By Jurisic A.

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**Example text**

40 1. First we look for a primitive root in Z∗7 . The possible orders of a unit 3 = 1 are 2, 3, 6. The residue class 2 is not because of 2 = 1, but 3 is, 2 3 since both 3 = 9 = 1 and 3 = −1 = 1. 2. Let us now take 3 as a candidate for a primitive root in Z49 . We have 6 to check that 3 = 1 - indeed Z49 6 3 = 81 · 9 = −17 · 9 = −153 = −6 = 1. Now 3 ∈ Z∗343 is a primitive root as well. 17. Let R be a ring with finite cyclic group of units, |R∗ | = q, and a ∈ R∗ , ∈ N>0 . The following conditions are equivalent: 1.

2. The element 2 ∈ Z∗11 has order 10, its sucessive powers being 2, 4, 8, 5, 10 = −1, −2 = 9, −4 = 7, −8 = 3, −5 = 6, 1. Furthermore ord(4) = 5 and ord(10) = 2. 4. For an element a ∈ R∗ of order d := ord(a) we have: 1. , ad−1 are pairwise different, and 35 2. ak = 1 ⇐⇒ d|k. In particular d|q. Proof. We have ak = ar , if k = sd + r with 0 ≤ r < d. This gives the nontrivial inclusion in the first part of the statement as well as the second part: If ak = 1, we have as well ar = 1 and that is possible only for r = 0, since d = ord(a).

In order to see that there is such an element, take any d ∈ Zpn above b ∈ Zp . Its order is a multiple of p − 1 (since ord(b) = p − 1) and divides the order pn−1 (p − 1) of the group of units Zpn , hence of the form ps (p − 1). Now s s we choose c := d p . Indeed c lies above b p = b. 13 that a = c(1+pt) ∈ Z∗pn is a primitive root iff p does not divide t - and the same criterion applies to σ(a) = σ(c)(1 + pt). 13 since Z∗2n = 1 + 2Z2n = (1 + 4Z2n ) ∪ (−1 + 4Z2n ) and −1 + 4Z2n = −(1 + 4Z2n ). 16.