2

Not a duplicate of this exquisite answer, the numbers in which I abide by here. Not querying the proof, hence please don't discourse on it. Proof blueprint:

Steps 1-2 in words. Left multiplication of every element in the group by a fixed element in the group constitutes a permutation of those elements.

Steps 1-2 in math. We must prove $\lambda_x(g) = xg$ for all $g, x \in G$ is a bijection $: G \to G$.
Or see Fraleigh p. 87 exercise 8.52: $p_x(g) = gx$ operates too.

Steps 3-4 in words. As the fixed elements used as multipliers $\color{brown}{\text{range over all elements in the group}}$, we get a family of permutations, one for each multiplier. The idea is to show that this family forms a group with the same fundamental structure as the original group.

Steps 3-4 in math. We must prove $\theta(x) = \lambda_x$ for all $x \in G$ is an injective homomorphism $: G \to S_G$. Or see Fraleigh p. 87 exercise 8.52: $u(x) = p_{x^{=1}}$ operates too.

Question (1.) Why left-multiply the elements in $g$ by $x$ in $\lambda_x(g) = xg$? To induce a permutation?

(2.) Why does $\lambda_x(g) = xg$ only need to be a bijection? Why not an isomorphism?

(3.) I condone $\phi \to Im(\phi)$ is a surjection because image = codomain, but still don't understand why Fraleigh doesn't prove $\theta$ is surjective? Related to this?

(4.) Why do we need $\theta(x) = \lambda_x$? It feels redundant. $\lambda_x(g) = xg$ is true for all $x \in G$, hence doesn't $\lambda_x(g) = xg$ $\color{brown}{\text{range over all elements in the group}}$ already ? My course doesn't cover Qiaochu Yuan's comment on currying on top of exquisite answer.

Community
  • 1

1 Answers1

2

Cayley's Theorem says that every group is isomorphic to a permutation group, that is, a subgroup of $S_n$ for some $n$.

1) Yes. We need to realize the elements of $G$ as permutations. This is an example of the left regular representation. That's why the permutation is denoted by $\lambda_x$: lambda is the Greek equivalent of "L", which stands for "left." (Also, that almost certainly a $\rho_x$ for the right regular representation, not a $p_x$, since rho is the equivalent of "R", which stands for "right.")

2) As I said above, we need to realize each element of $G$ as a permutation. And a permutation is simply an bijection $G \to G$, not necessarily an isomorphism.

3) In fact $\theta : G \to S_n$ is not surjective in general. It only maps $G$ onto $S_n$ if $G$ happens to be isomorphic to $S_n$. In general, $G$ is only isomorphic to a subgroup of $S_n$. Of course $\theta$ maps onto its image---that's tautologically true.

4) I think you're getting the maps mixed up. Given $x \in G$, we want to find a permutation that is "the same" as $x$. To do so, we examine how $x$ acts on $G$ by left multiplication by defining the map \begin{align*} \lambda_x : G &\to G\\ g & \mapsto xg \, . \end{align*} Just to emphasize, for each $x$, $\lambda_x$ is a map from $G$ to itself. (In fact, $\lambda_x$ is a bijection since its inverse is just $\lambda_{x^{-1}}$.) We have not yet involved a symmetric group. Finally, we define \begin{align*} \theta : G &\to S_G\\ x &\mapsto \lambda_x \end{align*} where $S_G$ is the symmetric group on $G$, i.e., the set of all bijections of $G$ with the binary operation of functional composition. That is, to each element $x \in G$ we associate the permutation of $G$ that is simply left multiplication by $x$.

Once we prove that $\theta$ is an injective homomorphism, then $\theta$ is an isomorphism onto its image, which is a subgroup of $S_G$. This shows that $G$ is isomorphic to a subgroup of a permutation group.

Does that answer all your questions?

Viktor Vaughn
  • 20,897