Help with defining a group

Question

Could anyone help me figure out exactly what does $\left(\mathbb Z^\times_{n^2 - 1}, \cdot\right)$ mean where $\mathbb Z^\times_{n^2 - 1} :=\lbrace[k]\in\mathbb Z_{n^2 - 1}: \gcd\left(k, n^2 - 1\right) = 1\,\text{in}\,\mathbb Z\rbrace$? My best guess is that it is the group under the property of multiplication acting on the set of integers modulo $n^2-1$ or the set containing the multiple equivalence classes $[k]$ that are generated by the integers that are relatively prime to $n^2-1$.

And this isn't even the proof I have to do using this hahaha.

Answer 1

The group in your question is just the multiplicative group of integers modulo k $\mathbb{Z}_k^{\times}$ for $k=n^2-1$. This is the set of all integers smaller than and relatively prime to $k$ with the operation of multiplication mod k- particularly, note that the condition of being relatively prime to $k$ is defined so we may have inverses.

Answer 2

The answer is your best guess.

Integers modulo $k$ are the residue classes modulo $k$:

$$\mathbb{Z}_k=\{\bar0,\bar1,\dots,\overline{k-1}\} \tag 1$$

The multiplication in $\mathbb{Z}_k$ is defined by:

$$(\bar a,\bar b) \mapsto \bar a\cdot\bar b:=\overline{ab} \tag 2$$

With this definition, $\bar1$ plays the role of unit, since $\forall \bar a\in \mathbb{Z}_k, \bar1\cdot\bar a=\overline{1a}=\bar a$.

Now, not every $\bar a \in \mathbb{Z}_k$ has inverse with respect to the operation $(2)$. In fact, given $\bar a\in \mathbb{Z}_k$:

\begin{alignat}{1} \exists \bar b\in \mathbb{Z}_k \mid\bar a\cdot\bar b=\bar1 &\iff \exists b \in \mathbb{Z} \mid ab\equiv 1 \pmod k \\ &\iff \exists b,m \in \mathbb{Z} \mid ab=mk+1 \\ &\iff \operatorname{gcd}(a,k)=1 \wedge \operatorname{gcd}(b,k)=1 \end{alignat}

Therefore, only the subset:

$$\mathbb{Z}_k^\times=\{\bar a \in \mathbb{Z}_k \mid \operatorname{gcd}(a,k)=1\}\subseteq \mathbb{Z}_k \tag 3$$

is a group with respect to the multiplication $(2)$.