Proving Group Homomorphism: P(G1) is a Subgroup of G2 | Permutation Group Help

[millwallcrazy]

i'm having trouble to show that if P: G1 --> G2 is a group homomorphism, then the image, P(G1) = {g belongs to G2 , s.t. there exists h belonging G1 , P(h) = g}, is a subgroup of G2

Also:

Let G be a group, and Perm(G) be the permutation group of G. Show that the
map Q : G --> Perm(G) g --> Qg (g is a subscript of Q) such that Qg(h) = gh (g is a subscript of Q) is well-defined, 1-1 and a group homomorphism, where g, h belong to G.

Suppose that G = Z3 = {e, a, a^2}, a^3 = e. Labelling the points of Z3 as {1, 2, 3},
with e = 1, a = 2 and a^2 = 3, give the permutations Qa and Qa^2 , explicitly. (a and a^2 are subscripts of Q)For the first part: Do i have to show the closure, identity, inverse and associativity

For the 2nd part: How do i show that the map is well defined?

For the third part: I'm not sure where to start?

 

[ninty]

Consider the definition of a homomorphism itself, then the elements on the image should form a group(which is a subgroup of G2)
Try applying the subgroup test to P(G1), whichever one you've learned.

My group theory is kinda rusty, so I'll leave the rest to others.