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I am finding a contradiction between those two theorems and I do not know what I am doing wrong. First theorem is:

The group $\Bbb Z_{m_1} \times \Bbb Z_{m_2} \times \dotsm \times \Bbb Z_{m_n}$ is cyclic and isomorphic to $\Bbb Z_{m_1 m_2 \dotsm m_n}$ if and only if any of $m_i$'s are relatively prime.

And the second theorem is the Fundamental Theorem of Finitely Generated Abelian Groups, which says:

Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $\Bbb Z_{p_1^{k_1}} \times \Bbb Z_{p_2^{k_2}} \times \dotsm \times \Bbb Z_{p_n^{k_n}}$ where $p_i$'s are not necessarily distinct primes.

For example, by the second theorem, we have $\Bbb Z_{360}$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_5$, but by the first theorem, it should not be, since $2$ and $2$ are not relatively prime. What am I doing wrong here?

Thanks

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  • $\begingroup$ Note: e.g. $\mathbb{Z}$ is a cyclic group (and hence a finitely generate abelian group), and does not occur in your version of the Fundamental Theorem. It is, however, correct if you restrict to finite (as opposed to finitely generated) abelian groups. $\endgroup$ – Douglas S. Stones Apr 14 '13 at 23:09
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The fundamental theorem of finitely generated abelian groups does not say that you can split up $360$ in any way you want as a product of primes and obtain an isomorphism. It says that for any abelian group of order $360$, there is a way to write $360$ as a product of prime powers $360=p_1^{r_1}\cdot\cdots \cdot p_n^{r_n}$ such that $G\cong \mathbb Z_{p_1^{r_1}}\times Z_{p_n^{r_n}}$.

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  • $\begingroup$ Then, isn't the only way to do this is to use 1st theorem? Then we can deduce the same thing from theorem 1, why do we need theorem 2? I don't get it $\endgroup$ – Yasin Razlık Apr 14 '13 at 12:49
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    $\begingroup$ Theorem 1 gives you a condition under which a product of groups of the form $\mathbb Z_{p_n^{r_n}}$ is isomorphic to a cyclic group. But not all finite abelian groups are cyclic (if they were, then you won't need theorem 2). Theorem 2 tells you something in the other direction. It says that any finite abelian group is a product of group of the form $\mathbb Z_{p_n^{r_n}}$, but not necessarily with the condition on the primes that is stated in theorem 1. $\endgroup$ – Ittay Weiss Apr 14 '13 at 13:23
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The first theorem gives you that if a group is cyclic, it may be factored into a direct product of cyclic groups of relatively prime order, whereas the second theorem gives you that if a finite group is abelian, it is isomorphic to some direct product of cyclic groups. $\mathbb{Z}_{360}$ is not isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5$, but it is isomorphic to $\mathbb{Z}_8\times \times \mathbb{Z}_9\times\mathbb{Z}_5$. You can factor out relatively prime numbers, e.g. $\mathbb{Z}_{pq}\cong\mathbb{Z}_p\mathbb{Z}_q$ when $p\not= q$ and $p,q$ are prime, but you can't decompose numbers which aren't relatively prime, e.g. $\mathbb{Z}_8\not\cong\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$.

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You are misinterpreting the 2nd theorem. ${\bf Z}_{360}$ is isomorphic to ${\bf Z}_8\times{\bf Z}_9\times{\bf Z}_5$. Note that, for example, ${\bf Z}_4$ is not isomorphic to ${\bf Z}_2\times{\bf Z}_2$.

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  • $\begingroup$ To put it another way, $\mathbb Z_4 \cong \mathbb Z_{2^2} \not\cong \mathbb Z_2^2$ :) $\endgroup$ – Ben Millwood Apr 14 '13 at 12:35

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