What Are the Kernel and Image of a Group Homomorphism φ: ℝ^x -> ℝ^x?

[RJLiberator]

Homework Statement

φ is a homomorphism of groups.
φ: ℝ^x -> ℝ^x, where φ(α) = α^4, for all α ∈ ℝ^x. Note that ℝ^x is a group under multiplication.

Describe ker(φ) and Im(φ).

Homework Equations

The Attempt at a Solution

This is another one of those problems that has me scratching my head due to the way it is written.

What does ℝ^x mean? A real number exponentiated?
If all it does is take α and map it to α^4, then the kernal is just going to be the same as it would be without the mapping?
The image is just exponentiated 4 times.

er?

 

[LCKurtz]

Homework Statement

φ is a homomorphism of groups.
φ: ℝ^x -> ℝ^x, where φ(α) = α^4, for all α ∈ ℝ^x. Note that ℝ^x is a group under multiplication.

Describe ker(φ) and Im(φ).

Homework Equations

The Attempt at a Solution

This is another one of those problems that has me scratching my head due to the way it is written.

What does ℝ^x mean?

Good question. How should we know? It's your problem so you tell us. Look it up in your book or course notes and tell us. Or maybe when you find out what it is, you will know how to do the problem.

 

[andrewkirk]

For any two sets $A$ and $B$, the symbol $A^B$ denotes the set of all functions with $B$ as domain and $A$ as range. If you think about this, starting with things like $A=\{0,9\}$ and $B=\{1,2,3,4\}$ (the set of combinations of a lock with four barrels each marked with 0...9 and then moving on to cases like $A=\{0,1\}$ and $B=\mathbb{N}$ (the set of all binary representations of real numbers in $[0,1]$ ) you'll see why this definition works.

 

[LCKurtz]

For any two sets $A$ and $B$, the symbol $A^B$ denotes the set of all functions with $B$ as domain and $A$ as range. If you think about this, starting with things like $A=\{0,9\}$ and $B=\{1,2,3,4\}$ (the set of combinations of a lock with four barrels each marked with 0...9 and then moving on to cases like $A=\{0,1\}$ and $B=\mathbb{N}$ (the set of all binary representations of real numbers in $[0,1]$ ) you'll see why this definition works.

Are you addressing that to the OP or to me? If to me, I know that. How does that explain what set x is in ℝ^x if that's what it means?

 

[andrewkirk]

Are you addressing that to the OP or to me?

The OP. Sorry if that wasn't clear. My impression is that the default assumption is that a post in an internet forum that doesn't quote or name a specific poster is addressing the OP. But the trouble is that nobody ever spells out the etiquette for this, so different people may be operating under different default assumptions.

BTW I don't think it matters what x is. It does not need to be specified in order to answer the two questions asked.

 

[andrewkirk]

@RJLiberator
If $\alpha$ is a real-valued function with domain $x$ then the only workable interpretation I can think of $\alpha^4$ is that $\alpha^4:x\to\mathbb{R}$ such that, for $y\in x,\ (\alpha^4)(y)=(\alpha(y))^4$.

The identity element of $\mathbb{R}^x$ will be the function that maps all of $x$ to a single number (what number?). One can then work out what the kernel of $\phi$ must be.

To work out what Im$\phi$ is, think about what sorts of functions from $x$ to $\mathbb{R}$ cannot give the same results that are given by a function of the form $\alpha^4$.

I wonder why they chose $\phi=\alpha^4$ though, because I think the answers to the two questions would be the same if we had $\phi=\alpha^2$. If the range were $\mathbb{C}$ rather than $\mathbb{R}$ then the answers for $\alpha^4$ and $\alpha^2$ would differ.

 

[micromass]

Neither of that is what $\mathbb{R}^\times$ means in this context. I know what it means, but I consider it the responsability of the OP to tell us what his notation mean.