Question

Suppose G and H are groups and ϕ:G -> H is a homomorphism. Let N be a normal subgroup of G contained in ker(ϕ). Define a mapping ψ: G/N -> H by ψ (aN)= ϕ (a) for all a in G.

Prove that ψ is a well-defined homomorphism from G/N to H.

Is ψ always an isomorphism? Prove it or give a counterexample

Answer #1

Prove this statement: Let ϕ : A1 → A2 be a homomorphism and let
N = ker ϕ. Then A1/N is isomorphic to ϕ(A1). Further ψ : A1/N →
ϕ(A1) defined by ψ(aN) = ϕ(a) is an isomorphism.
You must use the following elements to prove:
- well-definedness
- one-to-one
- onto
- homomorphism

Let G and H be groups and f:G--->H be a surjective
homomorphism. Let J be a subgroup of H and define f^-1(J) ={x is an
element of G| f(x) is an element of J}
a. Show ker(f)⊂f^-1(J) and ker(f) is a normal subgroup of
f^-1(J)
b. Let p: f^-1(J) --> J be defined by p(x) = f(x). Show p is
a surjective homomorphism
c. Show the set kef(f) and ker(p) are equal
d. Show J is isomorphic to f^-1(J)/ker(f)

Let φ : G → G′ be an onto homomorphism and let N be a normal
subgroup of G. Prove that φ(N) is a normal subgroup of G′.

4. Let f : G→H be a group homomorphism. Suppose a∈G is an
element of finite order n.
(a) Prove that f(a) has finite order k, where k is a divisor of
n.
(b) If f is an isomorphism, prove that k=n.

Let N and H be groups, and here for a homomorphism f:
H --> Aut(N) = group automorphism,
let N x_f H be the corresponding semi-direct product.
Let g be in Aut(N), and k be in Aut(H), Let C_g:
Aut(N) --> Aut(N) be given by
conjugation by g.
Now let z := C_g * f * k: H --> Aut(N), where *
means composition.
Show that there is an isomorphism
from Nx_f H to Nx_z H, which takes the natural...

Suppose : phi :G -H is a group isomorphism . If N is a normal
subgroup of G then phi(N) is a normal subgroup of H. Prove it is a
subgroup and prove it is normal?

Let G be a cyclic group, and H be any group. (i) Prove that any
homomorphism ϕ : G → H is uniquely determined by where it maps a
generator of G. In other words, if G = <x> and h ∈ H, then
there is at most one homomorphism ϕ : G → H such that ϕ(x) = h.
(ii) Why is there ‘at most one’? Give an example where no such
homomorphism can exist.

Suppose G, H be groups and φ : G → H be a group homomorphism.
Then the for any subgroup K of G, the image φ (K) = {y ∈ H | y =
f(x) for some x ∈ G}
is a group a group in H.

Suppose H is a subgroup of G such that ϕ(H)=H for all
ϕ is an element in Aut(G). Prove H is a normal subgroup of G.

Suppose H is a normal subgroup of G where both H and G/H are
solvable groups. Prove that G is then a solvable group as well.

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