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Hi,

I have got a very natural question in group theory. Suppose you have two countable groups $G_1,G_2$, some action of $\mathbb Z$ on them such that the semi direct products are isomorphic $\phi:G_1\rtimes \mathbb Z\simeq G_2\rtimes \mathbb Z$. We suppose that $\phi(\mathbb Z)=\mathbb Z$. Do we have that $G_1\simeq G_2$? It looks silly but I have not been able to find a counterexample.

Arnaud

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  • $\begingroup$ Is phi(Z)=Z even sensical? How do you view Z as a subgroup of G_1 semi-direct Z? $\endgroup$
    – Makhalan Duff
    Commented Oct 21, 2010 at 3:42
  • $\begingroup$ @Makhalan: you view Z as a quotient, not as a subgroup. $\endgroup$
    – Hugh Thomas
    Commented Oct 21, 2010 at 4:09
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    $\begingroup$ I agree that phi(Z) = Z doesn't make much sense. But there are examples of direct products G_1 x Z ~ G_2 x Z where G_1 is not isomorphic to G_2. $\endgroup$
    – Steve D
    Commented Oct 21, 2010 at 4:10
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    $\begingroup$ The question may be phrased like this: Suppose that a group $G$ has an infinite cyclic subgroup $C$ and a normal subgroup $G_1$ such that every element of $G$ is uniquely of the form $cg_1$. Suppose $G$ also has a normal subgroup $G_2$ such that every element of $G$ is uniquely of the form $cg_2$. Does it follow that $G_1$ is isomorphic to $G_2$? $\endgroup$
    – Tom Goodwillie
    Commented Oct 21, 2010 at 4:29
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    $\begingroup$ @Makhalan Duff and @Hugh Thomas: if you present a group $K = G \rtimes H$ as a semidirect product, then your choice of presentation in particular picks out an embedding $H \hookrightarrow K$ as the elements of the form $1 \rtimes h$ for $1\in G$ the identity and $h\in H$ arbitrary. In general is not the only such subgroup: as $H$ is not normal in $K$, conjugating by $g\in G$ will give a different copy. So having a semidirect product is a lot more information than having an extension of groups. $\endgroup$
    – Theo Johnson-Freyd
    Commented Oct 21, 2010 at 5:54

4 Answers 4

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No.

Let $X$, $Y$, and $Z$ be infinite cyclic groups with generators $x,y,z$. Make a semidirect product $XY$ using the nontrivial action of $X$ on $Y$. Make the direct product of this with $Z$. In this group there is the free abelian group $XZ$, and inside that there is the infinite cyclic group generated by $xz$. This has two "normal complements" $XY$ and $YZ$. One is nonabelian and the other is abelian.

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  • $\begingroup$ That's nice, Tom. $\endgroup$
    – Autumn Kent
    Commented Oct 21, 2010 at 23:26
  • $\begingroup$ Wow! So simple and so pretty! $\endgroup$
    – Alex B.
    Commented Oct 23, 2010 at 7:49
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Take a compact 3-manifold $M$ with $b_1(M)\geq 2$. Then there are many homomorphisms $\pi_1(M)\to \mathbb{Z}$, since $\mathbb{Z}^{b_1(M)}\leq H_1(M)$. Further, if the manifold fibers over $S^1$ corresponding to a map $\phi:M\to \mathbb{Z}$, then $ker(\phi)$ is finitely generated. If $\phi:M\to \mathbb{Z}$ is not fibered, then a theorem of Stallings implies that the cohomology class is not dual to a fiber. For example, consider the link L4a1: alt text

The complement is a compact manifold $M$ with $H_1(M)=\mathbb{Z}^2$. Orienting the two circles of the link in two different ways (up to negation) gives two different homomorphisms to $\mathbb{Z}$ (via the linking number). One orientation corresponds to a fibering, while the other does not (there is an annulus running between the two components). Also, the intersection number with the meridian is the same (up to sign) for each choice of orientation, so the cyclic subgroup condition is satisfied. So the kernel of one map is finitely generated (in fact free), while the other is infinitely generated.

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    $\begingroup$ The OP put in the condition that $\phi(\mathbb{Z})=\mathbb{\Z}$. (See Tom Goodwillie's comment for a sensible interpretation of this.) It doesn't look like that happens in your example (or have I misunderstood?). $\endgroup$
    – HJRW
    Commented Oct 22, 2010 at 5:55
  • $\begingroup$ @ Henry: You can take the cyclic subgroup to be generated by either meridian. Then it will map onto both $\mathbb{Z}$ quotients, since it intersects a Seifert surface (for each of the two orientations) just once. In the example, the Seifert surfaces are actually just the two checkerboard surfaces - one is an annulus, and one is a twice-punctured torus. $\endgroup$
    – Ian Agol
    Commented Oct 22, 2010 at 18:29
  • $\begingroup$ Ah, right, I had indeed misunderstood. Thanks for explaining. $\endgroup$
    – HJRW
    Commented Oct 24, 2010 at 23:09
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If instead of $\phi(\mathbb{Z}) = \mathbb{Z}$, you know that $\phi$ commutes with the projections to $\mathbb{Z}$, then $\phi$ induces an isomorphism between the kernels of the maps to $\mathbb{Z}$, i.e. between $G_1$ and $G_2$.

But the semidirect products fit into split extensions of $\mathbb{Z}$ by $G_1$ and $G_2$, so it does make sense to ask that $\mathbb{Z}$ is a subgroup of the semidirect product. However I can't quite see how asking $\phi$ to preserve $\mathbb{Z}$ implies that it commutes with the quotient maps.

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  • $\begingroup$ The splitting isn't natural. $\endgroup$
    – Tilman
    Commented Oct 21, 2010 at 7:14
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I think this works, and is based on the interpretation of Tom Goodwillie above.

Let $G = \langle a,b,c\ |\ [a,b]=[a,c]=b^{11}=1,b^c=b^4\rangle$. Now let $H=\langle a^{-1}c^{-1}\rangle$, $G_1=\langle b,c\rangle$ and $G_2=\langle b,ac^2\rangle$. Then $G=HG_1=HG_2$, but $G_1$ is not isomorphic to $G_2$.

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