# Question on subgroup?

If k is a subgroup of G', then (f inverse)(k) is a subgroup of G

Relevance

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'.

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'.

Happy to help.....

Source(s): Major in Mathematics....