Proof of Subgroup Involving an IsomorphismDate: 11/04/2004 at 14:25:11 From: Bridget Subject: Isomorphisms acting on Groups Suppose ö is an isomorphism from a group G to a group G'. Prove that if K is a subgroup of G, then ö(K) = {ö(k) | k is in K} is a subgroup of G'. We know that K < G. I know to show that a group is a subgroup you must use the two step process of showing that it is closed under the operation ö and closed under the inverse, but I'm not sure how to start this. Could you help me out? Date: 11/05/2004 at 18:09:26 From: Doctor Jordan Subject: Re: Isomorphisms acting on Groups Hi Bridget, Let us express H = o(K). Recall that for H to be a subgroup of G', it must be a nonempty subset of G' such that H is closed under the operation of G', each h in H has an inverse contained in H, and the identity element of G' is contained in H. We are given that K is a subgroup of G, hence it contains the identity element of G, each element in it has an inverse, and it is closed under the operation of G. Since o is an isomorphism, it is a homomorphism, and for e, y in K a homomorphism must satisfy o(ey) = o(e)o(y). In particular, for e is the identity element of K and thus ey = e, then o(e) must also be the identity element of G', as otherwise o(e)o(y) would not equal o(y). Thus we know that H has the identity of G'. By assumption, all elements of H are of the form o(x) for x in G. Let us have two arbitrary elements of H, x and y. By definition of o, they are of the form o(x') = x, o(y') = y for x', y' in K. But since K is a subgroup it is closed and so x'y' in K, hence o(x'y') = o(x')o(y'), which shows that the product of any two elements of H is in H. Can you show that for any element x in H, its inverse is also in H? It is helpful to remember that since K is a subgroup, each element in K has a (unique) inverse in K. If you have any problems with this or any other questions, please write me back! - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ |
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