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.

**Answer**

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!

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