Mapping Homomorphisms to Commutative n-Tuples: A Bijection

[daswerth]

Homework Statement

Consider the set Hom of homomorphisms from \mathbb{Z}^n (the n-dimensional integer lattice) to a group G.

Also let S = \left\{ \, ( g_1, g_2, \dots, g_n ) \, | \, g_i g_k = g_k g_i, \text{where} \, 0 < i,k \leq n, g_i \in G \right\}, the set of n-tuples from G which consist only of elements that commute with each other.

Task: Produce a natural bijection from Hom to S.

Homework Equations

The Attempt at a Solution

An example of a homomorphism from \mathbb{Z}^n to G would be to take \phi (X) = \phi ( x_1 e_1 + x_2 e_2 + \dots + x_n e_n ) = \phi (e_1)^{x_1} \phi (e_2)^{x_2} \dots \phi (e_n)^{x_n}. For each e_i (unit vector) we associate an element of G, so \phi (e_i) = g_i. In order for this to be a homomorphism, we need to have \phi (X) \phi (Y) = \phi (X+Y). This means each of the g_i must commute with each other. In other words, we associate the unit vectors, e_1 \dots e_n, with the elements of an n-tuple from S.

I just don't see how to uniquely assign a homomorphism from Hom to an n-tuple from S. That's what I'm stuck on. Once that light clicks on, I'm confident I can show that it's bijective. So, your hints will be very much appreciated!

This problem is from an undergraduate Algebra class. Thanks!

 

[daswerth]

It just occurred to me that this may be simpler than I thought. The obvious mapping would be simply to send the homomorphism to the n-tuple that it assigns its unit vectors to. I had assumed the map would be more complicated/interesting.

I'm going to investigate whether this map is bijective.