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I solved the following exercise:

Exercise: Show that $U(20) \neq \langle k \rangle$ for any $k \in U(20)$.

Here $\langle k \rangle $ is defined to be the set $\{k^i \mid i \in \mathbb Z \}$.

My answer: I calculated $\langle 3 \rangle = \{3,9,7,1\}$. Then I noted that for $x \in \langle 3 \rangle $ the set $\langle x \rangle $ is a subgroup of $\langle 3 \rangle $ hence it must be that $|x|\le 3$. Therefore none of the elements in $\langle 3 \rangle $ can generate $U(20)$.

Next I calculated $\langle 13 \rangle = \{13, 9, 17,1\}$ and by the same argument none of these elements can generate $U(20)$ either.

The last remaining elements in $U(20)$ are $19$ and $11$ but $\langle 19 \rangle =\{19, 1\}$ and $\langle 11\rangle = \{1,11\}$.

This concludes my proof. $\Box$

But the solution given to this question goes like this:

Use brute force to show that $k^4 = 1$ for all $k$.

So I was wondering:

Is the argument I use in my answer correct?

a student
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  • what is $U(20)$? – R_D Nov 04 '15 at 05:56
  • Yes your argument is correct and essentially the same as using 'brute force to show that $k^4=1$ for all $k$'. You have shown that $|3|=4$ by writing down the subgroup $\langle 3 \rangle$. Every element of $\langle 3 \rangle$ therefore has order at most 4 (in particular $k^4=1, \forall k\in\langle 3 \rangle$). Similarly, you have shown the same property for elements of $\langle 13 \rangle$. And then you show that $19$ and $11$ have order 2 (so, in particular $19^4=11^4=1$). Hence you have shown that $k^4=1$ for all $k$. – Sam Weatherhog Nov 04 '15 at 05:58
  • Your solution is fine. The brute force suggestion is not quite right, one can use simple theory instead of brute force, – André Nicolas Nov 04 '15 at 06:00
  • @Nizar, this must be the case as $\langle 3 \rangle$ is a group. If it had an element that could generate something outside of $\langle 3 \rangle$ it would not be a group. – Sam Weatherhog Nov 04 '15 at 06:00
  • @AndréNicolas By simple theory you mean what is said in the answer by P Vanchinathan? – a student Nov 04 '15 at 07:20
  • @SamWeatherhog Thank you for your comment. – a student Nov 04 '15 at 07:21
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    Sort of, but in greater generality. For example look at $U(124)$, same shape as $20$, $4$ times a prime. The group has $60$ elements, But by Fermat's Theorem we have $a^{30}\equiv 1\pmod{31}$, also $a^2\equiv 1\pmod{4}$, so $a^{30}\equiv 1\pmod{124}$. So all elements have order $\le 30$. The group has $60$ elements so cannot be cyclic. Very similar argument will work for all composite numbers not equal to $4$, $p^n$, or $2p^n$ where $p$ is an odd prime. So what happened with $20$ is what usually happens. – André Nicolas Nov 04 '15 at 07:34
  • @AndréNicolas Thank you very much, your comment is very helpful to me. If you had posted it as an answer I could have upvoted it better. It's okay though I understand if you don't care about gaining reputation. – a student Mar 21 '16 at 04:35
  • You are welcome. – André Nicolas Mar 21 '16 at 04:39

1 Answers1

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You must be knowing that the unit group $U(n)$ has order given by the Euler function, $\varphi(n)$. In this case $\varphi(20) = 8$. If this group is not cyclic then every element should generate a subgroup of order less than $8$, and dividing it.

So showing $x^4=1$ for all $x$ is a way of showing the group is not cyclic. There is nothing wrong in your approach, as other commenters have said.

P Vanchinathan
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