a correct table.
Correct?
Please help I can't figure out what I have done wrong or what the answer is to this problem
Thanks so much

Explanation / Answer

Normally the group denoted by Z/8Z is the cyclic group of order 8. The group operation is addition so the identity is 0 (not 1) since we need an additive identity. Thus 3^1 =3 3^2= 3+3=6 3^3=3+3+3=9=1(but 1 is not the identity in this group) Continuing in this manner we see the order of 3 is 8. It is not hard to verify that the order of any element coprime to 8 is also 8 as these element are all generators. Recall the order of any element in a group divides the order of the group, so the only possibilities for Z/8Z are 1,2,4,8 The elements of each order are as follows Order 1: 0 Order 2: 4 Order 4: 2, 6 Order 8: 1,3,5,7

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