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Problem statement:

Let $\mathbb{Z}_{77}^{*}$ be a multiplicative group of integers modulo $p=77$. Check if $g=17$ generates the group. If it does not, state the order of the generated subgroup.

Solution attempt:

We note that $p=77$ is not prime. I read that good ol' Gauss (1986) proved that a multiplicative group modulo $p$ is cyclic iff $p$ is of the form $p=2$, $p=4$, $p=q^k$, or $p=2q^k$, where $q$ is an odd prime and $k \geq 1$. I do not know why, but with that said we can conclude that $\mathbb{Z}_{77}^{*}$ is not cyclic as $p=77$ cannot be written in any of those forms. As such, there exists no primitive root mod $p$ that generates the group.

So, what remains is to figure out the size of the subgroup that $g=17$ generates.

I have understood it that the order of $g$ is defined to be the smallest positive integer $k$ such that $g^k \equiv 1 \mod p$. We also recall that according to Lagrange's Theorem, $H$ is a proper subgroup of a finite group $G$ iff the order of $H$ divides the order of $G$.

The order of $\mathbb{Z}_{77}^{*}$ is $\phi(77) = 60$. Possible subgroups orders are therefore the divisors of $60$ which are $\{1,2,3,4,5,6,10,12,15,20,30,60\}$. We ignore $1$ and $60$ as they correspond to the standard subgroups $\mathbb{Z}_{77}^{*}$ and $\{e\}$.

Evaluating $17^{d} \mod 77$ for each divisor $d$ yields $d=30$ as the correct answer. Thus, we conclude that the order of $g$ is $30$.

Questions:

Note that I have a very shallow background in modular arithmetics and Group Theory. As such, feel free to include in your answers more details rather than less.

  1. Where does the definition of the "order of $g$" come from, and how can I think of it intuitively? I don't understand what it means for $g^{d} \mod p$ to evaluate to $1$ or something other than that.

  2. I found this math.stackexchange answer which suggests that one does NOT need to try all the divisors, but rather only a subset of them. Is this true, and if so, why?

KSI
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    Gauss 1986?$~~~$ – Anne Bauval Nov 25 '24 at 16:59
  • @AnneBauval The book "Disquisitiones Arithmeticae" published in 1986. – KSI Nov 25 '24 at 17:00
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    @KSI I think you are referring to a 1986 English translation, but the original text was published in 1801. – angryavian Nov 25 '24 at 17:03
  • @angryavian Yes. – KSI Nov 25 '24 at 17:03
  • Do you know the Chinese remainder theorem? It could make the calculations easier. Also, to answer question 2, yes, if $17^{20}\not\equiv1\bmod77$ then $17^d\not\equiv1\bmod77$ for all $d\mid20$, so proper divisors of $20$ need not be tested – J. W. Tanner Nov 25 '24 at 17:03
  • For your first question: you seem to already know that "order of a [finite] group" refers to its cardinality. The order of an element of a group is the order of the cyclic group that it generates. In the particular case of your multiplicative group modulo $p$, this is equivalent to "the smallest $k$ such that $g^k \mod p \equiv 1$." – angryavian Nov 25 '24 at 17:05
  • As for your 2nd question: your link adresses a different question, which was to prove that a given element $g$ IS a primitive root mod $p$ (in that case, the method was to prove that the order of $g$ is not less than $\varphi(p)$, i.e. no proper divisor $d$ of $\varphi(p)$ satisfies $g^d\equiv1\bmod p$. In your post, the question is to calculate the order of $g\bmod{77}$. – Anne Bauval Nov 25 '24 at 17:18
  • @angryavian Perhaps I am just overthinking it and it's actually straight forward. If such a $k$ exists, then it simply means that we've found a cycle and if we keep going then we can always reduce as we move forward, e.g. $g^{k+1} = g mod p$, $g^{k+2} = g^{2} mod p$ etc. If we can find such a $k$ then we have found a valid subgroup of order $k$. In the case where the form of $p$ is "proper" so to yield a cyclic group, if $k = \phi(p)$, then we've found a generator for the group. Have I understood things correctly? – KSI Nov 25 '24 at 17:28
  • @AnneBauval Yes, I understand. However, my understanding is that in both cases (that question and this), the stated fact (need not use ALL divisors) is useful for doing the work efficiently. In both situations we end up only needing to try a smaller subset of divisors instead of all divisors. I want to understand why we can reduce the set of divisors in the way that person did. Like I said, my background is very shallow. – KSI Nov 25 '24 at 17:28
  • @J.W.Tanner I know of it, but I need to powerwash the dust off it, so to say. Feel free to demonstrate how the CRT can be applied to make the calculations easier. – KSI Nov 25 '24 at 17:29
  • In the answer of your link, in order (pun not intended) to prove that the order is not less that $\varphi(p)$, they just had to prove that no maximal proper divisor $d$ of $\varphi(p)$ satisfies $g^d\equiv1\bmod p$. In your problem, in order to calculate the order, you'll have to test all divisors (starting from the lowests) until you reach a power equal to $1$. And your search can be simplified using the CRT: calculate separately the order $\bmod7$ (which divides $6$) and the order $\bmod{11}$ (which divides $10$). – Anne Bauval Nov 25 '24 at 17:36
  • @AnneBauval You are right! – KSI Nov 25 '24 at 17:39
  • As @AnneBauval said, find the order of $17$ (or $3$) mod $7$ and the order of $17$ (or $6$) mod $11$ and then take the least common multiple – J. W. Tanner Nov 25 '24 at 18:03
  • Same method as here in the linked dupe (and many other answers). – Bill Dubuque Nov 26 '24 at 00:51

2 Answers2

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By Chinese remainder theorem, we know that $\mathbf{Z}/(77) \cong \mathbf{Z}/(7) \times \mathbf{Z}/(11)$ and the element $17$ corresponds to the pair $(3,6)$ in the direct product. Also, the corresponding unit groups are isomorphic, i.e. $U(\mathbf{Z}/(77)) \cong U(\mathbf{Z}/(7)) \times U(\mathbf{Z}/(11))$.
Now note that the elements $3$ and $6$ are respectively generators of the unit groups $U(\mathbf{Z}/(7))$ and $U(\mathbf{Z}/(11))$, which are of orders respectively $6$ and $10$. Therefore, the order of the pair $(3,6)$ is equal to ${\rm LCM}(6,10) =30$. But the element $17 \in U(\mathbf{Z}/(77))$ corresponds to the pair $(3,6)$ in the isomorphic direct product and hence has order $30$.

Anuradha N.
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I would get the orders $\bmod7$ and $\bmod11$ separately, then since $77$ is squarefree the order $\bmod77$ must be the least common multiple of these.

Then $17\equiv3\bmod7$ which is clearly a primitive root, so that contributes $6$ to the overall order. We also have $17\equiv6\bmod11$, which again is a primitive root (not a quadratic residue and with $11=$ twice a prime plus $1$, the only nonzero higher-power residues are $\pm1$); therefore that factor is $10$.

The LCM of $6$ and $10$ is $30$, so that must be the order of $17$ in $(\mathbb{Z}/77\mathbb{Z})^*$.

Since this order is only $30$, you ask what element has order $60$ and generates the entire group. There is none. The numbers $6$ and $10$ are the largest possible orders in the prime moduli $7$ and $11$ respectively, and their least common multiple is only $30$; therefore we can't get order more than $30$. The group $(\mathbb{Z}/77\mathbb{Z})^*$ cannot be reproduced fully from a single generator because it isn't cyclic. More generally, for $n\ge2$ the group $(\mathbb{Z}/n\mathbb{Z})^*$ is cyclic only if $n$ is $2,4,$ or $1$ or $2$ times a power (including first powers) of some odd prime.

The maximal order $30$ found here is the Carmichael function of $77$. The order of any element in the multiplicative group divides the Carmichael function of the modulus; and in most cases, as here, the Carmichael function is less than the Euler totient function. So you have fewer orders to choose from than just divisors of the Euler totient function.

Oscar Lanzi
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