---- > [!definition] Definition. ([[symmetric group]]) > The [[automorphism|set]] $\text{Aut}_{\mathsf{Set}}(A)$ of [[permutation|permutations]] on a set with $n$ elements forms a [[group]] (under function composition), denoted $S_{n}$. ^a02430 > [!example] Example. ($S_{3}$) > Let's construct the [[Cayley table]] for $S_{3}$. Let $\sigma$ be the transposition $(12)$ and $\tau$ be the 3-cycle $(123)$. Then we have > | * | $e$ | $\sigma$ | $\tau$ | $\tau^{2}$ | $\sigma \tau$ | $\sigma \tau^{2}$ | |---|---|---|---|---|---|---| | $e$ | $e$ | $\sigma$ | $\tau$ | $\tau^{2}$ | $\sigma \tau$ | $\sigma \tau^{2}$ | | $\sigma$ | $\sigma$ | $e$|$\sigma \tau$ | $\sigma$ $\tau^{2}$ | $\tau$ | $\tau^{2}$ | | $\tau$ | $\tau$ | $\sigma \tau^{2}$ | $\tau^{2}$ | $e$ | $\sigma$ | $\sigma \tau$ | | $\tau^{2}$ | $\tau^{2}$ | $\sigma \tau^{2}$ | $e$ | $\tau$ | $\sigma \tau^{2}$ | $\sigma$ | | $\sigma \tau$ | $\sigma \tau$ |$\sigma$ |$\sigma \tau^{2}$ | $\sigma$ | $e$ | $\tau$ | | $\sigma \tau^{2}$ | $\sigma \tau^{2}$ | $\sigma$ | $\tau$ | $\sigma \tau$ | $\tau^{2}$ | $e$| ^6e305f ![[CleanShot 2023-08-31 at 15.40.50.jpg]] The key observation here is that $\tau \sigma = \sigma \tau^{2}$ and of course that $\tau^3=e=\sigma^{2}$. Applying $\sigma$ sort of 'changes the direction' that $\tau$ rotates— $\tau \sigma \tau = \sigma$, for example. That observation lets one easily notice that $\tau \sigma \tau^{2}=\sigma \tau$, etc. > [!justification] > This is indeed a [[group]]: the identity element is the [[identity map]], inverses all exist by the definition of [[permutation]], and function composition is [[associative]]. > [!note] Remark. > $S_{3}$ is [[group isomorphis|isomorphic to]] the [[group]] $D_{3}$ of *rigid similarities of an equilaterial triangle*: ![[CleanShot 2023-08-31 at 17.34.30.jpg]] as well as to the [[subgroup]] of $GL_{2}$ consisting of the corresponding rotation and reflection matrices (imagining the triangle above as being embedded in $\mathbb{R}^{2}$ and viewing its image under reflection/rotation linear maps) . ^03dd93 This isn't the only [[group isomorphism|isomorphism]] possible; in fact, there are $6$ possible in total (3 valid manners of cycling $l_{2},l_{3},l_{1}$ per 2 ways to permute $\tau$ and $\tau^{2}$ for $3 \cdot 2=6$ total possibilities). These are:![[CleanShot 2023-09-05 at [email protected]]] we will denote the maps as $\phi_{1},\dots \phi_{6}$ respectively, and represent the rotations by $0$, $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ as $r_{0}$, $r_{\frac{2\pi}{3}}$ and $r_{\frac{4\pi}{3}}$ hereon. As explicitly depicted above, each map is a [[bijection]]. To verify they are all [[group homomorphism|homomorphisms]], it suffices to show for all $i \in [6]$ that $\begin{align} \phi_{i}(\tau \cdot \tau)=&\phi_{i}(\tau) \cdot \phi_{i}(\tau) ,\\ \phi_{i}(\tau \cdot \sigma)= & \phi_{i}(\tau)\cdot \phi _{i}(\sigma), \\ \phi_{i}(\sigma \cdot \tau)= & \phi_{i}(\sigma) \cdot \phi _{i}(\sigma), \text{ and } \\ \phi_{i}(\sigma \cdot \sigma)= & \phi_{i}(\sigma) \cdot \phi _{i}(\sigma), \end{align}$ since any other input product is a word in ($\tau$ and $\sigma$), and $\cdot$ is [[associative]]. # For $\phi_{1}$ We have $\begin{align} \phi_{1}(\tau \cdot \tau)= & \phi_{1}(\tau^{2})=r_\frac{4\pi}{3}=r_\frac{2\pi}{3} \cdot r_\frac{2\pi}{3}=\phi_{1}(\tau)\cdot \phi_{1}(\tau) \\ \phi_{1}(\tau \cdot \sigma)= & \phi_{1}(\sigma \tau^{2})=\ell_{1}= \ell_{2} \cdot r_{\frac{4\pi}{3}}=\phi_{1}(\sigma) \cdot \phi_{1}(\tau^{2}) \\ \phi_{1}(\sigma \cdot \tau)= & \phi_{1}(\sigma \tau)= \ell_{3} = \ell_{2} \cdot r_{\frac{2\pi}{3}} = \phi_{1}(\sigma) \cdot \phi_{1}(\tau) \\ \phi_{1}(\sigma \cdot \sigma) = & \phi_{1}(e)= r_{0} = \ell_{2} \cdot \ell_{2} = \phi_{1}(\sigma) \cdot \phi_{1}(\sigma). \end{align}$ # For $\phi_{2}$ We have $\begin{align} \phi_{2}(\tau \cdot \tau)= & \phi_{2}(\tau^{2})=r_\frac{4\pi}{3}=r_\frac{2\pi}{3} \cdot r_\frac{2\pi}{3}=\phi_{2}(\tau)\cdot \phi_{2}(\tau) \\ \phi_{2}(\tau \cdot \sigma)= & \phi_{2}(\sigma \tau^{2})=\ell_{3}= \ell_{1} \cdot r_{\frac{4\pi}{3}} =\phi_{2}(\sigma) \cdot \phi_{2}(\tau^{2}) \\ \phi_{2}(\sigma \cdot \tau)= & \phi_{2}(\sigma \tau)= \ell_{2} = \ell_{1} \cdot r_{\frac{2\pi}{3}} = \phi_{2}(\sigma) \cdot \phi_{2}(\tau) \\ \phi_{2}(\sigma \cdot \sigma) = & \phi_{2}(e)= r_{0} = \ell_{1} \cdot \ell_{1} = \phi_{2}(\sigma) \cdot \phi_{2}(\sigma). \end{align}$ # For $\phi_{3}$ We have $\begin{align} \phi_{3}(\tau \cdot \tau)= & \phi_{3}(\tau^{2})=r_\frac{4\pi}{3}=r_\frac{2\pi}{3} \cdot r_\frac{2\pi}{3}=\phi_{3}(\tau)\cdot \phi_{3}(\tau) \\ \phi_{3}(\tau \cdot \sigma)= & \phi_{3}(\sigma \tau^{2})=\ell_{2}= \ell_{3} \cdot r_{\frac{4\pi}{3}} =\phi_{3}(\sigma) \cdot \phi_{3}(\tau^{2}) \\ \phi_{3}(\sigma \cdot \tau)= & \phi_{3}(\sigma \tau)= \ell_{1} = \ell_{3} \cdot r_{\frac{2\pi}{3}} = \phi_{3}(\sigma) \cdot \phi_{3}(\tau) \\ \phi_{3}(\sigma \cdot \sigma) = & \phi_{3}(e)= r_{0} = \ell_{3} \cdot \ell_{3} = \phi_{3}(\sigma) \cdot \phi_{3}(\sigma). \end{align}$ For $\phi_{4},\phi_{5},\phi_{6}$, we repeat the above work, this time swapping $\tau$ for $\tau^{2}$. This entails counting the $3$-cycles from the first three maps *backward*. # For $\phi_{4}$ (compare to $\phi_{1}$) We have $\begin{align} \phi_{4}(\tau \cdot \tau)= & \phi_{4}(\tau^{2})=r_\frac{2\pi}{3}=r_\frac{4\pi}{3} \cdot r_\frac{4\pi}{3}=\phi_{4}(\tau)\cdot \phi_{4}(\tau) \\ \phi_{4}(\tau \cdot \sigma)= & \phi_{4}(\sigma \tau^{2})=\ell_{3}= \ell_{2} \cdot r_{\frac{2\pi}{3}}=\phi_{1}(\sigma) \cdot \phi_{1}(\tau^{2}) \\ \phi_{4}(\sigma \cdot \tau)= & \phi_{4}(\sigma \tau)= \ell_{1} = \ell_{2} \cdot r_{\frac{4\pi}{3}} = \phi_{4}(\sigma) \cdot \phi_{4}(\tau) \\ \phi_{4}(\sigma \cdot \sigma) = & \phi_{1}(e)= r_{0} = \ell_{2} \cdot \ell_{2} = \phi_{4}(\sigma) \cdot \phi_{4}(\sigma). \end{align}$ # For $\phi_{5}$ (compare to $\phi_{2}$) We have $\begin{align} \phi_{5}(\tau \cdot \tau)= & \phi_{5}(\tau^{2})=r_\frac{2\pi}{3}=r_\frac{4\pi}{3} \cdot r_\frac{4\pi}{3}=\phi_{5}(\tau)\cdot \phi_{5}(\tau) \\ \phi_{5}(\tau \cdot \sigma)= & \phi_{5}(\sigma \tau^{2})=\ell_{1}= \ell_{3} \cdot r_{\frac{2\pi}{3}}=\phi_{5}(\sigma) \cdot \phi_{5}(\tau^{2}) \\ \phi_{5}(\sigma \cdot \tau)= & \phi_{5}(\sigma \tau)= \ell_{2} = \ell_{3} \cdot r_{\frac{4\pi}{3}} = \phi_{5}(\sigma) \cdot \phi_{5}(\tau) \\ \phi_{5}(\sigma \cdot \sigma) = & \phi_{5}(e)= r_{0} = \ell_{3} \cdot \ell_{3} = \phi_{5}(\sigma) \cdot \phi_{5}(\sigma). \end{align}$ # For $\phi_{6}$ (compare to $\phi_{3}$) We have $\begin{align} \phi_{6}(\tau \cdot \tau)= & \phi_{6}(\tau^{2})=r_\frac{2\pi}{3}=r_\frac{4\pi}{3} \cdot r_\frac{4\pi}{3}=\phi_{6}(\tau)\cdot \phi_{6}(\tau) \\ \phi_{6}(\tau \cdot \sigma)= & \phi_{6}(\sigma \tau^{2})=\ell_{2}= \ell_{1} \cdot r_{\frac{2\pi}{3}}=\phi_{5}(\sigma) \cdot \phi_{5}(\tau^{2}) \\ \phi_{6}(\sigma \cdot \tau)= & \phi_{6}(\sigma \tau)= \ell_{3} = \ell_{1} \cdot r_{\frac{4\pi}{3}} = \phi_{6}(\sigma) \cdot \phi_{6}(\tau) \\ \phi_{6}(\sigma \cdot \sigma) = & \phi_{6}(e)= r_{0} = \ell_{1} \cdot \ell_{1} = \phi_{6}(\sigma) \cdot \phi_{6}(\sigma). \end{align}$ To see that these are the only [[group isomorphism|isomorphisms]], we note: ![[CleanShot 2023-09-05 at [email protected]]] ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```