Put $C_{12}\times C_{35} \times C_{45}$ is canonical product

Question

$C_{12} \cong C_3 \times C_4$

$C_{35}\cong C_5\times C_7$

$C_{45} \cong C_5 \times C_9$

Then you combine these and rearrange into factors according to the prime involved but there is no prime that divides $4$ for $C_4$.

Answer 1

You can write for example:

$$4\cdot5\cdot7\cdot9\;,\;\;\text{and}\;\;\;3\cdot5\;\;\implies G\cong C_{15}\times C_{4\cdot5\cdot7\cdot9}$$

Answer 2

You have already noted that $$C_{12}\times C_{35}\times C_{45}=C_3\times C_4\times C_5\times C_5\times C_7\times C_9.$$ Now for each prime $p$ take the product of the $C_{p^k}$ with $k$ maximal, without repeats. So in this case we get $$C_4\times C_5\times C_7\times C_9\cong C_{4\times5\times7\times9}=C_{1260}.$$ Now repeat the process with the remaining cyclic factors. We're left with $C_3$ and $C_5$, so we get $C_3\times C_5=C_{15}$, and were done: $$C_{12}\times C_{35}\times C_{45}=C_{15}\times C_{1260}.$$

Answer 3

Into the last factor, we should also place the largest power of each prime for similar reasons, which in this case is $C_9$ and one copy of $C_5$, $C_7$ and $C_4$ since if we didn't then we break the condition that $n_i|n_{i+1}$. For example, if we put in $C_3$ instead of $C_9$, then we would earlier get that $9$ divided $3$ which is wrong.

This gives the last factor as $C_4 \times C_5 \times C_7 \times C_9 \cong C_{1260}$ by the Chinese Remainder Theorem.

We now remove those groups from our list and repeat with the remaining ones. The only ones left are $C_3$ and $C_5$ which are the largest of their respective prime powers so we get $C_{15}$ for our penultimate factor.

There are now no groups left to deal with so we are left with the isomorphism $$C_{12} \times C_{35} \times C_{45} \cong C_{15} \times C_{1260}. $$