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So I just want to get some of my questions answered:

For the theorem: Every Cyclic group is abelian leads me to be confused a bit.

So, If a group is not cyclic, it can still be abelian? For example, I've seen $Z_7^{*}$ and $Z_{12}^{*}$ but i'm not sure why because I thought <2> fully generates $Z_7^{*}$ so isn't it cyclic? I'm not sure how to confirm it is abelian but all the cases I've seen involving integers under the operation multiplication and addition appear to be abelian.

Also, I am not sure if the inverse works:

Every abelian group is cyclic?

Justin
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    Consider the additive group of all rational numbers. Addition is commutative, so the group is abelian, but it cannot be generated by a single element, or even by finitely many elements. –  Mar 19 '15 at 04:38
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    $\Bbb{Z}7^*$ is cyclic (generated by $\overline{3}$ or $\overline{5}$, but not by $\overline{2}$ because its power only give you $\overline{1}$, $\overline{2}$ and $\overline{4}$). But you are right in that $\Bbb{Z}{12}^*$ is not cyclic. Also, what Bungo says. – Jyrki Lahtonen Mar 19 '15 at 05:18

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There are Abelian groups that are not cyclic. For example, $\Bbb{Z}_2 \times \Bbb{Z}_2$ (the Klein four group) is Abelian but not cyclic. $\Bbb{Z}_p^*$ is always cyclic for $p$ prime, so in particular, $\Bbb{Z}_7^*$ is cyclic. All of the four elements of $\Bbb{Z}_{12}^*$ have order $2$, so it is not cyclic. In fact, $\Bbb{Z}_{12}^* \cong \Bbb{Z}_2 \times \Bbb{Z}_2$.

More generally, the structure theorem for finitely generated Abelian groups tells us that every such group $G$ can be written as a direct product of cyclic groups:

$$G \cong \Bbb{Z}_{d_1} \times \cdots \times \Bbb{Z}_{d_k} \times \Bbb{Z}^m$$

where each $d_i \in \Bbb{N}$, $d_i \mid d_{i + 1}$, and $m \geq 0$. Moreover, this decomposition is unique up reordering.

Qudit
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