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What elements of $\Bbb Z_{24}$ are order $2$? Order $3$? Order $4$? Order $6$?

Order $2: 12$

Order $3: 8,16$

Order $4: 6,18$

Order $6: 4$

I had listed element $20$ as order $6$, but erased it late at night. A week later, I can't for the life of me remember why I didn't include it.

Am I correct or incorrect (and why) that element $20$ is order $6$?

balddraz
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tau
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    Testing multiples of $20$ up to the sixth should be very fast, and probably take less time than composing this post. Another possibility is to observe that in this group $20=-4$, and to use the fact that a multiple of $4$ is divisible by $24$ if and only if the same multiple of $-4$ is divisible by $24$. – Jyrki Lahtonen Jul 31 '19 at 06:31
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    As a check $,20 = 8 + 12,$. The summands have orders $,3,2,$ hence $20$ has order $3\cdot 2\ \ $ – Bill Dubuque Jul 31 '19 at 15:27

3 Answers3

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You can brute force calculate: \begin{align*} 20 \cdot 1 \equiv 20 \mod 24 \\ 20 \cdot 2 \equiv 16 \mod 24 \\ 20 \cdot 3 \equiv 12 \mod 24 \\ 20 \cdot 4 \equiv 8 \mod 24 \\ 20 \cdot 5 \equiv 4 \mod 24 \\ 20 \cdot 6 \equiv 0 \mod 24 \\ \end{align*}

So the first (positive) multiple of $20$ that produces the group identity $0$ is $6$. Therefore that is the order of $20$.

For bigger calculations in the future, you can instead remember this formula:

The order of any element $k \in \Bbb Z_n$ of the cyclic group with $n \in \Bbb N$ elements is $$ \boxed{\text{ord}(k) = \frac{n}{\gcd(k, n)}} $$ In this case, $n = 24$ and $k = 20$ and the greatest common divisor of $20$ and $24$ is $\gcd(20, 24) = 4$. So $\text{ord}(k) = \frac{24}{4} = 6$ as expected.

balddraz
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Let the order of $20$ in $\Bbb{Z}_{24}$ be $k$. Then $$\underbrace{20+20+\dotsb+20}_{k}=20k \equiv 0 \pmod{24}.$$ This is same as saying $5k \equiv 0 \pmod{6}$. Since $\gcd(5,6)=1$, thus $k \equiv 0 \pmod{6}$, which gives $k=6$.

Anurag A
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Definition. Let $\bar x \in \mathbb Z_A$. Then $\langle \bar x \rangle = \{n \bar x: n \in \mathbb Z \}$ is the subgroup of $\mathbb Z_A$ generated by $\bar x$.

The following statements are true.

If $g \mid A$, then $\langle \bar g \rangle = \left\{\bar 0, \ \bar g, \ 2\bar g, \ 3\bar g, \ \dots , \ \left( \dfrac Ag - 1\right) \bar g \right\}$. Hence the order of $g$ is $\dfrac Ag$.

Let $\bar x \in \mathbb Z_A$ and let $g = \gcd\{x, A\}$. Then $\langle \bar x \rangle = \langle \bar g \rangle$. Hence the order of $x$ is $\dfrac{A}{\gcd\{x,A\}}.$

Proof. Let $y \in \langle \bar x \rangle$. Then $y = nx$ for some integer $n$. Since $g \mid x$, then $x = mg$ for some integer $m$. So $y = nx = (nm)g$. Hence $\bar y \in \langle \bar g \rangle$. So $\langle \bar x \rangle \subseteq \langle \bar g \rangle$.

Let $\bar y \in \langle \bar g \rangle$. Then $y = mg$ for some integer $m$. Since $g = \gcd\{x, A\}$, then there exists integers $u$ and $v$ such that $g = ux + vA$. S0 $y = mg = (mu)x + (mv)A$. It follows that $y \equiv (mu)x \pmod A$. Hence $y \in \langle \bar x \rangle$ and $\langle \bar g \rangle \subseteq \langle \bar x \rangle$. $\blacksquare$

The divisors of $24$ are $\{1, 2, 3, 4, 6, 8, 12, 24\}$.

\begin{array}{c|c} \operatorname{ord} \bar x & x \pmod{24}\\ \hline 1 & 0 \\ 2 & 12\\ 3 & 8, 16\\ 4 & 6, 18\\ 6 & 4, 20\\ 8 & 3, 9, 15, 21\\ 12 & 2, 10, 14, 22\\ 24 & 1, 5, 7, 11, 13, 17, 19, 23 \\ \hline \end{array}

$$\gcd(0,24)=24$$ $$ \operatorname{ord}(\bar 0) = \dfrac{24}{24}=1$$ $$ \langle \bar 0 \rangle = \{ \bar 0 \}$$ $$(\text{$1$ element})$$

$$ \gcd(12,24)=12$$ $$ \operatorname{ord}(\bar{12}) = \dfrac{24}{12}=2 $$ $$ \langle \bar{12} \rangle = \{ \bar 0, \bar{12} \}$$ $$(\text{$2$ elements})$$

$$\gcd(8,24)=\gcd(16,24)=8$$ $$\operatorname{ord}(\bar{8}) = \operatorname{ord}(\bar{16}) = \dfrac{24}{8}=3$$ $$\langle \bar{8} \rangle = \langle \bar{16} \rangle = \{ \bar 0, \bar{8}, \bar{16} \}$$ $$(\text{$3$ elements})$$

$$\gcd(6,24)=\gcd(18,24)=6$$ $$\operatorname{ord}(\bar{6}) = \operatorname{ord}(\bar{18}) = \dfrac{24}{6}=4$$ $$\langle \bar{6} \rangle = \langle \bar{18} \rangle = \{ \bar 0, \bar{6}, \bar{12}, \bar{18} \}$$ $$(\text{$4$ elements})$$

$$\gcd(4,24)=\gcd(20,24)=4$$ $$\operatorname{ord}(\bar{4}) = \operatorname{ord}(\bar{20}) = \dfrac{24}{4}=6$$ $$\langle \bar{4} \rangle = \langle \bar{20} \rangle = \{ \bar 0, \bar{4}, \bar{8}, \bar{12}, \bar{16}, \bar{20} \}$$ $$(\text{$6$ elements})$$

$$\gcd(3,24)=\gcd(9,24)=\gcd(15,24)=\gcd(21,24)=3$$ $$\operatorname{ord}(\bar{3}) = \operatorname{ord}(\bar{9}) = \operatorname{ord}(\bar{15}) = \operatorname{ord}(\bar{21}) = \dfrac{24}{3}=8$$ $$\langle \bar{3} \rangle = \langle \bar{9} \rangle = \langle \bar{15} \rangle = \langle \bar{21} \rangle = \{ \bar 0, \bar{3}, \bar{6}, \bar{9}, \bar{12}, \bar{15}, \bar{18}, \bar{21} \}$$ $$(\text{$8$ elements})$$

$$\gcd(2,24)=\gcd(10,24)=\gcd(14,24)=\gcd(22,24)=2$$ $$\operatorname{ord}(\bar{2}) = \operatorname{ord}(\bar{10}) = \operatorname{ord}(\bar{14}) = \operatorname{ord}(\bar{22}) = \dfrac{24}{2}=12$$ $$\langle \bar{2} \rangle = \langle \bar{10} \rangle = \langle \bar{14} \rangle = \langle \bar{22} \rangle = \{ \bar 0, \bar{2}, \bar{4}, \bar{6}, \bar{8}, \bar{10}, \bar{12}, \bar{14}, \bar{16}, \bar{18}, \bar{20}, \bar{22} \}$$ $$(\text{$12$ elements})$$

$$\gcd(1,24)=\gcd(5,24)=\gcd(7,24)= \gcd(11,24)=\gcd(13,24)=\gcd(17,24)=\gcd(19,24)=1$$ $$\operatorname{ord}(\bar{1}) = \operatorname{ord}(\bar{5}) = \operatorname{ord}(\bar{7}) = \operatorname{ord}(\bar{11}) = \operatorname{ord}(\bar{13}) = \operatorname{ord}(\bar{17}) = \operatorname{ord}(\bar{19}) = \operatorname{ord}(\bar{23}) = \dfrac{24}{1}=24$$ $$\langle \bar{1} \rangle = \langle \bar{5} \rangle = \langle \bar{7} \rangle = \langle \bar{11} \rangle = \langle \bar{13} \rangle = \langle \bar{17} \rangle = \langle \bar{19} \rangle = \langle \bar{23} \rangle =$$ $$\{ \bar 0, \bar 1 \bar{2}, \bar 3, \bar{4}, \bar 5, \bar{6}, \bar 7, \bar{8}, \bar 9, \bar{10}, \bar{11}, \bar{12}, \bar{13}, \bar{14}, \bar{15}, \bar{16}, \bar{17}, \bar{18}, \bar{19}, \bar{20}, \bar{21}, \bar{22}, \bar{23} \}$$ $$(\text{$24$ elements})$$

Steven Alexis Gregory
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