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The symmetric group is the set of all permutations.

My question addresses the representability of the symmetric group using only additions. I am guessing that on the finite field $\mathbb{Z}/n \mathbb{Z}$ by defining a "Cartesian" basis $(1,0,,0,...),(0,1,0...),...$

Could we then say that the symmetric group is isomorphic to the abelian group?

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    $\begingroup$ I'm not sure I understand your question. The symmetric group on a set $A$ is abelian if and only if $|A| \le 2$. $\endgroup$ – user61527 Jul 17 '14 at 6:31
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You are mixing up a few concepts here.

  • The symmetric group $S_n$ is the group of permutations on $[n]=\{1,2,\dots,n\}$, i.e. $$S_n = \{ \varphi : [n]\to[n] \,|\, \text{$\varphi$ is bijective}\}.$$ The group operation on $S_n$ is given by composition of maps.
  • The group $\mathbb Z/n\mathbb Z$ consists of cosets of $n\mathbb Z$ in $\mathbb Z$, i.e. $$\mathbb Z/n\mathbb Z = \{ \overline 0, \overline 1, \dots,\overline{n-1} \},$$ where $\overline k = k + n\mathbb Z$. The group operation is induced by the addition on $\mathbb Z$, so $\overline k + \overline l = \overline{k+l}$.
  • An abelian group, is a group $(G,\bullet)$ where $g\bullet h=h\bullet g$ holds for all $g,h\in G$. In particular, there is not the abelian group, there are many!
  • Finite tuples of the form $(1,2,3,4),(0,1,0,0),\dots$ with integer entries, are elements of $\mathbb Z^n$, in this case $\mathbb Z^4$, which is the direct product $\mathbb Z\times\mathbb Z\times\mathbb Z\times\mathbb Z$ and is in no way related to $\mathbb Z/n\mathbb Z$, which consists of cosets, not tuples. The group operation on direct products is given component-wise, for example $$(1,2,3,4)+(0,1,0,0)=(1+0,2+1,3+0,4+0)=(1,3,3,4)$$ in $\mathbb Z^4$.
  • Infinite tuples $(1,2,3,4,5,\dots),(1,0,0,\dots),(0,1,0,0,\dots),\dots$ are elements of $\prod_{i=1}^\infty \mathbb Z$, which is the direct product of countably many copies of $\mathbb Z$.

All of the groups $\mathbb Z$, $\mathbb Z/n\mathbb Z$, $\mathbb Z^n$ and $\prod_{i=1}^\infty \mathbb Z$ are abelian, since the addition in $\mathbb Z$ is commutative.

The symmetric group $S_n$ is abelian only for $n=1,2$. As soon as you have at least 3 elements, you can define bijective maps like \begin{align} \sigma : [3] &\to [3] & \rho : [3] &\to [3]\\ 1 &\mapsto 2 & 1&\mapsto 1 \\ 2 &\mapsto 1 & 2&\mapsto 3 \\ 3 &\mapsto 3 & 3&\mapsto 2, \end{align} where $\sigma$ swaps $1$ and $2$ (often written $\sigma=(12)$, using cycle notation), and $\rho$ swaps $2$ and $3$ ($\rho=(23)$). I leave it to you to check that $\sigma\circ\rho \neq\rho\circ\sigma$. This shows that $S_n$ is not abelian when $n\ge 3$. In particular it is not isomorphic to any of the abelian groups mentioned above.

The only non-trivial abelian symmetric group is $S_2$, it is in fact isomorphic to $\mathbb Z/2\mathbb Z$, since there is only one possible group operation on a set with $2$ elements.

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