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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/   
    
Associated Topics:
College Logic
College Modern Algebra

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