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Given the Euler group $U_8= \{1,3,5,7\}$ , I wanted to check the followings :

  • if it is Abelian
  • if it is Cyclic

Let's check:

Abelian : each $x,y \in G$ : $xy=yx$ , hence indeed abelian.

Cyclic : if and only if $U_8$ has an element of order 8 :

  • $o(3)$: $3^2 = 9$, $9 \bmod 8 = 1$ , hence $o(3)=2$
  • $o(5)$: $5^2 = 25$, $25 \bmod 8 = 1$, hence $o(5)=2$
  • $o(7)$: $7^2 = 49$, $49 \bmod 8 = 1$ , then $o(7)=2$

then $U_8$ is not cyclic but abelian . Does it mean that:

  • given a cyclic group $G$, then $G$ must be abelian
  • given an abelian group $G$ , it doesn't mean that $G$ is cyclic ?

Regards

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    @Ron: [twice now](http://math.stackexchange.com/a/118276/742) (yes, [twice](http://math.stackexchange.com/a/117952/742)) you asked if you really had to check all elements in a group to see if the group was cyclic, and twice I pointed out that this was not the case and why; as soon as you find that the order of $3$ in $U_8$ is $2$, you are done, the group is not cyclic.2012-03-10

1 Answers 1

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An abelian group doesn't have to be cyclic. Take, for example, $(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$, which has the elements

$\{(0,0),(1,0),(0,1),(1,1)\}$

with multiplication defined by $(a,b) \times (c,d) = (ac,bd)$, where the products $ac$ and $bd$ are taken as they would be in $\mathbb{Z}/2\mathbb{Z}$. That this group is abelian follows from the fact that $\mathbb{Z}/2\mathbb{Z}$ is abelian.

As you can check, no element generates the whole group, so it is not cyclic.

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    A$n$d I've $n$ow realized that this group is the same one the OP described...2012-03-11