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For $d \leq n$ define an injective function $\mathbb{Z} /d \mathbb{Z} \rightarrow S_n$ such that group operations are preserved.

I think that I have an idea of how to define such a function. However, I don't know if it is the correct one. So I would like some help with that.

Essentially what I want to create is a function that would send say $[1]_d$ to $\begin{pmatrix} 1 & 2 & \dots & d & d+1 & \dots & n\\ 2 & 3 & \dots & 1 & d+1 & \dots & n \end{pmatrix}$.

I came up with the following "definition": Define $\phi: \mathbb{Z} /d \mathbb{Z} \rightarrow S_n$ to be the function such that $\phi([i]_d) = \sigma_i$, for all $i=0,1,\dots,d-1$. Here $\sigma_i$ is defined as the permutation on $\{ 1,2,\dots,n \}$ such that: $$\sigma_i(j) = \begin{cases} j+i \text{ if } j+i \leq d \\ j + i - d \text{ if } j + i > d \end{cases}$$ when $j \leq d$ and $\sigma_i(j) = j$ otherwise.

I don't believe what I have written is well defined. Even if it is, I'm not even sure how to show that the operations are preserved. Is this "definition" correct? Can we show that $\phi$ indeed preserves group operations?

Note: Sorry for my possible bad latex writing skills. I hope that the way I defined $\phi$ and the motivation for how I defined it is clear. Thank you.

Dinis P.
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  • You can embed $\mathbb Z_d$ into $S_d$ via Cayley's theorem, and then $S_d$ into $S_n$ by fixing $n-d$ elements. – Kan't Jul 27 '24 at 18:38
  • Hint: every group acts on itself by left multiplication (or addition, if the group happens to be abelian). – lulu Jul 27 '24 at 18:38

1 Answers1

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The function $\phi : \mathbb{Z}/d\mathbb{Z} \rightarrow S_n$ you defined is indeed injective and preserves the group operation. This can be seen as follows. The group $\mathbb{Z}/d\mathbb{Z}$ is generated by $[1]_d$ and your function can equivalently be defined by

$$ \phi([1]_d) = \begin{pmatrix} 1 & 2 & \dots & d & d+1 & \dots & n\\ 2 & 3 & \dots & 1 & d+1 & \dots & n \end{pmatrix} $$ and $$ \phi([i]_d) = \phi([1]_d)^i \, . $$ This is well-defined because $\phi([1]_d)^d = \text{id}$ and it is now easy to see that the group operation is preserved, i.e. $\phi(a \circ b) = \phi(a) \circ \phi(b)$. Injectivity is also easy to see, e.g. because the permutations $\phi([0]_d), ..., \phi([d-1]_d)$ map $1$ to $1, ..., d$, respectively. Writing out the action of $\phi([1]_d)^i$ on $j$ one recovers your definition.

user1363745
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