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How can you show that if $p$ is prime then the numbers with maximum possible order modulo $p^2$ is $\phi(\phi(p^2))$.

I tried finding order(a) modulo 9, and obtained the following: $1$ is $1$, $2$ is $6$,$3$ doesn't exist, $4$ is $3$, $5$ is $6$, $6$ doesn't exist, $7$ is $3$, and $8$ is $2$.

So maximum possible order is 6, and 2 and 5 have maximum possible order. Also $$\phi(\phi(3^2)) = \phi(6) = \phi(2)\phi(3) = 2$$ and so I verified that this is true. Now how do I go about proving the statement? Where do I even start?

Aaron Maroja
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Rohit
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  • I don't understand the statement. Especially "the numbers .... is ...." – mercio Apr 15 '15 at 11:12
  • @mercio It indeed is a little confusing, but I think it means: the number of elements in $;\Bbb Z_{p^2}=\Bbb Z/p^2\Bbb Z;$ which are generators of the group of units $;\Bbb Z^*_{p^2};$ is $;\varphi(\varphi(p^2));$ . – Timbuc Apr 15 '15 at 11:22
  • Yeah, sorry I'm an engineering student taking a math course on number theory and we don't talk about generators and groups and all that since that is deemed to be too complex for the course, so I have a hard time understanding the surface level of number theory without the group theory aspect of it. – Rohit Apr 16 '15 at 23:26

1 Answers1

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(1) The group of units $\;\Bbb Z_{p^2}\;$ is cyclic and it has order $\;\varphi(p^2)=p(p-1)\;$

(2) A cyclic group of order $\;n\;$ has exactly $\;\varphi(n)\;$ generators .

Now put together (1)+(2) and deduce what you want.

Timbuc
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