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We know from the fundamental theorem that every finite group $G$ can be written as $G\cong Z_{n_1}\times\cdots\times Z_{n_t}$, where $n_i| n_{i+1}$.

What is happening in the following example.

$Z_6\cong Z_2\times Z_3$, but 2 does not divide 3.

ted
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  • $\mathbb{Z}_6 \cong \mathbb{Z}_6$. Since there is only one $n_i$, the statement of division is vacuously true. – Greebo Aug 19 '13 at 11:39
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    I think you mean finite abelian group. Also, $\Bbb Z_6$ is the factorization given by the theorem. In principle you can write $\Bbb Z_1 \times \Bbb Z_6$ if you want it to look like a factorization too, but that might make people cringe, much the same way as saying $1$ is a prime would. – Arthur Aug 19 '13 at 11:40

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The fundamental theorem tells you that there is at least one decomposition of $G$ into cyclic groups $\Bbb Z_{n_i}$ such that $n_i\mid n_{i+1}$, it doesn't say that every decomposition into cyclic groups is of that form. Since $\Bbb Z_6$ is already a decomposition of itself into cyclic groups, there is no contradiction to the theorem.

walcher
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