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Hello I have a difficulty with this exercise the idea is to show $$\begin{align} \phi: \mathbb{Z}_{30} &\to \mathbb{Z}_2\times \mathbb{Z}_3\times \mathbb{Z}_5 \\ [x]_{30} &\mapsto ([x]_2,[x]_3,[x]_5) \end{align}$$ is surjective but what I was able to show is that $$\begin{align} \phi_1: \mathbb{Z}_{30} &\to \mathbb{Z}_2\times \mathbb{Z}_{15} \\ [x]_{30} & \mapsto ([x]_2,[x]_{15}) \end{align} $$ is surjective I wanted to consider $$\begin{align} \phi_2 : \mathbb{Z}_2\times \mathbb{Z}_{15} & \to \mathbb{Z}_2\times \mathbb{Z}_3\times \mathbb{Z}_5 \\ ([x]_2,[x]_{15}) & \mapsto ([x]_2,[x]_3,[x]_5) \end{align}$$ and show that is surjective but but I can't show $\phi_2$ is surjective I need help to tackle this exercise

Shaun
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Virostake
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    Hint: $\ \ 3\cdot 5\to (1,0,0),\ 2\cdot 5\to (0,1,0),\ 2\cdot 3\to (0,0,1),,$ thus $,15a + 10b+6c\to (a,b,c),,$ by linearity. See here for the idea (CRT formula). – Bill Dubuque Mar 10 '22 at 17:45
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    The domain and codomain both have cardinality thirty. Therefore bijectivity (and also surjectivity) is automatic once you show injectivity, i.e. that the kernel is trivial. To get an explicit inverse you need to work a bit harder. Bezout's identity and all that. – Jyrki Lahtonen Mar 10 '22 at 18:23

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Let $(b_1,b_2)\in \mathbb{Z}_{2}\times \mathbb{Z}_{15} $ and consider $ a=-14b_2+15b_1$ with have $ \phi(a)=(b_1,b_2)$ since $ [a]_2=b_1, [a]_{15}=b_2$

Virostake
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