Let $U(N)$ is a group under multiplication modulo $N$ called
the group of units of $\mathbb{Z}/N$.
$U(N) = \{a \in Z/N: (a,N) = 1 \} $, a and N are relative prime.
$U(3) = \{1,2\}$
$U(5) = \{1,2,3,4\}$
$U(15) = \{1,2,4,7,8,11,13,14\}$
I did not understand the group direct product between $U(3)$ and $U(5)$.
It is claimed (in a book) that $U(15) = U(3) \times U(5)$.
How the group direct product is calculated?
Best Answer
For sure your book says that $$U(\mathbb{Z}/ 15\mathbb{Z}) \simeq U(\mathbb{Z}/ 3\mathbb{Z}) \times U(\mathbb{Z}/ 5\mathbb{Z}), $$ where $\simeq$ means "isomorphic", because they are not equal.
That is, you can find a function $f:U(\mathbb{Z}/ 15\mathbb{Z}) \to U(\mathbb{Z}/ 3\mathbb{Z}) \times U(\mathbb{Z}/ 5\mathbb{Z})$ such that $f$ is bijective and $f$ is a homomorphism i.e. $f(xy) = f(x)f(y), \forall x ,y\in U(\mathbb{Z}/ 15\mathbb{Z})$ (where the LHS multiplication is in $U(\mathbb{Z}/ 15\mathbb{Z})$ and the RHS multiplication is in $ U(\mathbb{Z}/ 3\mathbb{Z}) \times U(\mathbb{Z}/ 5\mathbb{Z}))$. Here, a function that satisfies the conditions mentioned is $f(x \pmod {15}) = (x \pmod 3, x \pmod 5).$
The direct product of two groups $(G,*)$ and $(H, @)$ is just the cartesian product $G \times H$ with the operation $\cdot: (G \times H) \times (G \times H) \to G \times H$, that makes it a group, defined as following: $$(g_1,h_1) \cdot (g_2,h_2) = (g_1 * g_2, h_1 @ h_2), \forall g_1,g_2 \in G \text{ and } \forall h_1,h_2 \in H, $$ I'll leave you to check the group axioms.