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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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    The fact you have an abelian non-cyclic group does indeed mean that an abelian group isn't necessarily cyclic.2012-03-10
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    @ron I am very sad you had to ask this question. In the previous questions, you were hinted at these facts several times. Please make an effort to learn from the previous questions and answers. Please understand that this site exists to help you in the extreme situation, when you have thought about it, say for a week, or so... Please start from the definitions and ask if you can solve that difficulty. Group theory is better learnt by thinking over and over again than by discussing with several others.2012-03-10
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    @Kannappan , I think you should be very proud of me . I came to that conclusion on my own , just wanted to make sure with you guys . 10x , you're great , all of you !2012-03-10
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    Ron, you gave one example of an abelian group that isn't cyclic. You might be interested to know that there are very many such examples, some of them very familiar. Try thinking about $(\mathbb R,+),$ $(\mathbb R\setminus\{0\},\times),$ $(\mathbb Q,+),$ $(\mathbb Q\setminus\{0\},\times).$ These are infinite abelian groups. Can you prove that they're not cyclic?2012-03-10
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    @Ron: *Please* enclose mathematical expressions in dollar signs so that they render correctly, and don't use quotation marks. It makes your posts hard to read.2012-03-10
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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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    And I've now realized that this group is the same one the OP described...2012-03-11