- #1

Samuelb88

- 162

- 0

## Homework Statement

Let [itex]f : G \rightarrow H[/itex] be a surjective homomorphism. Prove that if G is cyclic, then H is cyclic, and if G is abelian, then H is abelian.

## Homework Equations

Notation: Let it be understood that [itex]f^n(x) = [f(x)]^n[/itex].

## The Attempt at a Solution

Just would like to see if I am right.

*Proof.*(of G cyclic [itex]\Rightarrow[/itex] H cyclic)Suppose that [itex]f[/itex] is a surjective homomorphism, and suppose that [itex]G[/itex] is cyclic. Let [itex]G[/itex] be generated by [itex]x[/itex]. I want to show that there exists [itex]y[/itex] such that [itex]y[/itex] generates [itex]H[/itex].

First I will prove that [itex]\forall x^n \in G[/itex] [itex]\Rightarrow[/itex] [itex]f(x^n) = f^n(x)[/itex]. As our base case for [itex]n=1[/itex] we cite that [itex]f(x e_G) = f(x) f(e_G) = f(x)[/itex]. Next suppose that [itex]f(x^n) = f^n(x)[/itex] (Inductive hypothesis). Then [itex]f(x^{n+1}) = f(x^n x) = f(x^n) f(x) = f^n(x) f(x) =f^{n+1}(x)[/itex], which from our inductive hypothesis.

The proof that [itex]\forall x^{-n} \in G[/itex] [itex]\Rightarrow[/itex] [itex]f(x^{-n}) = f^{-n}(x)[/itex] is similar.

Now since [itex]f[/itex] is surjective [itex]\Rightarrow[/itex] [itex]\mathrm{im}(f) = H[/itex] and therefore [itex]H[/itex] is generated by [itex]f(x)[/itex] and is therefore cyclic.

*Proof.*(of G abelian [itex]\Rightarrow[/itex] H abelian) Suppose now that [itex]G[/itex] is abelian and let [itex]x, y \in G[/itex]. I want to show that [itex]\forall f(x) , f(y)[/itex] [itex]\Rightarrow[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex]. Then [itex]f(xy) = f(x)f(y)[/itex] and [itex]f(yx) = f(y)f(x)[/itex], but [itex]f(xy) = f(yx)[/itex] [itex]\Rightarrow[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex]. Since [itex]f[/itex] is surjective [itex]\mathrm{im}(f) = H[/itex] [itex]\Rightarrow[/itex] [itex]\forall f(x), f(y) \in H[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex] [itex]\Rightarrow[/itex] [itex]H[/itex] is abelian.

My intuition says I'm right, but it seems my intuition is wrong more than right this quarter.