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I have come across this question in a course about Cayley graphs recently. I don't really have a clue how to answer this question. Here it is.

Consider two groups $A,B$. Let $K< A*B$ be the kernel of the homomorphism $\phi: A * B \to A \times B$ that extends the inclusions $A \hookrightarrow A \times B$ and $B \hookrightarrow A\times B$. Show that $K$ is free.

As usual, thanks in advance :)

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A free set of generators of $K$ is $a*b*a^{-1}*b^{-1}$, $1\neq a\in A$, $1\neq b\in B$. It can be seen combinatorially (using reduced words). There should however be a more intelligent geometrical proof (as the course is about Cayley graphs :) –  user8268 Mar 15 '11 at 8:57
    
This also follows immediately from the Kurosh subgroup theorem, if you are familiar with that. –  user641 Mar 15 '11 at 22:14
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up vote 3 down vote accepted

I don't know if Bass--Serre theory is covered in your course, but it gives a slick proof.

Let $T$ be the Bass--Serre tree of the splitting $A*B$ and consider the action of $K$ on $T$. Then $K$ acts freely. For, if $g\in K$ fixes a point in $T$ then, by definition, $g$ is conjugate into $A$ or $B$, and so $\phi(g)\neq 1$ unless $g=1$.

Now, we have constructed a free action of $K$ on a tree $T$, so $K$ is the fundamental group of $T/K$, which is a graph. Therefore $K$ is free.

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