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!