Can someone explain Cayley’s Theorem step by step?

This is from Fraleigh’s First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map \lambda_x (g) = xg. Can someone explain this proof step by step? Thank you so much! Here is the proof:

Let G be a group. We show that G is isomorphic to a subgroup of S_G.

Define a one-to-one function \phi: G \to S_G such that \phi(xy)=\phi(x) \phi(y) for all x,y \in G. For x \in G, let \lambda_x: G \to G be defined by \lambda_x (g) = xg for all g \in G. The equation \lambda_x (x^{-1} c) = x(x^{-1} c) = c for all c \in G shows that \lambda_x maps G onto G. If \lambda_x (a) = \lambda_x (b), then xa=xb so a=b by cancellation. Thus \lambda_x is also one to one, and is a permutation of G. We now define \phi: G \to S_G by defining \phi(x) = \lambda_x for all x \in G.

To show that \phi is one to one, suppose that \phi(x) = \phi(y). Then \lambda_x = \lambda_y as functions mapping G into G. In particular \lambda_x (e) = \lambda_y (e), so xe=ye and x=y. Thus \phi is one to one. It only remains to show that \phi(xy) = \phi(x) \phi(y), that is, that \phi_{xy} = \lambda_x \lambda_y. Now for any g \in G, we have \lambda_{xy} (g) = (xy)g. Permutation multiplication is function composition, so (\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg). Thus by associativity, \lambda_{xy} = \lambda_x \lambda_y.


First let’s think about what Cayley’s theorem is trying to do. The desired conclusion is that every finite group is isomorphic to a subgroup of the symmetric group. In order to do this, we prove that the group operation defines permutations of the elements of the group. How do we do this?

Step 1. Suppose we line all of the elements of the G up in some arbitrary order and number them from left to right, like so: \begin{array}{ccccc}1 & 2 & 3 & \cdots & n \\ a_1 & a_2 & a_3 & \cdots & a_n \end{array}
Step 2. Now pick an element x\in G. Let’s left multiply all of the elements of G by x.
\begin{array}{ccccc}1 & 2 & 3 & \cdots & n \\ xa_1 & xa_2 & xa_3 & \cdots & xa_n \end{array}

Step 3. Multiplying on the left by x rearranges the elements of G. Recall that in step 1, we assigned all the elements of G a number. That means we can look up which element xa_i was in step 1 and write xa_i=a_j. This is not very far from saying x sends the element at position i to position j.” So let’s define a function from \lambda_x:\{1,\ldots,n \}\rightarrow \{1,\ldots,n \} by \lambda_x(i)=j when x sends a_i to a_j. This is just another way of writing what we did in step 2: \begin{array}{ccccc}1 & 2 & 3 & \cdots & n \\ a_{\lambda_x(1)} & a_{\lambda_x(2)} & a_{\lambda_x(3)} & \cdots & a_{\lambda_x(n)} \end{array}

Step 4. Finally we define \phi:G\rightarrow S_n by \phi(x)=\lambda_x. In other words, we’re associating the element x with the permutation of \{1,\ldots, n\} that x induces – that is, the permutation \lambda_x. It’s easy to verify that \phi is an injective homomorphism. So, we’re done.

Note that there are a couple details missing from the above proof which you’ll find in Fraleigh’s (e.g. that \lambda_x is a bijection for any x). I recommend you comb through the proof again and prove these subtle points to make sure you understand everything. Good luck!

Source : Link , Question Author : user72625 , Answer Author : Alexander Gruber

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