Find the order of 22 and 55 module 101101. Futhermore, find all the elements of order 2020 in (Z∖101Z)×(\mathbb{Z}\backslash 101\mathbb{Z})^\times.

Since ϕ(101)=100=2252\phi(101)=100=2^25^2 then 22 and 55 only can have order 2,5,22,2⋅52,5,2^2,2\cdot 5 or ,22⋅5,2^2\cdot 5. After evaluating every power of 22 and 55 I got that 22 and 55 are primitive roots module 101101. Is there any shorter path to find the answers?

There exists ϕ(20)=8\phi(20)=8 elements of order 2020. I know that 2i2^i is a primitive root if and only if (i,100)=1(i,100)=1, so (2i)5(2^i)^5 and (5i)(5^i) are elements of order 20. Is there any shorter path to find all these elements?

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Do you mean (Z/101Z)(\mathbb{Z} / 101 \mathbb{Z})*? Cause otherwise, 2∗51mod101=12 * 51 \mod 101 = 1 and 2 is not a generator

– Nikolas Wojtalewicz

Oct 20 at 19:31

@NikolasWojtalewicz every nonzero element of (Z/101Z)+(\Bbb Z/101\Bbb Z)^+ is a generator, including 22.

– Greg Martin

Oct 20 at 20:02

1

@YotasTrejos 2 is indeed a primitive root modulo 101. However, note that this contradicts your first sentence after the statement of the problem; so you might want to check your thinking on that sentence. Also, 5 is not a primitive root modulo 101 (for one thing, it’s a quadratic residue); in fact, its order is 25.

– Greg Martin

Oct 20 at 20:05

Why is there a contradiction?

– YTS

Oct 20 at 20:21

@GregMartin Right, since 101 is prime, and the order of an element must divide the order of the group, every non-zero element is a generator. Whoops 😀

– Nikolas Wojtalewicz

2 days ago

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1 Answer

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Since 22 is a generator, 22 has order 100100 and so 252^5 has order 2020.

Therefore, the elements of order 2020 are (25)k(2^5)^k where gcd(20,k)=1\gcd(20,k)=1.