Group theory : inverse of a map.

In summary, the question is asking about the preimage of the map f, given that H and K are subgroups of a finite group G and f is defined as f : H \times K \rightarrow HK : (h,k)\rightarrow hk. The preimage, f^{-1}(hk), can be described in terms of h, k, and the elements of H\cap K. However, there is an issue with the definition of f^{-1} for individual members of the group, and it may be necessary to use f^{-1}(\{hk\}) instead.
  • #1
josh146
6
0

Homework Statement


Let [tex]H[/tex], [tex]K[/tex] be subgroups of a finite group [tex]G[/tex]. Consider the map, [tex]f : H \times K \rightarrow HK : (h,k)\rightarrow hk[/tex]. Describe [tex]f^{-1}(hk)[/tex] in terms of [tex]h[/tex], [tex]k[/tex] and the elements of [tex]H\cap K[/tex].


Homework Equations


[tex]HK = \{hk : h \in H, k \in K \}[/tex]

[tex]f^{-1}(hk)=\{ (h',k') : f(h',k')=hk \}[/tex]

The Attempt at a Solution


I should be able to get somewhere with this but I can't. Can someone give me one or two hints to start me off?
 
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  • #2
I think you'd better ask your professor about this and here's why: Based on what you've written, [itex]f[/itex] shouldn't even have an inverse. Here's a simple example to show you what I mean. Let [itex]G=\mathbb{Z}_4[/itex] and let [itex]H=K=<2>[/itex], the subgroup generated by [itex]2\in\mathbb{Z}_2[/itex]. Your map does the following.

[tex]f(0,2)=0\cdot 2 =0[/tex]
[tex]f(2,0)=2\cdot 0 =0[/tex]

Since [itex]f(0,2)=f(2,0)[/itex] the map is not 1-1 and is therefore not invertible. Something is wrong here.
 
  • #3
Sorry, I shouldn't have written 'inverse' in the title. For this map, [tex]f^{-1}[/tex] is defined as the 'preimage' of the map.
 
  • #4
Then your problem still makes no sense. You cannot prove anything about "[tex]f^{-1}(hk)[/tex]" because [tex]f^{-1}[/tex] is not defined for individual members of the group, only subsets. Did you mean [tex]f^{-1}(\{hk\})[/tex]?

In the example Tom Mattson gave, [tex]f^{-1}(0)[/tex] is not defined but [tex]f^{-1}(\{0\})= \{(0,2),(2,0)\}[/tex].
 
  • #5
I'm pretty sure that [tex]f^{-1}(a)[/tex] is used as shorthand for [tex]f^{-1}({a})[/tex].

I'm still having trouble doing the question though. I'm sure it's simple but I'm not seeing how to define the set.
 

1. What is a group in group theory?

A group in group theory is a mathematical concept that consists of a set of elements and a binary operation that combines any two elements to form a third element. This binary operation must also follow four basic axioms: closure, associativity, identity, and inverse.

2. What does the inverse of a map mean in group theory?

In group theory, the inverse of a map is the function that reverses the operation of a given map. It essentially "undoes" the original operation. For example, if the original map is multiplication by 2, the inverse map would be division by 2.

3. How is the inverse of a map found in group theory?

The inverse of a map can be found by solving for the unknown variable in the equation. In group theory, this means finding the element that, when combined with the original element using the binary operation, yields the identity element. This element is known as the inverse element.

4. What is the importance of the inverse of a map in group theory?

The inverse of a map is important in group theory because it allows for the cancellation of elements. This means that if an element and its inverse are combined using the binary operation, they will cancel each other out and result in the identity element. This is a crucial property in solving equations and understanding the structure of a group.

5. Can every element in a group have an inverse?

In a group, every element must have an inverse. This is one of the axioms of a group in group theory. If an element does not have an inverse, it cannot be part of the group and the structure would not follow the group axioms. Therefore, every element in a group must have an inverse.

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