To prove that U(20) is not cyclic.

Question

A similar question already has an answer elsewhere on this site, though this question is a bit different from those.

When I first saw the formulation of question, namely:

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

I thought to myself, "this should be doable by contradiction". Noting that I don't want to invoke any theorems other than the group axioms, subgroup definition and respective subgroup tests (as that is all that the book (Gallian) I'm following deals with so far).

So, I assumed that let there be some element $k\in U(20)$ such that $U(20) = \langle k \rangle$. Now, I tried to arrive at contradiction by considering that since $3\in U(20)$ and $11\in U(20)$ so $3\in \langle k \rangle$ and $11\in \langle k \rangle$. Though that goes only so far, and I am stuck. Is it even possible to proceed in this manner? If, so any guidance will be appreciated!

Here $$U(n)= \{x: (\gcd(x,n)=1)\land (1\leq x<n) \land (x,n \in \mathbb{N} )\}$$ is a group under multiplication modulo n.

Though, it would be nice if someone can write $U(n)$ definition a bit more neatly. This looks ugly as of yet. — Shyam Tripathi, Sep 13 '23 at 02:56
$U(20)$ has $8$ elements, but $4$ and $5$ divide $k^4-1$, so $k^4\equiv1\bmod20$, for all $k\in U(20)$ — J. W. Tanner, Sep 13 '23 at 03:00
In your definition of $U(n)$, did you mean $n$ where you typed $20$? — J. W. Tanner, Sep 13 '23 at 03:11
$(\mathbb Z / n \mathbb Z)^\times$ is more standard notation. — Eric, Sep 13 '23 at 03:23

Nicky Hekster · Accepted Answer · 2023-09-13T10:05:54.737

4

Why don't use the following fact: every finite cyclic group of even order has a unique element of order $2$. (I leave that to you to prove, it is not difficult (or see here (but try yourself first!)). Now calculate the order of $9$ and of $11$ mod $20$ in $U(20)$.

edited Sep 13 '23 at 10:05

answered Sep 13 '23 at 08:36

Nicky Hekster

49,281

To prove that U(20) is not cyclic.

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