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My textbook writes Cayley's theorem on groups as follows:

Every finite group $G$ of order $n$ is isomorphic to a permutation group. By permutation group is meant a subgroup of the group of permutations of $G$.

My question is why it gives the restricted statement? It seems like we could say

Every group $G$ is isomorphic to a transformation group.

That is, every group $G$ is isomorphic to a subgroup of the group of bijections on $G$. My book uses permutation to mean transformation for the case of a finite group.

Is the first statement somehow more useful? Am I incorrect with the second statement?

EE18
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    The map $g \mapsto (h \mapsto gh)$ injects $G$ into $\text{Bij}(G)$. Probably your text just wants to focus on finite groups and so doesn't try to state any results for infinite groups. – Charles Hudgins Mar 20 '23 at 00:51
  • @CharlesHudgins Is the result somehow more powerful/useful for finite groups? It's curious since this text otherwise gives the general statement where allowed. – EE18 Mar 20 '23 at 00:56
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    A common formulation of it is that any finite group $G$ embeds in some $S_n$, and many introductory group theory textbooks don't cover the idea of permutation groups attached to arbitrary sets. – anomaly Mar 20 '23 at 00:58
  • $S_n$ being the permutation group of the finite set ${1,...,n}$ in particular? @anomaly – EE18 Mar 20 '23 at 00:59
  • Right (or, in no greater generality, an arbitrary fixed set of cardinality $n$). There's no inherent difficult in jumping to the infinite case, but I suppose a fair number of textbooks concentrate on proving results (e.g., the Sylow theorems) by combinatorial methods that may be harder to generalize. – anomaly Mar 20 '23 at 01:02
  • Noted, thank you all! – EE18 Mar 20 '23 at 01:04
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    Historically, Cayley's Theorem was Cayley's way of letting everyone know that his "abstract" notion of a group was really no different than the "traditional" notion of a group, which was that a group was a collection of permutations on a finite set that was closed under composition. He was saying "it is clear everything that used to be called a "group" is a group under my definition, but don't worry, everything which is a group under my definition is also a "group" under the old definition." And the notion of infinite sets and working with infinite sets as completed objects was still new. – Arturo Magidin Mar 20 '23 at 01:08
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    See this previous question/answer. – Arturo Magidin Mar 20 '23 at 01:10
  • The finite case fomulation allows for the classical version of the theorem, namely: every finite group $G$ is isomorphic to a subgroup of $S_n$, where $n:=|G|$. – citadel Mar 20 '23 at 06:11

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You are right that Cayley's theorem applies to every group, assuming we talk about bijections. But what are permutations? Are they different from bijections? And why we mostly consider them on finite sets?

This is of course a matter of nomenclature. However typically when we think about permutations, we consider bijections over a finite set. Because when we think about permutations, we also think about their properties, for example every permutation has a sign. It is an important property, but it does not generalize to arbitrary bijections. If we want to keep good properties of permutations, we can define it as follows: a permutation is a bijection $f:X\to X$ such that $f(x)=x$ for all but finitely many $x\in X$. Such permutations have well defined sign, and we can say more about a group of permutations $S(X)$. For example it has a subgroup $A(X)$ of index $2$, which is simple. The more general group of all bijections $Bi(X)$ is waaay too complicated. The problem is that Cayley's theorem does not hold for $S(G)$, except for the special case when $G$ is finite.

And so, in infinite case it is a trade-off: either we have "good" bijections, but then Cayley doesn't hold or Cayley holds, but we have "bad" bijections, too wild for proper analysis.

freakish
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