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The book I'm reading defines the Fundamental Theorem of Finite Abelian Groups in the usual way that a finite abelian group is isomorphic to a direct product of cyclic groups $\mathbb{Z}_{{p_1}^{e_1}} \times \cdots \times \mathbb{Z}_{{p_r}^{e_r}}$, but then it also states that the group is isomorphic to $\mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_t}$ where for each $i = 1,...,t-1$ we have $m_i | m_{i+1}$. How can this be if the groups orders are supposed to be relatively prime.

If anyone can give an example of how this works I would greatly appreciate it.

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  • $\begingroup$ Well, for distinct primes $p_i$ and $p_j$, we have $\mathbb{Z}_{p_i^{e_i}}\times\mathbb{Z}_{p_j^{e_j}}\cong\mathbb{Z}_{p_i^{e_i}p_j^{e_j}}$. This is true in general for coprime integers. This means you can move around the $p_i$ in a way that will allow you to do this; I'm not sure exactly what the proof is but that's a reason why it's possible. $\endgroup$ Commented Apr 17, 2016 at 15:55
  • $\begingroup$ Nothing says the group orders are relatively prime. For any prime $p$ the direct sum may involve cyclic groups of order $p^k$ for several $k$. $\endgroup$ Commented Apr 17, 2016 at 15:57
  • $\begingroup$ @AndréNicolas So if for example the group was $\mathbb{Z}_{15*3}$ this would be isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_{15}$? $\endgroup$
    – Ash
    Commented Apr 17, 2016 at 15:59
  • $\begingroup$ @M.Smith no because $3$ is not coprime to $15$. But $\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{15}$ $\endgroup$ Commented Apr 17, 2016 at 16:00
  • $\begingroup$ @AndréNicolas That's where I'm getting confused because then how can $m_i | m_{i+1}$ if they have to be coprime? $\endgroup$
    – Ash
    Commented Apr 17, 2016 at 16:03

1 Answer 1

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That is because of the Chinese remainder theorem. If the group orders are pairwise coprime, the product is isomorphic to $\;\mathbf Z/p_1^{e_1}\dots p_r^{e_r}\mathbf Z$, and there is only one such $m_i$ (which, incidentally, are called the invariant factors of the abelian group).

Here is an example: suppose the primary decomposition of the group is: $$\mathbf Z/2\mathbf Z\times \mathbf Z/3^2\mathbf Z\times \mathbf Z/3^2\mathbf Z\times \mathbf Z/5\times\mathbf Z/5^2\mathbf Z\times \mathbf Z/5^3\mathbf Z. $$ First display the primary factors on different lines (one line per prime factor) in increasing order of the exponent and right-aligned: $$\begin{array}{r} \mathbf Z/2\mathbf Z^{\phantom{2}}\\ {}\times \mathbf Z/3^2\mathbf Z\times \mathbf Z/3^2\mathbf Z\\ {}\times \mathbf Z/5\mathbf Z\times\mathbf Z/5^2\mathbf Z\times \mathbf Z/5^3\mathbf Z \end{array}. $$ Then apply the C.r.t. to each column: $$ \mathbf Z/5\mathbf Z\times \mathbf Z/225\mathbf Z\times \mathbf Z/2250\mathbf Z.$$

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  • $\begingroup$ Thank you for your response. Can you give an example to illustrate how it's used? $\endgroup$
    – Ash
    Commented Apr 17, 2016 at 16:21
  • $\begingroup$ @M. Smith: I've added an example. Hope it is clear. $\endgroup$
    – Bernard
    Commented Apr 17, 2016 at 18:51

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