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Assume the group $G = \mathbb{Z}^*_{29} = \{1,2,\dots, 28 \}$ with operation of multiplication modulo $29$.

Since $G$ is cyclic, for every $d \in \{1,2,4,7,14,28 \}$, there exists a $H \leq G$ with $|H| = d$.

  • $d=1$: $H_1 = \{1\}$
  • $d = 2$: $H_2 = \{1, 28\}$, since $28^2 \mod 29 = 1$
  • $d= 4$: $H_3 = \{1,12,17,28 \}$, since $12 \cdot 17 \mod 29 = 1$ and we have closure.

Is there a way to yield the subgroups of order $7$ and $14$, without having to check all the possible combinations to assert closure?

Paris
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    Note: $2$ is a primitive root mod $29$ so generates the cyclic multiplicative group – J. W. Tanner Nov 04 '20 at 17:17
  • Also, as a side comment, it's not common to write $28^2 \mod 29 = 1$: this looks more like computer science notation, and the typesetting is awkward on account of the space. Instead, $28^2 \equiv 1 \pmod{29}$: $28^2 \equiv 1 \pmod{29}$. – Théophile Nov 04 '20 at 17:19
  • Look for elements of order $7$ and $14$. For example $7^7\equiv1\pmod{29}$. – Darsen Nov 04 '20 at 17:22
  • @J.W.Tanner Could you give a reference for that? – Darsen Nov 04 '20 at 17:46
  • @Darsen: just note that $2^{14},2^4\not\equiv1\bmod29$ – J. W. Tanner Nov 04 '20 at 18:26
  • @Théophile There is absolutely nothing wrong (or uncommon) by writing $,28^2!\bmod 29 = 1,,$ so please don't confuse students by claiming otherwise. Your claim is akin to saying mathematicians never use reduced fractions - which is obviously absurd. Normal form reductions are ubiquitous in both CS and math. See here for more on $!\bmod $ the operator vs. congruence. – Bill Dubuque Nov 04 '20 at 18:33
  • @BillDubuque "Uncommon", "wrong", and "never" are very different words. I didn't say it was wrong, and I didn't say that mathematicians never use it. Your reaction to my comment seems unnecessarily pointed. In my experience, elementary number theory textbooks most often use the congruence form, and in your own answer that you linked to, you speak of the preference for this form in theoretical contexts. – Théophile Nov 04 '20 at 18:55
  • @Théophile I stand strongly behind my remark. If it seems pointed it is because far too often I see here misguided denigrations of mod the operator. I meant to write "uncommon (or wrong)". – Bill Dubuque Nov 04 '20 at 18:59

2 Answers2

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The following table demonstrates that $(\mathbb Z/29\mathbb Z)^\times$ is cyclic with generator (primitive root) $2$:

$\begin{array} 1&n&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\ &2^n\bmod29&2&4&8&16&3&6&12&24&19&9&18&7&14&28\\ \\ &n&15&16&17&18&19&20&21&22&23&24&25&26&27&28\\ &2^n \bmod29&27&25&21&13&26&23&17&5&10&20&11&22&15&1\end{array}.$

(Note that $2^{14}\equiv-1\bmod29$, so $2^{14+n}\equiv-(2^n)\bmod29$.)

Now we can take advantage of the isomorphism (same structure) between $(\mathbb Z/29\mathbb Z)^\times$ and $\mathbb Z/28\mathbb Z$

to conclude (without checking all the possible products to assert closure),

that the subgroup of order $7$ is $\{16,24,7,25,23,20,1\}$,

and the subgroup of order $14$ is $\{4,16,6,24,9,7,28,25,13, 23,5,20,22,1\}$.

J. W. Tanner
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It's well known that $\Bbb Z_p^*=(\Bbb Z_p)^×\cong\Bbb Z_{p-1}$ for $p$ prime.

Once you find a primitive root $\bmod{29}$, you can use it to get your subgroups.

By Lagrange's theorem, the order of $2$ divides $28$. You can rule out each of $1,2,4,7$ and $14$, and then you have that $2$ is primitive.

Now $2^4$ has order $7$, and $2^2$ has order $14$, by basic theory of cyclic groups.