Under what conditions can the symmetric group be isomorphic to the abelian group?

Question

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?

Answer 1

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.