Projection Functions and Homomorphisms

[Bashyboy]

Homework Statement

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.

Homework Equations

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?

 

[pasmith]

Homework Statement

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.

Homework Equations

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?

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?

 

[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 $f(k_1 \star_1 k_2)$?

 

[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)?

 

[Bashyboy]

What is f(k1)f(k_1) in terms of the homomorphisms you are given? What is f(k2)f(k_2)?

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

What is $f(k_1k_2)$? Is it equal to $f(k_1)f(k_2)$?

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.

 

[Bashyboy]

So, does anyone have any new thoughts?

 

[Bashyboy]

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

 

[Stephen Tashi]

The Attempt at a Solution

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

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. Homework 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.