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Here is a question that I can't find the answer to in my notes or my textbook.

Is there any relation between $\text{Sym}_n$ and $\text{Sym}_k$ where $k < n$? Is $\text{Sym}_k \subset \text{Sym}_n$? Since $k! | n!\text{ } \forall k < n $ they satisfy Lagrange's theorem, but of course that doesnt guarantee the existence of a subgroup

I'm trying to use this result to show that that there is a transitive action of a finite group $G$ on a smaller finite set $S$ by showing that $G$ or a subgroup of $G$ is isomorphic to $\Sigma (S)$ (The permutation group of $S$) which has a natural transitive action.

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I'm not sure that I am familiar with $\Sigma(S)$ notation. –  Asaf Karagila Sep 21 '10 at 8:20
    
Sorry, $\Sigma(S)$ refers to the permutation group of S, I added this to the question –  crasic Sep 21 '10 at 8:30
    
More generally, every group is embeddable in a symmetric group. So, just consdier $\text{Sym}_n = G$ a generic group, and you have your answer. –  Andy Sep 21 '10 at 11:40
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2 Answers

up vote 6 down vote accepted

$S_n$ is not contained in $S_m$ when $n<m$ if you define $S_k$ to be the set of all bijections $\{1,\dots,k\}\to\{1,\dots,k\}$, simply because an element of $S_n$ has domain $\{1,\dots,n\}$ and elements of $S_m$ have domain $\{1,\dots,m\}$, and the two sets are different.

Now this is a silly problem, so surely there is some way out...

Suppose again that $n<m$. There is a map $\phi:S_n\to S_m$ such that whenever $\pi\in S_n$, so that $\pi:\{1,\dots,n\}\to\{1,\dots,n\}$ is a bijection, then $\phi(\pi):\{1,\dots,m\}\to\{1,\dots,m\}$ is the map given by $$\phi(\pi)(i)=\begin{cases}\pi(i),&\text{if $i\leq n$;}\\i,&\text{if $i>n$.}\end{cases}$$ You can easily check that $phi$ is well-defined (that is, that $\phi(\pi)$ is a bijection for all $\pi\in S_n$) and that moreover $\phi$ is a group homomorphism which is injective.

It follows from this that the image of $\phi$ is a subgroup of $S_m$ which is isomorphic to $S_n$. It is usual to identify $S_n$ with its image $\phi(S_n)\subseteq S_m$, and then we can say that «$S_n$ is a subgroup of $S_m$», but this is nothing but a façon de parler.

It should be noted, though, that there are many injective homomorphisms $S_n\to S_m$, so there are many ways to identify a subgroup of $S_m$ with $S_n$. The one I described above is nice because it looks very natural.

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Thanks, very detailed explanation. –  crasic Sep 21 '10 at 20:08
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Yes. $Sym_k < Sym_n$ for $n < k$ and it corresponds to a subgroup of all permutations with the same $k-n$ elements.

(That is, fix $k-n$ elements first, then look at all the transpositions that keep them in place. This subgroup of $S_k$ is isomorphic to $S_n$.)

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