# Torsion Coefficient in Group Theory

I have seen a calculation about torsion coefficient Determining torsion coefficients But I am now facing another similar question and not sure about is my answer correct.

The question: Find the torsion coefficients of $\mathbb{Z}_6\times\mathbb{Z}_{12}\times\mathbb{Z}_{20}$.

This is my approach:

\begin{align}\mathbb{Z}_6\times\mathbb{Z}_{12}\times\mathbb{Z}_{20}&\cong(\mathbb{Z}_2\times\mathbb{Z}_3)\times(\mathbb{Z}_3\times\mathbb{Z}_4)\times(\mathbb{Z}_4\times\mathbb{Z}_5)\\&\cong (\mathbb{Z}_5\times\mathbb{Z}_4\times\mathbb{Z}_3)\times(\mathbb{Z}_3\times\mathbb{Z}_4)\times(\mathbb{Z}_2)\\&\cong\mathbb{Z}_{60}\times\mathbb{Z}_{12}\times\mathbb{Z}_2\end{align} So the torsion coefficients are $60,12,2$. Am I correct?

Other question 1: If it is correct, can I say the torsion coefficients are $12,2,60$ (not following the order)? Because I notice this must follow that every subscript number must divisible by the next subscript number.

Other question 2: Is it a trick of how to rearrange the direct product? I found that I only have one $\mathbb{Z}_5$, so put it at the leftmost place to ensure it doesn't have to divisible by another $\mathbb{Z}_5$. But I'm not too sure about how to do it.

If anyone could help me, I will appreciate that because I really need someone to tell if I study abstract algebra the right way.

First question: The definition of torsion coefficient is given numbering them (which is used to say that the $i$-th one divides the $(i+1)$-th one), so if you find that $G\simeq \mathbb{Z}_2\times\mathbb{Z}_{12}\times\mathbb{Z}_{60}$, then you have that $m_1=2$, $m_2=12$, $m_3=60$ and $s=0$ (the torsion-free coefficient). So it is ok to shuffle them, but they should be listed with their orders.
Second question: That is the trick, basically, namely that of "moving to the right" the terms with the maximal prime power for each prime involved in the group. In your example one could also have noticed that the only problem in the triad $(6,12,20)$ was $12$ not dividing $20$. This means that the "culprit" is in the couple $(6,12)$ and the "culprit" is $3$. So, noticing that $\mathbb{Z}_6\simeq \mathbb{Z}_2\times\mathbb{Z}_{3}$ since $2$ and $3$ are coprime, you just "move" $\mathbb{Z}_3$ to the end, getting the final decomposition.