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I'm struggling with this. I know it is going to use Lagrange's Theorem. This is what I have so far.

Suppose $|U_n| = k.$ This implies $a^k = 1$ for all $a$ in $U_n$ and $|a|$ divides $k$.

Now, what can be said about the order of an element $a$? I haven't been able to conclude that it's even with what I have, therefore I continued so: $|\langle a\rangle|$ divides $k$ where $\langle a\rangle$ is the cyclic subgroup generated by $a.$

However, I haven't been able to conclude that $|\langle a\rangle|$ is even.

So I tried using this: an element $a$ is in $U_n$ iff gcd$(a, n) = 1$.

Any hints?

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Hint: Use a pairing argument. If $x$ is a unit, then so is $-x$. And if $n\gt 2$ they are distinct.

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  • $\begingroup$ Where is the $-x$ in the group, $U_4 = \{1, 3\}$? I'm trying to understand this in a small example first. $\endgroup$ – Dude Apr 16 '14 at 12:39
  • $\begingroup$ Sorry, typo. edited to $U_4$. $\endgroup$ – Dude Apr 16 '14 at 12:42
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    $\begingroup$ I think you mean $U_4$. If $x=1$, then $-x=3$, if $x=3$, then $-x=1$. The objects $1$ and $3$ are a couple. In today's business-oriented language, we can call them partners. $\endgroup$ – André Nicolas Apr 16 '14 at 12:42
  • $\begingroup$ The example you chose to ask about is perhaps too small to see what happens in general. Look at $U_{10}$ or $U_{11}$. $\endgroup$ – André Nicolas Apr 16 '14 at 12:45
  • $\begingroup$ Perhaps it's easier if written $x \mapsto n-x$. $\endgroup$ – lhf Apr 16 '14 at 13:01
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Yet another hint: Show that $\{+1,-1\}$ is a subgroup of $U_n$ of order $2$. Now apply Lagrange.

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Different hint: $|U_n|=\varphi(n)$, the Euler totient function. So you have to prove that for $n \gt 2$ this number is even.

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