Isomorphism with $Z_{10}$ and $Z_2 \times Z_5$ I want to write out explicitly the pairing between elements of $\mathbb{Z}_{10}$ and of $\mathbb{Z}_2 \times \mathbb{Z}_5$ which yields the isomorphism in the following theorem:
Theorem: If $m = m_1 .... m_r$, where the $m_i > 1$ are relatively prime pairs, then $\mathbb{Z}_m$ is isomorphic to $\mathbb{Z}_{m_1} + ..... + \mathbb{Z}_{m_r}$.
Would constructing a map from $\left \{1,3,7,9 \right \}$ to another set $\mathbb{Z}_2 \times  \mathbb{Z}_5$ be enough? If so, how would I create this map? (Or, if this is the totally wrong idea, how would I approach this exercise?)
 A: The map given by the canonical projections works:
$\mathbb{Z}_{10} \to \mathbb{Z}_2\times\mathbb{Z}_5$ given by $[x]_{10} \mapsto ([x]_{2}, [x]_{5})$.
You need to prove that this map is well-defined and injective.
The same map works for $n=ab$ with $\gcd(a,b)=1$ to prove that $\mathbb{Z}_{ab} \cong \mathbb{Z}_a\times\mathbb{Z}_b$.
A: Just consider the map$$\begin{array}{ccc}\mathbb{Z}_2\times\mathbb{Z}_5&\longrightarrow&\mathbb{Z}_{10}\\(a,b)&\mapsto&5a+2b\pmod{10}.\end{array}$$
A: It's almost easier to guess and be correct than to systematically solve.
But we need $(a,b)\to x \implies m*(a,b) \to m*x$.  Obviously $(0,0)$, the identity in $\mathbb Z_2\times \mathbb Z_5$ maps to $0$, the identity in $Z_{10}$.
$|(1,0)| = 2$ (that is $2(1,0) = (0,0)$ and $m(1,0)\ne (0,0); 0 < m < 2$) so $2*(1,0) = (0,0)$ and $2*x \equiv 0 \mod 5; x \not \equiv 0 \mod 5 \implies x \equiv 5 \mod 5$.  So $(1,0) \to 5$.
$|(0,1)| = 5$ and $5x \equiv 0\mod 5;x \not \equiv 0 \mod 5 \implies x$ is even so $x = 2k$.
If we map $(\{0|1\}, b)\to (\{0|5\} + 2kb)$, that should do it. $(a,c) \to 5a + 2mc$. It can easily be verified that $(a,b) + (c,d) \to (5a + 2kb) + (5c+2kd)$ and that  the map is injective.
