Question

Letφ:G→G′is a group homomorphism. Prove that φ is one-to-one if and only if Ker(φ) ={e}.

Answer #1

I am using f in place of phi.

•Injective ⟹ the kernel {e}

Suppose the homomorphism f:G→G' is injective.

Then since f is a group homomorphism, the identity element e of G
is mapped to the identity element e′ of G'. Namely, we have
f(e)=e′.

If g∈ker(f), then we have f(g)=e′ and thus we have

f(g)=f(e).

Since f is injective, we must have g=e. Thus we have ker(f)={e}.

•Kernel ={e} ⟹ f is injective

On the other hand, suppose that ker(f)={e}

If g_{1}, g_{2} are elements of G such that

f(g_{1})=f(g_{2}), then we have,

f(g_{1} g_{2}^{-1}) = f(g_{1}
g_{2}^{-1}) (since f is a homomorphism)

=f(g_{1})f(g_{2})^{−1} (since f is a
homomorphism)

=f(g_{1})f(g_{1})^{-1}

=e′

Thus the element g_{1} g _{2} -in the kernel
ker(f)={e}, and hence g_{1} g_{2}^{-1} =
e.

This implies that we have g_{1}= g_{2} and f is
injective.

Let φ : A → B be a group homomorphism. Prove that ker φ is a
normal subgroup of A.

Prove the following theorem: Let φ: G→G′ be a group
homomorphism, and let H=ker(φ). Let a∈G.Then the set
(φ)^{-1}[{φ(a)}] ={x∈G|φ(x)} =φ(a)
is the left coset aH of H, and is also the right coset Ha of H.
Consequently, the two partitions of G into left cosets and into
right cosets of H are the same

Let
φ:G ——H be a group homomorphism and K=ker(φ). Assume that xK=yK.
Prove that φ(x)=φ(y)

Let G be an Abelian group. Let k ∈ Z be nonzero. Define φ : G →
G by φ(x) = x^ k . (a) Prove that φ is a group homomorphism. (b)
Assume that G is finite and |G| is relatively prime to k. Prove
that Ker φ = {e}.

Please explain it in detail.
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must be abelian.

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Then the for any subgroup K of G, the image φ (K) = {y ∈ H | y =
f(x) for some x ∈ G}
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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
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(a) Prove that φ(Z(G)) = Z(G′).
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