What is the number of generators of \(\mathbb{Z}_n^*\)?
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What is the number of generators of \(\mathbb{Z}_n^*\)?
1. verify that \(\mathbb{Z}_n^*\)is cyclic (iff n = 2, 4, \(p^e\), \(2p^e\), with \(e \ge 1\) and \(p\) is an odd prime)
2. if \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by lemma)
3. the number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of coprime elements of the group
2. if \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by lemma)
3. the number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of coprime elements of the group
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | What is the number of generators of \(\mathbb{Z}_n^*\)? | |
| Back | 1. verify that \(\mathbb{Z}_n^*\)is cyclic (iff n = 2, 4, \(p^e\), \(2p^e\), with \(e \ge 1\) and \(p\) is an odd prime)<br>2. if \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by lemma) <br>3. the number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of coprime elements of the group |
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What is the number of generators of \(\mathbb{Z}_{25}^* \)?
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What is the number of generators of \(\mathbb{Z}_{25}^* \)?
it is \(\varphi(\varphi(25)) = |\mathbb{Z}_{\varphi(25)}^*| = |\mathbb{Z}_{20}^*| = 8\) ( 1, 3, 7, 9, 11, 13, 17, 19 )
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | What is the number of generators of \(\mathbb{Z}_{25}^* \)? | |
| Back | it is \(\varphi(\varphi(25)) = |\mathbb{Z}_{\varphi(25)}^*| = |\mathbb{Z}_{20}^*| = 8\) ( 1, 3, 7, 9, 11, 13, 17, 19 ) |
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What is the number of subgroups of \(\mathbb{Z}_n\)?
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What is the number of subgroups of \(\mathbb{Z}_n\)?
it is the number of divisors of \(n\)
if \(n\) is written \(n = p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k}\) then it is \(\prod_{i=1}^k (e_i+1)\)
if \(n\) is written \(n = p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k}\) then it is \(\prod_{i=1}^k (e_i+1)\)
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | What is the number of subgroups of \(\mathbb{Z}_n\)? | |
| Back | it is the number of divisors of \(n\)<br>if \(n\) is written \(n = p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k}\) then it is \(\prod_{i=1}^k (e_i+1)\) |