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I need to find the list of all units in the ring $\mathbb Z_{12}$.

I know that there must be a multiplicative inverse, $a_{-1}$, for there to be units, $a$, and that

$$a\cdot a^{-1} = a^{-1} \cdot a = 1.$$

I think the multiplicative inverse here is $1$ and the answer tells me that the units are $\{1, 5, 7, 11\}$, but I am having trouble understanding.

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The multiplicative identity is $1$, as (I think) you meant. Each number is allowed to have its own inverse, so we check. $1$ clearly divides itself, so $1$ is always a unit. $5 \cdot 5 = 25 = 1$, so we see that $5$ is a unit. $7 \cdot 7 = 49 = 1$, so $7$ is a unit. And $11 \cdot 11 = 121 = 1$, so it's also a unit.

You might have noticed a few things - we happen to have squared each of these numbers, so that they are each their own inverse. Also, the primes that divide $12$ are $2$ and $3$, and $5,7,11$ are exactly the numbers that are coprime to $12$. I don't know what you've learned so far, but neither of these are coincidences (really, it's a multiplicative group now too, so perhaps your group theory can kick in).

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Suppose that $xy = 1\pmod{12}.\:$ Then $\:3\ |\ 12\ |\ xy-1$ and this implies $x$ is coprime to $3\:$ (else $3\ |\ x, xy-1\ \Rightarrow\ 3\ |xy-(xy-1) = 1,\:$ contradiction). Similarly $x$ is coprime to $2$. This leaves only $\{\pm1,\:\pm5\}\pmod{12}$ as potential units. $\:\pm1$ are units. Ditto for $\pm5$ since $5^2\equiv 1\pmod{12}.$

Generally the Bezout identity for the GCD implies $n$ is a unit mod $m$ iff $\gcd(n,m)=1$. In a finite ring an element $a$ is a unit iff it is not a zero-divisor since the map $\:x\mapsto ax$ is onto iff it is $1$-to-$1$.

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