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For example, $\mathbb{Z}_7$ has at least two units, since $2\cdot 4 = 4\cdot 2=1$.

But $\mathbb{Z}_6$ has no units (since nothing multiplies to $7$).

But $\mathbb{Z}_8$ has a unit: $3\cdot 3=1$

So it seems that for $\mathbb{Z}_m$ $($ where $m\in \mathbb{Z})$, if $m+1$ is not prime then $\mathbb{Z}_m$ has units, and if $m+1$ is prime then $\mathbb{Z}_m$ does not have units.

Is this a correct characterization (does it always hold)?

James Ronald
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    $\ n\ $ is a unit in $\ \mathbb Z_m\ $ if and only if $\ \gcd(m,n)=1\ $ – Peter Mar 12 '20 at 16:21
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    It's more subtle than that. $1$ is a unit in $\mathbb{Z}_6$, or any $\mathbb{Z}_m$. So is $m-1$. For instance, $5^2 = 1$ in $\mathbb{Z}_6$. – Matthew Leingang Mar 12 '20 at 16:23
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    "Nothing multiplies to $7$", but since you are working modulo $6$, you just want to find two integers that multiply to something of the form $1+6k$ with $k\in\mathbb{Z}$. Furthermore, it is not true that nothing multiplies to $7$, e.g. you have $1\cdot 7=7$. This implies that the equivalence class $[1]$ is a unit in $\mathbb{Z}_6$ (it is the residue of both $1$ and $7$ mod $6$). – Thorgott Mar 12 '20 at 18:37
  • See in particular this answer in the linked dupe. – Bill Dubuque Mar 13 '20 at 00:32

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As per the comments, there are always precisely $\varphi(m)$ units, where $\varphi$ is Euler's totient function.

To take your example, $\Bbb Z_6$ has $\varphi(6)=2$ units. They are $1$ and $5$.