# Homomorphisms of a group

• daswerth
In summary, the task is to find a natural bijection between the set of homomorphisms from the n-dimensional integer lattice to a group G and the set of n-tuples from G where the elements commute with each other. The suggested solution is to map each homomorphism to the n-tuple that it assigns its unit vectors to. This mapping will be investigated for bijectivity.

## 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 $$S$$.

## 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!

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.

## 1. What is a homomorphism of a group?

A homomorphism of a group is a function that preserves the group structure, meaning that it maps the group's operation to itself. In other words, if we have two elements x and y in a group G, a homomorphism f will map their product to the product of their images, f(x*y) = f(x) * f(y).

## 2. What is the difference between a homomorphism and an isomorphism?

A homomorphism is a function that preserves the group structure, while an isomorphism is a bijective homomorphism, meaning it is both one-to-one and onto. In other words, an isomorphism not only preserves the group operation, but also preserves the elements and their relationships within the group.

## 3. How do you determine if a function is a homomorphism?

To determine if a function is a homomorphism, you can check if it preserves the group operation. This means that for any two elements x and y in the group, the function should map their product to the product of their images. Additionally, you can also check if the function is well-defined, meaning it produces a unique output for every input in the group.

## 4. Can a homomorphism be an identity function?

Yes, a homomorphism can be an identity function. This means that the function maps every element in the group to itself, and therefore preserves the group operation. However, not all identity functions are homomorphisms, as they must also be well-defined.

## 5. How are homomorphisms used in group theory?

Homomorphisms play a crucial role in group theory as they allow us to compare and classify groups. By studying the homomorphisms between groups, we can identify common structures and properties, and determine if two groups are isomorphic. Homomorphisms also help us understand the subgroups and quotient groups of a given group.