Determine the torsion coefficients of $\mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10}$.

Now I know that if I rearrange $\mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10}$ to the form $\mathbb{Z}_{m1} \times \mathbb{Z}_{m2} \times . . . \times \mathbb{Z}_{mk} \times \mathbb{Z}^s$ for $s \in \mathbb{N}$, where $s$ is the rank of $G$, with $G \cong \mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10}$. Then the torsion coefficients are $m_1, m_2,. . . , m_k$ (by the classification theorem).

But how do I rearrange this?

  • $\begingroup$ If your statement was true then you have $\mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10} = \mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}^0$, so the torsions coefficients would be $6$, $6$ and $10$, which seems pretty strange to me. Could you provide us with the exact definition of torsion coefficients that you are given? $\endgroup$ Jan 22 '16 at 5:36
  • $\begingroup$ @JendrikStelzner I'm being given them via the classification theorem. For which I have as: Any finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups$\mathbb{Z}_{m1} \times \mathbb{Z}_{m2} \times . . . \times \mathbb{Z}_{mk} \times \mathbb{Z}^s$ for $s \in \mathbb{N}$ where $m_1|m_2, m_2|m_3, . . . , m_{k-1}|m_k$, $s = $ rank of $G$ and $m_1, m_2, . . . , m_k$ are the torsion coefficients of $G$. $\endgroup$
    – Nique
    Jan 22 '16 at 5:41

I am not familiar with this, but given the definition you provided in the comments you can do the following:

If $n$ and $m$ are coprime then $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$. Therefore $$ \mathbb{Z}_6 \times \mathbb{Z}_6 \times \mathbb{Z}_{10} \cong \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_5 \cong \mathbb{Z}_2^3 \times \mathbb{Z}_3^2 \times \mathbb{Z}_5. $$ By rearranging the factors (or rather summands) we get $$ \mathbb{Z}_2^3 \times \mathbb{Z}_3^2 \times \mathbb{Z}_5 \cong \mathbb{Z}_2 \times (\mathbb{Z}_2 \times \mathbb{Z}_3) \times (\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5) \cong \mathbb{Z}_2 \times \mathbb{Z}_6 \times \mathbb{Z}_{30}. $$ So using the definition you provided in the comments the torsion coefficients are $2, 6, 30$ (as this is an example of a finite groups the rank is $0$).

  • $\begingroup$ This approach should actually work for all finite abelian groups if I am not mistaken. $\endgroup$ Jan 22 '16 at 5:54

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