What's the order of $2$ in $U_{143}$

Question

Kindly read my comments, proving that $2^{60}=1 \pmod{143}$ doesn't prove that it's the order of 2 as we need to prove that 60 is the smallest number to achieve this result...

I know that in $U_{11}$, $o(2)=10$ and that in $U_{13}$, $o(2)=12$

I need to find $o(2)$ in $U_{143}$. How can I find that?

I think this has something to do with the fact that both $11$ and $13$ are prime numbers and that $143 = 11\cdot 13$

I tried to calculate $\mathrm{lcm}(10,12)$ which is $60$. I chose those 2 numbers in specific because they are orders of 2 as mentioned above.

Is there any mathematical law which is relative to the facts mentioned above?

I think your $U_n$ is the multiplicative group of residue classes modulo $n$ which are relatively prime to $n$, correct? — Dustan Levenstein, Nov 30 '22 at 21:28
if $2^{10}\equiv1\bmod11$ and $2^{12}\equiv1\bmod13$, then $2^{60}\equiv1\bmod11$ and $13$, so $2^{60}\equiv1\bmod143$ by the C.R.T. — J. W. Tanner, Nov 30 '22 at 22:26
@J.W.Tanner this doesn't prove it's the smallest number - which is needed for order (referring to 60) — zoro, Nov 30 '22 at 22:30
if $2^L\equiv1\bmod143$, then by the C.R.T. $2^L\equiv1\bmod11$ and $13$, so $10$ and $12$ divide $L$, so $L$ is a multiple of $10$ and $12$, and the LCM is the smallest positive multiple of $10$ and $12$ — J. W. Tanner, Nov 30 '22 at 22:59
@J.W.Tanner thanks, but according to wikipedia I don't see that C.R.T says or concludes what you have written — zoro, Nov 30 '22 at 23:10
You don't need the Chinese Remainder Theorem to see that if $143$ divides $2^L-1$ then $11$ divides $2^L-1$. — Gerry Myerson, Nov 30 '22 at 23:41

Chris · Answer 1 · 2022-12-01T00:58:13.793

Even though this may not answers precisely the question of the OP, it could be useful to be posted.

Let's make clear that $U_{143}$ is the group of units of the commutative ring $\Bbb Z_{143}$. In addition, $$U_{143}=\{a\in \Bbb Z_{143}:\gcd(a,143)=1,\ 1\leq a \leq 142\}.$$ As $\gcd(2,143)=1$, it is clear that $2\in U_{143}$. The order of this group is $|U_{143}|=\varphi(143)=120$. But using a result which yields from Lagrange's Theorem, $\mathrm{ord}(2)\mid |U_{143}|$ and therefore $$\mathrm{ord}(2)\in \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}.$$ So, it suffices to calculate the powers $$2^{a}\bmod{143}$$ where $a=1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60$ or $120$. The smallest such $a$ which gives us $1\bmod{143}$ is the order of $2$ modulo $143$.

Hope that helps.

calc ll · Answer 2 · 2022-12-01T01:29:45.067

0

Yes. Pretty easily $\pmod {143}$, we have $2^{20}\ \not\equiv 1, 2^{30}\not\equiv 1$. Thus the conclusion, since no divisors of $20,30$ (hence $60$) can be the order. (Just check $\pmod {11}$ and $\pmod {13}$.)

(For instance, since you computed that the order of $2$ in $U_{13}$ is $12$, we have it, since neither $20$ nor $30$ is a multiple of $12$.)

edited Dec 01 '22 at 01:29

answered Dec 01 '22 at 00:41

calc ll

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What's the order of $2$ in $U_{143}$

2 Answers2