The multiplicative group module $n$, $(\mathbb Z/n\mathbb Z)^\ast$, is the group of reduced residue class modulo $n$. It has $\phi(n)$ elements.
Unlike the group $\mathbb Z/n\mathbb Z \cong C_n$, which can can be understood by drawing a regular polygon with n vertices, $(\mathbb Z/n\mathbb Z)^\ast$ is somehow less intitutive.
For $(m, n) = 1$, we have $(\mathbb Z/mn\mathbb Z)^\ast \cong (\mathbb Z/m\mathbb Z)^\ast \oplus (\mathbb Z/n\mathbb Z)^\ast$ The isomorphism is explicitly given by \(a \to ([a]\_m, [a]\_n)\) \(([x]\_m, [y]\_n) \to [n[n]^{-1}\_m x + m[m]^{-1}\_n y]\_{mn}\) where $[a]_n$ is the image of a in the homomorhpism $\mathbb Z \to \mathbb Z/n \mathbb Z$.
When $n$ has primitive roots, the this group becomes a cyclic group. But how can we find out a connection between $(\mathbb Z/p\mathbb Z)^\ast$ and a regular polygon of $p-1$ vertices?
On the other hand, when $(m, n) = 1$ the direct sum of $C_m$ and $C_n$ has a strong geometric sense. Remember that the product of two circles as topological space is a torus. Image we have a grid on a torus, with $m$ horizontal circles and $n$ vertical circles, when we walk along the direction $(1, 1)$, we can traverse all grid points $(x, y)$ on then torus and the path will become a circle of length $mn$.
For $(\mathbb Z/n\mathbb Z)^\ast$, an elegant result is that, every finite abelian group can be embeded into such a group for some integer n. Assume that abelian group G has cyclic decomposition $\bigoplus C_i$, we pick up distinct prime numbers $p_i$ such that $|C_i| \vert p_i - 1$, then there is an embeding $C_i \to \mathbb Z/p\mathbb Z$. The existence of $p_i$ is granted by the Dirichlet theorem on arithmetic progress.
How every, there is not aways an embeding of $(\mathbb Z/n\mathbb Z)^\ast$ into the unit circle $S^1$, which means not all abelian group admits a faithful one-dimensional representation.
Is there an intitutive way to express the group $(\mathbb Z/n \mathbb Z)^\ast$ analog to $\mathbb Z/n\mathbb Z$?