Group theory : inverse of a map.

[josh146]

Homework Statement

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

Homework Equations

HK = \{hk : h \in H, k \in K \}

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

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?

 

[quantumdude]

I think you'd better ask your professor about this and here's why: Based on what you've written, f shouldn't even have an inverse. Here's a simple example to show you what I mean. Let G=\mathbb{Z}_4 and let H=K=<2>, the subgroup generated by 2\in\mathbb{Z}_2. Your map does the following.

f(0,2)=0\cdot 2 =0
f(2,0)=2\cdot 0 =0

Since f(0,2)=f(2,0) the map is not 1-1 and is therefore not invertible. Something is wrong here.

 

[josh146]

Sorry, I shouldn't have written 'inverse' in the title. For this map, f^{-1} is defined as the 'preimage' of the map.

 

[HallsofIvy]

Then your problem still makes no sense. You cannot prove anything about "f^{-1}(hk)" because f^{-1} is not defined for individual members of the group, only subsets. Did you mean f^{-1}(\{hk\})?

In the example Tom Mattson gave, f^{-1}(0) is not defined but f^{-1}(\{0\})= \{(0,2),(2,0)\}.

 

[josh146]

I'm pretty sure that f^{-1}(a) is used as shorthand for f^{-1}({a}).

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.