Question on subgroup?

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

1 Answer

  • 8 years ago
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    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....
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