# Is the group of units of a ring an additive group?

My university's textbook on Abstract Algebra states:

Let $$n\in \mathbb{N}_0$$. We note the group of units of the quotient ring $$\langle\mathbb{Z}_n,+,\cdot\rangle$$ as $$\langle\mathbb{Z}^\times_n,+,\cdot\rangle$$, where $$\mathbb{Z}^\times_n = \{\bar x\mid x\in \mathbb{Z}\;\wedge\; \gcd(x,n)=1\}$$.

Given that $$\gcd(x,n)=1$$ is equivalent to $$x$$ being invertible in $$\langle\mathbb{Z}_n,\cdot\rangle$$, it's easy to prove that such $$x$$ form a multiplicative group (for any $$x_1$$ and $$x_2$$, $$x_1x_2$$ is invertible as well with inverse $$x_2^{-1}x_1^{-1}$$ and for any $$x$$, $$x^{-1}$$ is also invertible with inverse $$x$$).

However, I am doubting the textbook's use of $$+$$ in the context of $$\mathbb{Z}^\times_n$$ being a group. Of course, addition is still a valid algebraic operator, but I don't think it's closed in $$\mathbb{Z}^\times_n$$, letalone a group operator. For example, $$1$$ is always in $$\mathbb{Z}^\times_n$$, but $$1+1$$ is not.

Am I missing something here, or is their notation/phrasing sloppy?

• Probably a copy-paste-modify error, they forgot to remove the "$+$". For every unit $u$, $-u$ is also a unit, and $0 = u + (-u)$ isn't. May 21 '20 at 15:55

No it's definitely not a group because the identity $$0$$ is not there.
It should read $$\langle {\mathbb Z^\times_n, \cdot}\rangle$$, not $$\langle {\mathbb Z^\times_n, +, \cdot}\rangle$$ which is the symbolism for a ring, not a group.
Incidentally, what is meant by $$\bar x$$?
• Thanks for clarifying. This is our university's homebrew textbook (ISBN: 90-5350-771-X). $\bar x$ is the residue class with representative $x$. You made me wonder, though: is there certain notation to indicate that a ring $G$ is a group w.r.t. both its multiplication (not counting 0, then) and addition (so really, notation that it's a field)?
• Not as far as I know -- that is there's no notation to distinguish $\langle{S, +, \cdot}\rangle$ being a ring from it being a field. What you'd do is state whether $\langle{S^*, \cdot}\rangle$ was a semigroup or a group. May 21 '20 at 16:14