Subgroups of $\mathbb{Z}^*_{29}$

Question

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?

Answer 1

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 $\{4,16,6,24,9,7,28,25,13, 23,5,20,22,1\}$.

Answer 2

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.