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What is the number of abelian groups of order 40? I thought the number is just $3$. More specifically, they are

$$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_5$$ $$\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_5$$ $$\mathbb{Z}_8\times\mathbb{Z}_5$$

However, my answer says:enter image description here

Am I missing anything?

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Your book is wrong, $\mathbb Z_{10}$ is congruent to $\mathbb Z_5\times \mathbb Z_2$.

So $\mathbb Z_{10}\times \mathbb Z_2\times \mathbb Z_2$ is congruent to $\mathbb Z_5\times \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2$

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    $\begingroup$ In general if $n=p_1^{\alpha_1}\times p_2^{\alpha_2}\times \dots \times p_s^{\alpha_s}$ then there are $\prod\limits_{i=1}^s p(\alpha_i)$ abelian groups of order $n$. Where $p$ is the partition function. $\endgroup$ – Jorge Fernández Hidalgo Oct 18 '16 at 2:08
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The ones listed in the book are not distinct.

$$\Bbb Z_{40} \cong \Bbb Z_{5} \times \Bbb Z_8 $$

$$\Bbb Z_{20} \times \Bbb Z_2 \cong \Bbb Z_{10} \times \Bbb Z_4 \cong \Bbb Z_{5} \times \Bbb Z_4 \times \Bbb Z_2 $$

$$\Bbb Z_{10} \times \Bbb Z_2 \times \Bbb Z_2 \cong \Bbb Z_{5} \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$$

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