How does one work out which elements map from, say, $(\mathbb{Z}/n\mathbb{Z},+)$ to $S_n$, i.e. if $\theta(1)$ is some permutation in $S_n$, for example?

I understand that ($\mathbb Z/n\mathbb Z,+$) is additive and $S_n$ is multiplicative, and thus $$\theta(u+v) = \theta(u)*\theta(v).$$ My problem is understanding what $u$ and $v$ are in $S_n$.

For an example with numbers: $G$ is the $group ($\mathbb Z/18\mathbb Z,+$), $H$ is the group $S_6$ and $\theta(1) = (13)(56)$.

Why does $[1]$ map to $(13)(56)$ and how could I use this to work out the kernel? I've worked out that $[1]$ has order $18$ in $G$ and $(13)(56)$ has order $2$ in $H$.

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    – Shaun
    Nov 26 '20 at 15:51
  • $\begingroup$ Note that $\Bbb Z/n\BbbZ$ uses additive notation but $S_n$ uses multiplicative. So we'll have $\theta(k)=\theta(1)^k$ $\endgroup$
    – Berci
    Nov 26 '20 at 17:50
  • 1
    $\begingroup$ $\theta(1)$ can be any permutation whose order divides $n$. $\endgroup$
    – Berci
    Nov 26 '20 at 18:17
  • $\begingroup$ The order of the image of an element under a homomorphism must divide the order of the element. $\endgroup$
    – user810157
    Nov 27 '20 at 17:56

Here we are in the lucky case where "order of the group $=$ degree of symmetric group", and Cayley theorem gives us the chance to show at least a particular homomorphism among all those sought for, namely the embedding by left multiplication $\Bbb Z/n\Bbb Z\stackrel{\epsilon}{\hookrightarrow} S_{\Bbb Z/n\Bbb Z}\to S_{n}$ (for a given bijection $f\colon \Bbb Z/n\Bbb Z\to\{1,\dots,n\}$):

\begin{alignat}{1} (f\epsilon(m)f^{-1})(i) &= f(\epsilon(m)(f^{-1}(i))) \\ &= f(m+ f^{-1}(i)) \\ \tag 1 \end{alignat}

For example, take $n=3$ and the bijection $f(0)=1, f(1)=2, f(2)=3$. Then:

\begin{alignat}{1} &(f\epsilon(0)f^{-1})(1)=f(0+f^{-1}(1))=f(f^{-1}(1))=1 \\ &(f\epsilon(0)f^{-1})(2)=f(0+f^{-1}(2))=f(f^{-1}(2))=2 \\ &(f\epsilon(0)f^{-1})(3)=f(0+f^{-1}(3))=f(f^{-1}(3))=3 \\ \end{alignat}

whence the homomorphic image of $0\in\Bbb Z/3\Bbb Z$ in $S_3$ is given by $()$ (the identity of $S_3$). Similarly:

\begin{alignat}{1} &(f\epsilon(1)f^{-1})(1)=f(1+f^{-1}(1))=f(1+0)=f(1)=2 \\ &(f\epsilon(1)f^{-1})(2)=f(1+f^{-1}(2))=f(1+1)=f(2)=3 \\ &(f\epsilon(1)f^{-1})(3)=f(1+f^{-1}(3))=f(1+2)=f(0)=1 \\ \end{alignat}

whence $1\mapsto (123)$, and finally:

\begin{alignat}{1} &(f\epsilon(2)f^{-1})(1)=f(2+f^{-1}(1))=f(2+0)=f(2)=3 \\ &(f\epsilon(2)f^{-1})(2)=f(2+f^{-1}(2))=f(2+1)=f(0)=1 \\ &(f\epsilon(2)f^{-1})(3)=f(2+f^{-1}(3))=f(2+2)=f(1)=2 \\ \end{alignat}

whence $2\mapsto (132)$. To sum up, this particular (injective) homomorphism sends $\Bbb Z/3\Bbb Z$ to $\langle (123)\rangle\le S_3$.


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