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Find the order of $2$ and $5$ module $101$. Futhermore, find all the elements of order $20$ in $(\mathbb{Z}\backslash 101\mathbb{Z})^\times$.

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

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

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  • $\begingroup$ Do you mean $(\mathbb{Z} / 101 \mathbb{Z})$*? Cause otherwise, $2 * 51 \mod 101 = 1$ and 2 is not a generator $\endgroup$ – Nikolas Wojtalewicz Oct 20 '16 at 19:31
  • $\begingroup$ @NikolasWojtalewicz every nonzero element of $(\Bbb Z/101\Bbb Z)^+$ is a generator, including $2$. $\endgroup$ – Greg Martin Oct 20 '16 at 20:02
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    $\begingroup$ @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. $\endgroup$ – Greg Martin Oct 20 '16 at 20:05
  • $\begingroup$ Why is there a contradiction? $\endgroup$ – YTS Oct 20 '16 at 20:21
  • $\begingroup$ @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 :D $\endgroup$ – Nikolas Wojtalewicz Oct 21 '16 at 6:51
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Since $2$ is a generator, $2$ has order $100$ and so $2^5$ has order $20$.

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

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