OpenCart Templates
Tuesday , August 22 2017
Home / Modern Abstract Algebra / Every group is isomorphic to a group of permutations.

Every group is isomorphic to a group of permutations.

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 Tg from G to G by Tg (x) = gx for all x in G.

Tg is just multiplication by g on the left. We have to prove that Tg is a permutation on the set of elements of G. Now, let G ={Tg |  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  TgTh (x) = Th(Th(x)) = Tg(hx) = g(hx) =(gh)x = Tgh(x), so that Tg Th = Tgh. From this it follows that Te is the identity and (Tg)-1 = Tg-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) = Tg. If Tg = Th, then Tg(e) = Th(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) = Tab = TaTb = ∅ (a) ∅ (b).

Check Also

Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the ...

Leave a Reply

Your email address will not be published. Required fields are marked *