SOLUTION: This for a Abstract Algebra class Assume that φ is a homomorphism from the group G to G′. Prove that if K is any subgroup of G′, then φ−1(K) = {a ∈ G | φ(a) ∈ K} is

Algebra ->  Test -> SOLUTION: This for a Abstract Algebra class Assume that φ is a homomorphism from the group G to G′. Prove that if K is any subgroup of G′, then φ−1(K) = {a ∈ G | φ(a) ∈ K} is      Log On


   

Question 1186054: This for a Abstract Algebra class
Assume that φ is a homomorphism from the group G to G′. Prove that if K is any subgroup of G′, then φ−1(K) = {a ∈ G | φ(a) ∈ K} is a subgroup of G.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let a, b ∈ phi%5E%28-1%29%28K%29
===> phi%28a%29, phi%28b%29 ∈ K.
Now phi%28a%29%2Aphi%28b%29+=+phi%28a%2Ab%29 ∈ K because phi is a homomorphism.
===> a%2Ab ∈ phi%5E%28-1%29%28K%29.
Now let a ∈ phi%5E%28-1%29%28K%29. ===> phi%28a%29 ∈ K.
===> since phi is a homomorphism.

===> %28phi%28a%29%29%5E%28-1%29+=+phi%28a%5E%28-1%29%29 ∈ K by the uniqueness of the inverse element of G'.

===> a%5E%28-1%29 ∈ phi%5E%28-1%29%28K%29.

Since for any element a, b ∈ phi%5E%28-1%29%28K%29, a*b and a%5E%28-1%29∈ phi%5E%28-1%29%28K%29, it follows that phi%5E%28-1%29%28K%29 is a subgroup of G.
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The conclusion can also be obtained by showing that if a, b ∈ phi%5E%28-1%29%28K%29, then a%2Ab%5E%28-1%29 ∈ phi%5E%28-1%29%28K%29. The arguments are similar to above.