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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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  • $\begingroup$ Welcome to math.SE. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – 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
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    $\begingroup$ $\theta(1)$ 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
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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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