Exercise 4.2

Proof. $\text{∙}$ $p=11$. Then $p-1=10=2\times 5$.

Since $2^2=4\not\equiv 1(\mathit{mod}11)$, and $2^5=32\equiv -1\not\equiv 1(\mathit{mod}11)$, $2$ is a primitive element modulo $11$.

The other primitive elements modulo $11$ are congruent to the powers $2^i,i\wedge 10=1,1\le i<10$, namely $2,2^3,2^7,2^9$.

$2^7\equiv 7(\mathit{mod}11),2^9\equiv 6(mod11)$, so

$\{\bar{2},\bar{8},\bar{7},\bar{6}\}$ is the set of the generators of $U(\mathbb{Z}∕11\mathbb{Z})$.

Similarly :

$\text{∙}$ $p=13$ : $\{2,6,11,7\}$ is the set of the generators of $U(\mathbb{Z}∕13\mathbb{Z})$.

$\text{∙}$ $p=17$ : $\{3,10,5,11,14,7,12,6\}$ is the set of the generators of $U(\mathbb{Z}∕17\mathbb{Z})$.

$\text{∙}$ $p=19$ : $\{2,13,14,15,3,10\}$ is the set of the generators of $U(\mathbb{Z}∕19\mathbb{Z})$.

I obtain these results with the direct orders in S.A.G.E. :

p = 19; Fp = GF(p); a = Fp.multiplicative_generator()
print([a^k for k in range(1,p) if gcd(k,p-1) == 1])

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