0
$\begingroup$

I'm studying deeply, starting from different POVs, group actions, stabilizers, etc. Quoting from Rotman's Theory of Groups, p.49:

"We have encoutered several situations in which elements of a group may be regarded as permutations of a set."

Actually I was wondering why this is true just for "several situations" and not for "all situations". Try to be more precise with my doubt.

Actually Cayley stated "every group G is isomorphic to a subgroup of a given symmetric group acting on G", so roughly speaking "everything is a permutation group (where "is" is up to isomorphisms, of course).

More than this, a group action can be seen as a homorhism between a group and the Sym(X), i.e. the group of permutation on the set X (having X coinciding with G itself, eventually).

Of course, the action can be faithful or not, but my feeling every element of every group can be seen as a permutation of something.

Am I right in stating so? I guess this is also the same view Galois & Co had initially, before introducing "modern" abstract group", where every group was seen as a permutation of something.

May you help where my sentence could be eventually wrong, please?

thanks in advance

Ricky

$\endgroup$
  • $\begingroup$ "We have encoutered several situations in which elements of a group may be regarded as permutations of a set." This sentence is rather inprecise. $\endgroup$ – Wuestenfux Jul 18 '17 at 12:49
  • 1
    $\begingroup$ Your sentence "my feeling (is that) every element of every group can be seen as a permutation of something" is true: this is Cayley's theorem! $\endgroup$ – lokodiz Jul 18 '17 at 12:51
  • $\begingroup$ For an explicit way to associate a permutation to an element of a group $G$, look at the multiplication map $h \rightarrow gh$ for $g, h \in G$. This induces a permutation in $\text{Sym}(G)$ by left-multiplication with $g$. $\endgroup$ – Tob Ernack Jul 18 '17 at 12:54
  • $\begingroup$ @lokodiz and Tob, yes that's Cayley theorem for sure, but this theorem refers of course to Sym(G), i.e. to permutations of elements of the same group G. My doubt was more about "for a given group G, is there always a set X onto which it can act without being G itself?". Does this question make sense? $\endgroup$ – riccardoventrella Jul 18 '17 at 15:29
  • $\begingroup$ Supposing $|G| = n < \infty$, enumerate the elements of $G$ as $G = \{ g_1, \dots, g_n \}$, and let $G$ act on $X = \{ 1, \dots , n \}$ via $g \cdot i = j$ where $gg_i = g_j$. Of course, now we're just acting on $G$ in disguise. But we could set $X' = X \cup \{ n+1 \}$ and let $G$ act on $X'$ by acting on $1, \dots, n$ as above, and acting on $n+1$ trivially. So now we have a faithful action on a set which is different to the set $G$ (i.e., not isomorphic as a set). $\endgroup$ – lokodiz Jul 18 '17 at 16:06
0
$\begingroup$

It's true that every group can be seen as a permutation of SOMEthing.

But the case we're often interested in is where we have some object $X$, and seek a group $G$ that acts on $X$. An example is where $X$ is a surface, with the group being the group of isometries of the surface. Another example is the set of all possible configurations of a Rubik's cube, with the group being the "operations" one can perform on a cube.

If I give you a Rubik's cube and the group $\Bbb Z/37\Bbb Z$, you're not going to be able to make the latter act on the former (except trivially) --- every element of the group corresponds to "do nothing to the cube."

So the interest isn't "goes a group act on something?" but rather, "When I've got a group acting (nontrivially) on a set $X$, what can I say in general?"

In particular, you might be able to say that the group is not very large (e.g., there are no nontrivial self-isometries of this particular surface), or that it's abelian, or that it has certain subgroups, etc.

To answer your explicit question about what Rotman is saying, let me rephrase. I think he means

We've seen several $(G, X)$ pairs in which $G$ can be said to be acting on $X$,

rather than

We've seen several situations in which for some group $G$, there's a set $X$ with $G$ acting on $X$,

since the latter is (as you observe) always true if we pick $X = G$ and use the action by left-multiplication, say.

$\endgroup$
  • $\begingroup$ I see your point @John and I agree, this is what I meant. That said I think there's a general misconcetpion, at least as far as I'm googling the net, to identify a group according to the way in which it acts. A group can acts differently on different sets. For example D8 is the symmetry group of the square, which acts both on the whole object (square) and on the set of its vertices. But if I pick A4, it still acts on the vertices, but can't be applied to the whole square, since it would break it withsome permutations. This of course implies A4 is not the automorphism group of the square. $\endgroup$ – riccardoventrella Jul 18 '17 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.