Listing the elements of $U(\mathbb{Z}_{54})$.

Question

The set of all integers modulo $q$ is denoted by $\mathbb{Z}_q$. When equipped with multiplication modulo $q$, has the structure of a commutative monoid, the identity element being equivalence class $1$. $Q$ is a natural number, $≥ 2$.

Really struggling with this question, Im fairly sure the elements are all the odd numbers, less the divisors of $54 (2,3,6,9,18,27)$? But am unsure how to write this properly. Also I seem to think that the elements will stop at $17$? Any help would be very much appreciated.

Answer 1

HINT:

Consider the set $S_{n}=\{1\le m\le n\vert \gcd(m,n)=1\}$.

Note that if $\gcd(m,n)=1,\,\,\exists\text{ integers }p,q$ such that.

$mq+pn=1$ (as gcd can be expressed as a linear combination)

Reduce both sides modulo $n$. Can you see where to go from here?

EDIT:

First, consider the elements of $\mathbb{Z}_{54}$. If $i\in \mathbb{Z}_{54}$ such that $\gcd(i,54)=d>1$, then $i$ is a zero divisor $\Big($$\because\,\,\, i\cdot\frac nd=0$ $\Big)$ and hence $i$ cannot be a unit (why?)

Important: If $i$ had a multiplicative inverse then this would imply $\frac nd\text=\,0$ which is clearly not the case.

Now, claim that if $\gcd(i,54)=1$, then $i$ is a unit.

Suppose $\gcd(i,54)=1$. Then $\exists\,\, p\,,\,q\in\mathbb{Z}$ such that $ip+54q=1$ or $(ip)\pmod{54}=1$ and hence, $i^{-1}=p$ in $\mathbb{Z}_{54}$.

Thus we have i is a unit $\iff$ gcd(i,54)=1

So the set of units is $S_{54}$, where $S_n$ is as defined in the first step.