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I am learning some basic group theory, and just getting my head around permutations. In particular, I want to consider the counter-example noted in the title.

Let $X = \{1, 2, 3\}$, then why is $\sigma$, where $$ \sigma = \left(\begin{matrix} 1 & 2 \\ 2 & 3\end{matrix}\right),$$ not a permutation?

The reason I have in my notes, which I happily accepted during class, is that the above permutation is not "closed"; that is, it does not map $Y \mapsto Y$ where $Y = \{1, 2\}$. But it does $X \mapsto X$, if I define $X$ as above.

So is it not a permutation on $X$?

We know that all permutations are bijective. For injectivity, we can checik if $\sigma$ has an inverse, and it does:

$\sigma^{-1} = \left(\begin{matrix}2 & 3 \\ 1 & 2\end{matrix}\right).$

However, I think that $\sigma$ is not surjective, since it does not map every element of $X$ to some other element in $X$. So, $\sigma$ is not bijective.

Am I understanding the counter-example correctly?

bzm3r
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    Yes permutations are bijections by definition. Yours is not sending 3 anywhere. You must define were 3 is going. And there is really only one choice ... – Nicky Hekster Sep 27 '17 at 18:26
  • @AlvinLepik sorry, I meant, we can check if it has a left inverse – bzm3r Sep 30 '17 at 01:29
  • Should you succeed in defining $\tau$ such that $\sigma\tau$ and $\tau\sigma$ are identity mappings, then you have shown $\sigma$ is bijective, therefore injective AND surjective. But you must make sure your mapping $\tau$ is well-defined. – AlvinL Sep 30 '17 at 06:55

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A permutation is a type of function, and a function defined on a domain has to act on every element of the domain. The function you are defining is not a mapping from $X$ to $X$, but rather from a subset of $X$ into $X$. To be a permutation, a function must first be a mapping from a set to itself, and that means from the entire set to itself.

G Tony Jacobs
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