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?
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.
Which textbook is this?
Incidentally, what is meant by $\bar x$?