**PROOF**: To prove this, let G be any group. We must find a group of permutations that we believe is isomorphic to G. Since G is all we have to work with, we will have to use it to construct . For any g in G, define a function T_{g} from G to G by T_{g }(x) = g_{x} for all x in G.

T_{g} is just multiplication by g on the left. We have to prove that T_{g} is a permutation on the set of elements of G. Now, let __G__ ={T_{g} | g ∈G}. Then, is a group under the operation of function composition. To verify this, we first observe that for any g and h in G we have T_{g}T_{h} (x) = T_{h}(T_{h}(x)) = T_{g}(hx) = g(hx) =(gh)x = T_{gh}(x), so that T_{g} T_{h} = T_{gh}. From this it follows that T_{e} is the identity and (T_{g})^{-1} = T_{g}^{-1} is associative, we have verified all the conditions for to be a group.

The isomorphism ∅ between G and is now ready-made. For every g in __G__, define ∅(g) = T_{g}. If T_{g} = T_{h}, then T_{g}(e) = T_{h}(e) or ge = he.

Thus, g = h and ∅ is one-to-one. By the way was constructed, we see that f is onto. The only condition that remains to be checked is that ∅ is operation-preserving. To this end, let a and b belong to G. Then

∅ (ab) = T_{ab} = T_{a}T_{b} = ∅ (a) ∅ (b).