Proof that f(K) is a Subgroup of G'Date: 11/26/2001 at 22:48:32 From: Jennifer Subject: Modern Algebra-Isomorphisms If G and G' are groups, f is an isomorphism from G into G', and K is a subgroup of G, then show that the set f(K) = {f(k)\k is a member of K} is a subgroup of G'. I know this isn't too terribly hard, but I'm stuck. Please see if you can help. Date: 11/27/2001 at 08:57:24 From: Doctor Paul Subject: Re: Modern Algebra-Isomorphisms We need to show: f(K) is nonempty. This is easy: K is a subgroup of G so it contains the identity element of G, written 1_G. f is an isomorphism, hence a homomorphism, and we know that homomorphisms take the identity from the first set to the identity of the second set. Thus f(1_G) = 1_G' and it follows that 1_G' is an element of f(K). Now we use the "One Step Subgroup Test" to complete the proof We want to show: If x and y are elements of f(K) then x*y^(-1) is an element of f(K). Here's how to do it: Let x and y be elements of f(K). Then, by the definition of what it means to be in f(K), there exist a and b in K such that f(a) = x and f(b) = y. K is a subgroup, hence closed under multiplication and taking inverses, so we know that a*b^(-1) is an element of K as well. Thus f(a*b^(-1)) is an element of f(K). But f is a homomorphism, so we can rewrite f(a*b^(-1)) as: f(a*b^(-1)) = f(a) * f(b^(-1)) = f(a) * f(b)^(-1) = x*y^(-1) Thus x*y^(-1) is an element of f(K), which completes the proof that f(K) is a subgroup of G'. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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