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which HNN-extensions are free products? this question is related with another still unsolved about Nielsen-Thruston-reducibility and connected-sum-irreducibility of 3d-torus- bundles...

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up vote 10 down vote accepted

This might help.

Lemma If $A$ does not split freely and $C$ is a non-trivial subgroup of $A$ then the HNN extension $G=A*_C$ does not split freely.

The proof uses Bass--Serre theory---see Serre's book Trees.

Proof. Let $T$ be the Bass--Serre tree of a free splitting of $G$. Because $A$ does not split freely, $A$ stabilizes some unique vertex $v$. But $C$ is non-trivial, so $C$ also stabilizes a unique vertex, which must be $v$. Therefore, $G$ stabilizes $v$, which means the free splitting was trivial. QED

A similar argument shows the following.

Lemma If $ A \*\_C $ splits non-trivially as an amalgamated free product $ A' \*\_{C'} B'$ then either $A$ splits over $C'$ or $C$ is conjugate into $C'$.

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i must admit that i didn't seek until find, but lot of thanks H.W. i feel that MO put the advance of all "mathemagizians" in a better pace... –  janmarqz Dec 9 '09 at 16:28
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juan, are you referring to the Phantom Tollbooth? –  HJRW Dec 10 '09 at 6:29
    
i just check on Phantom Toollbooth (google) but i still don't see how my comments make you -Prof- think what i am refering... –  janmarqz Dec 10 '09 at 17:53
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There's a character in the book called "The Mathemagician". –  HJRW Dec 10 '09 at 18:06
    
we all are mathemagiZians :) –  janmarqz Dec 11 '09 at 1:44
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Let me add an explicit partial solution to the question above: for a torus bundle $E$ over the circle, the fundamental group of $E$ can't be a free product of groups, because if it were, the fundamental subgroup of the fiber would be a free product (by the Kurosch's theorem) which is impossible for the torus, then $E$ isn't a connected sum, hence irreducible.

At least the HNN extensions which are free products can't be torus bundles, and in fact, no other surface bundles unless the surface be the 2-sphere

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Another way of seeing this is to look at $\pi_2$. A non-trivial free product has an essential 2-sphere, whereas it's easy to see that the universal cover of a torus bundle is homeomorphic to $\mathbb{R}^3$. –  HJRW Dec 11 '09 at 5:02
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