# Homework Help: Projection Functions and Homomorphisms

1. Nov 18, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Let $G$, $H$, and $K$ be groups with homomorphisms $\sigma_1 : K \rightarrow G$ and $\sigma_2 : K \rightarrow H$. Does there exist a homomorphism $f: K \rightarrow G \times H$ such that $\pi_G \circ f = \sigma_1$ and $\pi_H \circ f = \sigma_2$? Is this function unique?

If either $\sigma_1$ or $\sigma_2$ are monomorphisms, then $f$ will also be a monomorphism.

2. Relevant equations

3. The attempt at a solution

Define $f$ to be the mapping $f : K \rightarrow G \times H$.

$\pi_G ((g,h)) = g$ and $\pi_H((g,h)) = h$, where $(g,h) \in G \times H$.

$\pi_G \circ f = \pi_G (f(k))$, where $k \in K$.

$\pi_G \circ f = \pi_G(f(k)) = \pi((g,h)) = g \in G$

There exists an $k_1 \in K$ such that $\sigma_1 (k_1) = g$. Thus,

$\pi_G \circ f = \sigma_1(k_1)$ or

$\pi_G \circ f = \sigma_1$.

I did something similar for $\pi_H \circ f$. However, this feels unsettling. Also, how would I even verify that $f$ is a homomorphism, if I do not know the rule of its mapping? Would I have to contrive a rule?

2. Nov 19, 2014

### pasmith

You are asked to show whether or not such a homomorphism exists. If you think it does, then you can either try to show that the non-existence of such a homomorphism leads to a contradiction, or you can exhibit a specific $f$.

Unless otherwise specified, the group operation on $G \times H$ is $(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2)$. You are given homomorphisms $\sigma_1 : K \to G$ and $\sigma_2 : K \to H$. Can you think of a way to use those to construct a map $K \to G \times H$? Is that map a homomorphism? Does it satisfy the other conditions given in the problem?

3. Nov 20, 2014

### Bashyboy

Would the rule be as simple as $f(k) = (g,h)$? I tried this, but I ran into some difficulty when trying to verify if it is a homomorphism:

First of all, let me define some conventions: Let $\star_1$ be the operator associated with the group $K$#, $\star_2$ with $G$, $\star_3$ with $H$, and $\star$ with $G \times H$.

$f(k_1 \star_1 k_2)= f(k_1) \star f(k_2)$

$f(k_1 \star_1 k_2) = (g_1,h_1) \star (g_2,h_2)$

$f(k_1 \star_1 k_2) = (g_1 \star_2 g_2 , h_1 \star_3 h_2)$

The problem I am facing is, how do I evaluate $f(k_1 \star_1 k_2)$?

4. Nov 20, 2014

### pasmith

What is $f(k_1)$ in terms of the homomorphisms you are given? What is $f(k_2)$? What is $f(k_1k_2)$? Is it equal to $f(k_1)f(k_2)$?

5. Nov 21, 2014

### Bashyboy

Well, I suppose that $f(k_1) = (g_1,h_1)$ and $f(k_2) = (g_2,h_2)$.

That is what I am trying to demonstrate. If the equality $f(k_1 k_2) = f(k_1) f(k_2)$ is true, then I know that $f$ is a homomorphism. However, I am not sure how to evaluate $f(k_1 k_2)$, which leads me to suspect that the rule I have constructed is not specific enough.

6. Nov 29, 2014

### Bashyboy

So, does anyone have any new thoughts?

7. Nov 29, 2014

### Bashyboy

I figured it out. Define the mapping $f: K \rightarrow G \times H$ to have the rule $f(k) = \bigg(f_1(k),f_2(k) \bigg)$

8. Nov 29, 2014

### Stephen Tashi

If there are preliminary definitions needed to define $f$, you should give them before you say "Define $f$". Aren't projections and their notation already defined in your course matherials? If so, defining $\pi_G$ and $\pi_H$ could be part of "2. Relevant equations", but you don't need to state those definitions in your proof.

To define $f$, all you must do is (for each $k \in K$ ) is to define the element $f(k)$. What element of $G \times H$ will it be?

Once you have defined $f$ as a function, you don't need to define things like $\pi_G \circ f$. The definition of that follows from the standard definition for composing functions.