Homomorphism Problem - Need work checked

[Samuelb88]

Homework Statement

Let f : G \rightarrow H 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 f^n(x) = [f(x)]^n.

The Attempt at a Solution

Just would like to see if I am right.

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

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

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

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

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

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

 

[micromass]

Hi Samuelb88!

Homework Statement

Let f : G \rightarrow H 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 f^n(x) = [f(x)]^n.

The Attempt at a Solution

Just would like to see if I am right.

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

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

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

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

Could you explain in more detail why H is generated by f(x). I'm sure you know it, but it's a crucial detail that should be mentioned.

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

That proof is correct! But it's perhaps a bit chaotic, here's a proposal to clean it:

Take h and h' in H. We want to show that hh'=h'h. Because of surjectivity, we know that there are g and g' in G such that f(g)=h and f(g')=h'. Since G is abelian, we know that gg'=g'g. But then it follows that hh'=f(g)f(g')=f(gg')=f(g'g)=f(g')f(g)=h'h.

I'm not saying your proof is wrong, though.

 

[Samuelb88]

Since f is surjective \Rightarrow \mathrm{im}(f)=\{ ..., f^{-1}(x), e_H , f(x) , f^2(x) , ... \} = H. Doesn't this imply that H is generated by f?

 

[micromass]

Since f is surjective \Rightarrow \mathrm{im}(f)=\{ ..., f^{-1}(x), e_H , f(x) , f^2(x) , ... \} = H. Doesn't this imply that H is generated by f?

Yes, of course, I just wanted some more explanation on that part
Well, it seems it is all correct!

 

[Samuelb88]

Great to hear! Hehe. Thanks much, micromass!