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I am trying to show that if $a$ is any integer relatively prime to $n$, then $a^{\phi(n)}\equiv 1\pmod n$, where $\phi(n)$ is Euler's totient function whose value is equal to the number of numbers less than $n$ that are relatively prime to $n$.

This seems number-theoretic, but given the context, is meant to be solved with group theory. I know that the order of $U(n)$ (the group of all numbers less than and relatively prime to $n$ under multiplication) is $\phi(n)$. Therefore, for any $g \in U(n)$ we know that $g^{\phi(n)}=1$. I have been trying to use this fact in my proof. Clearly $a$ need not be in $U(n)$, but I thought perhaps if it is congruent to a member of $U(n)$ I can get the desired result. For this reason, I applied the division algorithm to write $a=nm+r$ where $m$ is some integer and $1 \le r \le n-1$, and tried to show that $r$ is relatively prime to $n$, so that $a \equiv r \mod n$. I do not know that this is the best approach, but no others have borne fruit either.

I'd really appreciate a HINT, as always, on how to prove this. Thanks.

Alex Petzke
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  • Hint: $\gcd(a,n) = gcd(a + k\cdot n, n)$ for all $a,,k,,n$. – Daniel Fischer Jul 23 '13 at 12:11
  • If $d$ is a common divisor of $r$ and $n$ then $d$ is a divisor of $nm+r=a$, so that $n$ and $a$ cannot be relatively prime. In conclusion, it is true that $r$ and $n$ are relatively prime. – Quimey Jul 23 '13 at 12:13
  • If $d$ is a divisor of both $n$ and $r$, you should be able to show from your equation that $d$ is a divisor of $a$ too. But $a$ and $n$ are coprime, so $d$ must be $1$. – Mark Bennet Jul 23 '13 at 12:13
  • $U(n)$ should not be thought of as the set of integers smaller than $n$ and coprime to $n$, but rather as the group of units in $\mathbb{Z}/n\mathbb{Z}$. – Tobias Kildetoft Jul 23 '13 at 12:14

2 Answers2

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The order of any element of a finite group divides the order of the group itself. This follows directly from Lagrange's theorem and considering the cyclic subgroup generated by the element. Therefore, if $a$ is any element of a finite group $G$, then $a^{|G|} = 1$. If you consider the group of integers that are relatively prime to $n$ under multiplication, then this group has $\phi(n)$ elements. Therefore we know that $a^{\phi(n)}\equiv 1\pmod n$ when $a$ and $n$ are relatively prime.

Gino P
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  • I wish I knew a proof of this that does not depend on Lagrange's theorem. See http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory. – lhf Jul 26 '13 at 11:26
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The easiest proof that does not rely on group theory is to consider the map $x \mapsto ax$ on $U(n)$. Since $a$ and $n$ are relatively prime, this map is injective and hence bijective. Therefore $x_1 \cdots x_{\phi(n)}$ is fixed by the map. Now, rearrange and cancel.

lhf
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