# Group theory : inverse of a map.

1. Feb 14, 2009

### josh146

1. The problem statement, all variables and given/known data
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 $$H\cap K$$.

2. Relevant equations
$$HK = \{hk : h \in H, k \in K \}$$

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

3. 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?

2. Feb 14, 2009

### Tom Mattson

Staff Emeritus
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.

3. Feb 15, 2009

### josh146

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

4. Feb 15, 2009

### HallsofIvy

Staff Emeritus
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 $$f^{-1}(\{0\})= \{(0,2),(2,0)\}$$.

5. Feb 17, 2009

### 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.