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i'm watching these group theory lectures and i'm confused about minute 2:30 https://www.youtube.com/watch?v=WbzwNYN2dH4

Groups are being explained as a buch of simbols that produce permutations on another set (the symbol "1" moves every element on another set by 1, -1 does the opposite), and their composition is explained as doing one permutation after another.

at this point infinite cyclic groups are being explained, and the lecturer is trying to make a point into why the integers under addition are an infinite cyclic group, he tries to do this by counter example.

The idea is that the integers are a cyclic group that is infinite on both sides, if they weren't then the group would not be a permutation so it would not even be a group.

I hope this picture makes it a bit clearer, here the group is infinite to the right, I'm trying to abstract the symbols but this is supposed to be a cyclic group, the only way to get $C$, for example, is to compose $A *A*A$ or $A^3$ .

enter image description here

This means that there is no $A^n = A^1$ (as it would happen with a finite cyclic group) so the only operation that can get $A^1$ is $A^0 * A^1$.

So my question is, why is this arraingment not even a group, what group axioms are being broken?

Please let me know if any clarification is needed.

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  • $\begingroup$ There is no inverse. $\endgroup$ – Hagen von Eitzen Jun 1 '17 at 17:24
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Say $a$ is a generator of an infinite cyclic group,

Do $e,a,a^2,a^3,a^4\dots $ form a group??

No!!

Why? Because inverse of $a$ is not included! as the inverse of $a$ cannot be $a^n$ for some $n\in N$ as then $a^{n+1}=e$ and the group won't be infinite! So you got to take the inverses too in backward direction!

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  • $\begingroup$ The neutral element is also not included. $\endgroup$ – lhf Jun 1 '17 at 18:16
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The existence of inverses fails.

It's exactly the situation between $\mathbb Z$, which is a groups, and $\mathbb N$, which is not.

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  • $\begingroup$ so for example the inverse of $B$ could be $A$ but the inverse of $A$ itself does not exist? $\endgroup$ – Joaquin Brandan Jun 1 '17 at 17:28
  • $\begingroup$ @JoaquinBrandan, yes. $\endgroup$ – lhf Jun 1 '17 at 17:31

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