[Math] Why doesn’t the Chinese remainder theorem contradict the Fundamental Theorem of Finitely Generated Abelian Groups

abelian-groupsfinite-groupsgroup-theory

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

Best Answer

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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