Group theory : inverse of a map.

1. The problem statement, all variables and given/known data
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].


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

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

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?
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
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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.
 
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.
 

HallsofIvy

Science Advisor
41,626
821
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].
 
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.
 

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