0

Theorem: Every group is isomorphism to a group of permutation.

I took some time to revisit a theorem that I couldn't understand before. It seems that while I can now understand the entire proof, I cannot fully decipher the author's motivation behind the outset of his proof.

Proof:

We seek to construct an isomorphism from any arbitrary group G to a permutation group $\bar{G}$. To do so, the author argues that it is necessary to construct first a map from G to G by defining the map to be

$T_{g}: G\rightarrow G$

$x \mapsto T_{g}\left ( x \right )=gx$

Why a construction of a map from G to G necessary?

Any illumination of my doubt is greatly appreciated.

Kan't
  • 4,819
Mathematicing
  • 6,541
  • Who said it was necessary? It's quite sensible though: try your own hand at proving every $G$ is isomorphic to a permutation group without using $G$ itself to define the set that $G$ acts on! – anon Jun 25 '17 at 14:24
  • this theorem is more a relic of the past than a theorem – mercio Jun 25 '17 at 14:25
  • @arctictern I certainly didn't thought it was necessary. I was puzzled why the author thought it necessary to construct a map from G to G. – Mathematicing Jun 25 '17 at 14:43
  • @Mathematicing Where did the author say they thought it was necessary? – anon Jun 25 '17 at 14:44
  • He mentioned that to construct the group of permutation one has only the group G (domain) to work with. – Mathematicing Jun 25 '17 at 14:45
  • Yes, to construct a group action of $G$ on a set $X$, the only thing you have to start with is the group $G$, nobody tells you what set $X$ to use, you have to figure out a set $X$ to use - and the one set you already have on hand by hypothesis is $G$'s underlying set itself. Do you disagree with that? – anon Jun 25 '17 at 14:46
  • @arctictern I have to revisit Group action before I can answer that. My initial doubt was why wasn't it permissible to define a map from an element $g$ in the domain to a permutation in the range. But Riju raise a similar explanation concerning Group action so I might have to refresh my memory on Group action. I certainly wasn't expecting the author to make an argument from Group action. – Mathematicing Jun 25 '17 at 14:55
  • "why wasn't it permissible to define a map from an element $g$ in the domain to a permutation in the range" - do you mean a group homomorphism $G\to\mathrm{Perm}(G)$? Because that is what is being done. If you mean something else, then I don't know what you mean. – anon Jun 25 '17 at 15:43

6 Answers6

1

Note that there's not just one map from $G$ to $G$ being constructed, but rather a separate map $T_g$ for each element $g$ in $G$. A permutation is a map from a set to itself, so it's perhaps not surprising that it's necessary to construct some to establish an isomorphism to a permutation group.

aPaulT
  • 998
0

In order to show that a finite group is isomorphic to a group of permutations, it is necessary to find a finite set $X$ over which it acts by permutations. In general, the only natural construction is the following: since (left) multiplication by any $g \neq 0$ is a group automorphism, any element $g \in G$ induces by left multiplication a permutation of the set $G$.

The map from $G$ to $\mathrm{Perm}(G)$ that associates to any $g$ the corresponding permutation is a group monomorphism, and this proves Cayley's theorem.

Francesco Polizzi
  • 4,520
0

Perhaps it is worth to mention a bit more on the motivation. We have the following very interesting question, see MO, and MSE, or also here etc.:

Question: Given a finite group $G$, how do I find the smallest $n$ for which $G$ embeds in $S_n$?

This is a refinement of Cayley, although for some groups one cannot really do better than Cayley - for example, $Q_8$ embeds into $S_8$, but into no smaller $S_n$.

Dietrich Burde
  • 140,055
  • Perhaps $C_p$ ($p$ prime) is an easier example than $Q_8$, considered that $p\nmid( p-k)!$ for any $k=1,\dots,p-1$. – Kan't Sep 10 '24 at 19:58
0

The claim is that any group is isomorphic to a subgroup of $S_{n}$. Now you want to prove the claim but the problem is that what $"n"$, you want to choose. Now the injection or should I call embedding of $G$ into $S_{G}$ is what is popularly known as the "Cayley's theorem" , in most books is proved as you have written down. But a more conceptual way to look at this embedding is via group action.

We all know that what group action is right. It's G acting on some set say $X$ that is a map from $G\times X\to X$, satisfying some axioms.

What is a alternative way to look at a group action. You can easily derive from the previous know definition of G action on a set $X$, along with the axioms that it is nothing but a group homomorphism say

$$\phi:G\to S_{X}$$

where $S_{X}$ is the set of all bijections on X with composition as the binary operation. $S_{X}$ is basically a group. When $X$ is finite of size $n$, it is the popularly known Symmetric group $S_{n}$.

Now using this visualisation of group action, We can see the following group action G action on G itself by "left-multiplication" i.e

$$. :G\times G\to G $$ $$(a,g)\mapsto a.g$$

This is essentially a group action. And if you visualize this action via the other way you get a map $\phi:G\to S_{G}$, and that map is exactly the map you have written. This map eventually turns out to be injective and hence the result.

But the question about whether that $"n"$ can be made smaller than $|G|$, is tough question in general to ask! I don't know what's the answer to it. But in some cases $n$ cannot be made smaller. For example in case of $Q_{8}$, it is good exercise to prove that $Q_{8}$ cannot be embedded inside $S_{n}$ for $n\leq 7$.

Riju
  • 4,215
0

A permutation on a set $X$ is a function $f:X \to X$ which is bijective. This is equivalent to the statement that $f$ has an inverse $g$ (ie: $fg = gf = Id_x$)

Now look at the group axioms: we ask for each element to be invertible! So it works a lot like invertible functions (ie permutations)

Now suppose you want to prove that a group $G$ is indeed a permutation group. You have to actually find a set on which $G$ is permuting it's elements. The most natural choice, of course, is $G$ itself (since we're doing it abstractly)

Using the two previous remarks, this map $T_g$ seems to be actually quite natural

Thiago
  • 798
  • 4
  • 14
0

We ought to ask this to Mr. Cayley. It looks like he -soon after having set up the definition of abstract group precisely by mimicking the properties of $\operatorname{Sym}(X)$ (for whatever set $X$)- wanted to reassure everybody that every abstract group actually "lives back" as copy in $\operatorname{Sym}(X)$, for every set $X$ equipotent to the underlying one of $G$.

Kan't
  • 4,819