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How do I show that $\text{Hom}(\mathbb{Z},G)$ and $G$ are isomorphic?

The method suggested by the book is to define a map $k$: $\text(\mathbb{Z},G) \to G$ by $f \to f(1)$.

I am stuck on how exactly to show that it is an isomorphism. Could anyone shed some light into this? Is $f(1)$ an element of G?

(For those who wanted to know, this is not homework)

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Is $G$ abelian? – Jonas Meyer Dec 24 '10 at 7:33
yes. abelian group. – Seoral Dec 24 '10 at 7:41
up vote 7 down vote accepted

Any homomorphism $f$ from $\mathbb{Z}$ to $G$ is determined by its value at 1: indeed $f(n) = f(1)^n$ for any $n\in \mathbb{Z}$, since $f$ is a group homomorphism (I am using multiplicative notation for the group operation on $G$). So it is natural to consider the map suggested by the book: given $f\in \text{Hom}(\mathbb{Z},G)$, evaluate it at 1: $k:f\mapsto f(1)$. It sends $f$ to an element of $G$, namely $f(1)$.

To check that this is a group homomorphism, you need to understand how $\text{Hom}(\mathbb{Z},G)$ is a group in the first place. That's very simple: the group operation is pointwise multiplication: the product of two homs $f$ and $g$ is given by $fg(n) = f(n)g(n)$ (to explain what $fg$ is you need to evaluate it on an arbitrary input, since you are describing a homomorphism). So now, you should have no difficulties verifying that $k$ respects the group structure.

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Are you assuming that $G$ is abelian so that $fg$ is a homomorphism? Edit: Since I added this comment, Seoral has added that $G$ is abelian, so nevermind. – Jonas Meyer Dec 24 '10 at 7:40
does it also work for non-abelian group G? thanks. – Seoral Dec 24 '10 at 7:51
@Seoral: If $ab\neq ba$ in $G$, and $f$ and $g$ are the homomorphisms from $\mathbb{Z}$ to $G$ such that $f(1)=a$ and $g(1)=b$, then $f(2)g(2)\neq (f(1)g(1))^2$, so $fg$ is not a homomorphism. – Jonas Meyer Dec 24 '10 at 8:13
@Jonas: From what above it seems that G must be abelian. If that is the case why would we wanted to define $f(n)=f(1)^n$ this way. isnt it better if we define $f(n)=nf(1)$? – Seoral Dec 24 '10 at 8:47
@Seoral: Those are different notations for the same thing, depending on whether you are using additive or multiplicative notation for $G$. While it is pretty much universal not to use additive notation unless the group is abelian, there is nothing wrong with using multiplicative notation for an abstract abelian group. Also note that some abelian groups have multiplication as their natural operation, like the group of positive real numbers with multiplication. – Jonas Meyer Dec 24 '10 at 9:01

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