0
$\begingroup$

I can't get my head around the fact that for any finite abelian group $G$ of size $k$ there are $k^n$ homomorphisms $\mathbb{Z}^n \to G$. It's meant to be 'clear' but can someone please explain why?

2 Answers 2

5

$\mathbb{Z}^n$ is what is called a "free abelian group of rank $n$."

It has a very nice property: think of the elements as $n$-tuples, $(a_1,\ldots,a_n)$. Given any abelian group $G$, and any elements $g_1,\ldots,g_n\in G$, there exists one and only one group homomorphism $\varphi\colon \mathbb{Z}^n\to G$ such that $\varphi(e_i) = g_i$, where $e_i=(0,\ldots,0,1,0,\ldots,0)$ is the vector with a $1$ in the $i$th position and $0$s elsewhere.

This map is defined simply as $\varphi(a_1,\ldots,a_n) = g_1^{a_1}\cdots g_n^{a_n}$ (I'm writing $G$ multiplicatively; if you write $G$ additively, then the map is $\varphi(a_1,\ldots,a_n) = a_1g_1+\cdots+a_ng_n$).

You should verify that this is indeed a group homomorphism.

Now, this tells you that any choice of an $n$-tuple of elements of $G$ will determine a homomorphism. Conversely, given any homomorphism $\phi\colon \mathbb{Z}^n\to G$, $\phi$ will determine an $n$-tuple, namely $(\phi(e_1),\phi(e_2),\ldots,\phi(e_n))$. This tuple has the property that the map $\varphi$ we defined above that corresponds to this choice is in fact the homomorphism $\phi$ that we started with.

That is: there is a one-to-one correspondence between group homomorphisms $\mathbb{Z}^n\to G$, and $n$-tuples $(g_1,\ldots,g_n)$ of elements of $G$. This is the "universal property of the free (abelian) group".

That means that to count the number of homomorphism from $\mathbb{Z}^n$ to $G$, we only need to count the number $n$-tuples of elements of $G$. If $|G|=k$, how many $n$-tuples of elements of $G$ are there?

(There is a related concept that you obtain if you drop the requirement that $G$ be abelian; a group $F(n)$ with a distinguished set of $n$ elements, $x_1,\ldots,x_n$, with the property that given any $n$-tuple $(g_1,\ldots,g_n)$ of elements of a group $G$ there exists a unique group homomorphism $\varphi\colon F(n)\to G$ with $\varphi(x_i) = g_i$ is called a "free group of rank $n$". The description is more complicated than that of $\mathbb{Z}^n$, which is to be expected because groups in general are more complicated than abelian groups; the concept of free object is an interesting, important, and useful one in many areas of algebra. For example, the ring of polynomials is a special case of a free object)

2

A homomorphism is determined by its image on the generators of $\mathbb{Z}^n$. There are $n$ generators, and you can decide for each of them independently where you want to send it to. You have $k$ possible images for each of them, so $k^n$ possible maps altogether. Since you are specifying your map only on the generators and extending $\mathbb{Z}$-linearly, and since the generators satisfy no relations, other than that they commute, any such map gives a group homomorphism.

  • 0
    @Matt Fair enough. Of course any $n$ generators will form a $\mathbb{Z}$-basis, but it doesn't hurt stressing it.2012-04-05