Question

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

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

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 g1, g2 are elements of G such that

f(g1)=f(g2), then we have,

f(g1 g2-1) = f(g1 g2-1) (since f is a homomorphism)

=f(g1)f(g2)−1 (since f is a homomorphism)

=f(g1)f(g1)-1

=e′

Thus the element g1 g 2 -in the kernel ker(f)={e}, and hence g1 g2-1 = e.
This implies that we have g1= g2 and f is injective.

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