Thread: intersection of finitely generated groups

1. intersection of finitely generated groups

Hi! We have a free group F and two finitely generated subgroups G and H. I want to show that their intersection is also finitely generated. The hint is to consider two graphs whose fundamental groups are G and H respectively. Then with these two graphs one can construct a graph whose fundamental group is exactly the intersection of G and H.

Can anybody help me with this? I am completely stuck.

nice greetings
banach

2. What do you know already? Do you know about immersions (maps from a graph to another graph where the induced maps between edges starting at a given vertices is injective)?

Basically, as your subgroups are finitely generated they correspond to immersions. Pullbacks of immersions correspond to intersections, and so construct the pullback of the two given immersions (and draw the diagram!). This pullback is a finite graph, and a component of it corresponds to your intersection. This gives you an immersion, and so a finite basis of your intersection has been found.

I would say more, but I don't know what you already know...however, I think what I've said should be hint enough - it is quite a deep theorem so really you should know what a pullback etc. is if you're to tackle it...

3. Thanks for your answer but i unfortunately i am not familiar with these terms. However, i think i know how to work it out now:
Wlog we may assume that $\displaystyle F=F(a,b)$. Let $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ the associated graphs of G and H. Define the graph $\displaystyle \Gamma$ by $\displaystyle V(\Gamma):=V(\Gamma_1) \times V(\Gamma_2)$ and there is an edge between (a,x) and (b,y) iff there are edges from a to b in $\displaystyle \Gamma_1$ and x to y in $\displaystyle \Gamma_2$ having the same label. Since G and H are f.g. the graphs $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ are finite graphs, thus $\displaystyle \Gamma$ is also a finite graph whose fundamental group is $\displaystyle G \cap H$. Hence the intersection is f.g., qed.
What do you think of that proof?

Greetings
Banach

4. Originally Posted by Banach
Thanks for your answer but i unfortunately i am not familiar with these terms. However, i think i know how to work it out now:
Wlog we may assume that $\displaystyle F=F(a,b)$. Let $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ the associated graphs of G and H. Define the graph $\displaystyle \Gamma$ by $\displaystyle V(\Gamma):=V(\Gamma_1) \times V(\Gamma_2)$ and there is an edge between (a,x) and (b,y) iff there are edges from a to b in $\displaystyle \Gamma_1$ and x to y in $\displaystyle \Gamma_2$ having the same label. Since G and H are f.g. the graphs $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ are finite graphs, thus $\displaystyle \Gamma$ is also a finite graph whose fundamental group is $\displaystyle G \cap H$. Hence the intersection is f.g., qed.
What do you think of that proof?

Greetings
Banach
That seems right, and is a much neater way than the way I was thinking of...

5. Originally Posted by Banach
Thanks for your answer but i unfortunately i am not familiar with these terms. However, i think i know how to work it out now:
Wlog we may assume that $\displaystyle F=F(a,b)$. Let $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ the associated graphs of G and H. Define the graph $\displaystyle \Gamma$ by $\displaystyle V(\Gamma):=V(\Gamma_1) \times V(\Gamma_2)$ and there is an edge between (a,x) and (b,y) iff there are edges from a to b in $\displaystyle \Gamma_1$ and x to y in $\displaystyle \Gamma_2$ having the same label. Since G and H are f.g. the graphs $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ are finite graphs, thus $\displaystyle \Gamma$ is also a finite graph whose fundamental group is $\displaystyle G \cap H$. Hence the intersection is f.g., qed.
What do you think of that proof?

Greetings
Banach
Actually, you would need to prove that the fundamental group of this graph is $\displaystyle G \cap H$. I do not think that this is obvious.