Permutation group is a subgroup of another permutation group

Question

I'm trying to understand, how is it possible to know if $S_3= \{ (), (1,2), (2,3), (1,3), (1,2,3), (1,3,2)\}$ is a subgroup of $S_{27}$.

I tried to calculate $27!$ and see if it's divisible by $3! = 6$, to check if it satisfies Lagrange's Theorem, but $27!$ is a very large number.

( + Quick Second question : how to check associativity in permutation groups, as it's one of the condition for it to be a group ?)

Thank you in advance !

Answer 1

Recall that $S_n$ is the set, under composition, of all bijections from $\{1,\dots,n\}$ to $\{1,\dots,n\}$. Thus consider the map $\varphi: \{1,2,3\}\to\{1,\dots, 27\}$ given by $\varphi(x)=x$ for $x\in\{1,2,3\}$, viewing the output as an element of $\{1,\dots,27\}$. Can you see how to extend this to an embedding of $S_3$ into $S_{27}$?

---

Associativity of permutations follows from associativity of bijections.

Answer 2

Strictly speaking, $S_3\cap S_{27}=\emptyset$ (a map on a set of $3$ elements can't never be equal to one on a set of $27$ elements, isn't it?), and hence asking whether "$S_3\le S_{27}$" is an abuse of language. The correct closest question to that, rather, is whether $S_3\hookrightarrow S_{27}$. Now, consider the map ($X:=\{1,\dots,27\}$):

\begin{alignat}{2} \epsilon: S_3&\longrightarrow& S_{27} \\ \sigma&\longmapsto& \epsilon_\sigma: X &\longrightarrow X \\ &&i&\longmapsto \epsilon_\sigma(i) \\ \tag 1 \end{alignat}

defined by

\begin{alignat}{2} &\epsilon_\sigma(i):=\sigma(i), &&\space\space\text{for} \space i=1,2,3 \\ &\epsilon_\sigma(i):=i, &&\space\space\text{for} \space i=4,5,\dots,27 \\ \tag 2 \end{alignat}

You can prove that:

1. for every $\sigma\in S_3$, we have indeed $\epsilon_\sigma\in S_{27}$, so the writing $(1)$ makes sense;
2. $\epsilon_\sigma=\epsilon_\tau \Rightarrow \sigma=\tau$, so $\epsilon$ is injective;
3. for every $\sigma,\tau\in S_3$, we have that $\epsilon_{\sigma\tau}=\epsilon_\sigma\epsilon_\tau$, so $\epsilon$ is a group homomorphism.

Therefore, yes: $S_3\hookrightarrow S_{27}$. (Actually, with the same construction you get for free the generalization: $S_m\hookrightarrow S_n$ as soon as $n>m$; or even wider: $S_Y\hookrightarrow S_X$ for every set $X$ and $Y\subset X$.)