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This question is, in a way, a follow-up of this earlier question of mine.


Let $A$, $B$ and $F$ be finite groups and let $\alpha: A \to F$ and $\beta: B \to F$ be surjective homomorphisms.

Let $A \times_F B$ denote the fibered product of $A$ and $B$ over $F$, defined as the subgroup of $A\times B$ consisting of those elements $(a,b)$ such that $\alpha(a) = \beta(b)$. It is the categorical pullback of the following diagram $$\begin{matrix} & & A \cr & & \downarrow^\alpha \cr B & \stackrel{\beta}{\longrightarrow} & F \cr \end{matrix}$$

Now let $\tau \in \operatorname{Aut}(F)$ be an automorphism and consider the "twisted" fibered product $$A\times_{(F,\tau)} B = \lbrace (a,b) \in A \times B \mid \alpha(a) = \tau(\beta(b))\rbrace.$$ In other words, it is the pullback of the diagram $$\begin{matrix} & & A \cr & & \downarrow^\alpha \cr B & \stackrel{\tau \circ\beta}{\longrightarrow} & F \cr \end{matrix}$$

It follows from Robin Chapman's answer to my earlier question, that in the case where $A$ and $B$ are cyclic groups, $A\times_{(F,\tau)} B$ and $A\times_F B$ are abstractly isomorphic. In other words, the isomorphism type of the fibered product is impervious to twisting by automorphisms of $F$.

This situation is not exclusive to cyclic groups. In fact, in a paper I am writing at the moment, a large number of fibered products of ADE subgroups of $\operatorname{Sp}(1)$ arise and in all cases the fibered products do not see the automorphism of $F$, up to isomorphism. The key observation in all cases is that one can lift the automorphism $\tau$ of $F$ to an automorphism of either $A$ or $B$. This is trivial for inner automorphisms, since they lift via surjections, but $F$ often admits automorphisms which are not inner and they too happen to lift.

Naturally, one is always suspicious that something which can be shown to hold by a case-by-case analysis might in fact follow from some general result. Hence my question:


How general is this?

More precisely, let me ask two questions.

(1) Do automorphisms always lift via surjections?

If true, this would explain what I have observed, but I suspect this is not true: although inner automorphisms do indeed lift, normal subgroups (which define surjections) need not be preserved under outer automorphisms. And at any rate, this would perhaps be too strong a result. What I really want to know is the answer to this next question:

(2) Are twisted fibered products $A\times_{(F,\tau)} B$ corresponding to different automorphisms $\tau$ always abstractly isomorphic?

Thanks in advance!

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up vote 3 down vote accepted

(1) No. Let $C_n$ denote the cyclic group of order $n$, and since I will only consider abelian groups, I will write all groups additively. There is a surjection $C_9 \times C_3 \to C_3 \times C_3$ given on generators by $(1,0) \mapsto (1,0)$ and $(0,1) \mapsto (0,1)$. Then all preimages of $(1,0) \in C_3 \times C_3$ have order $9$ in $C_9 \times C_3$, whereas all preimages of $(0,1)$ have order $3$ in $C_9 \times C_3$. So the involution of $C_3 \times C_3$ given by switching the two generators does not lift to an automorphism of $C_9 \times C_3$.

(2) is the interesting question, and I don't see the answer immediately. My suspicion is "no".

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... or follow your nose from Theo's example ... – Tom Goodwillie Jul 14 '10 at 13:15
Yeah, my thought was that this would give a counterexample to the claim, but I didn't take the time to think it through. – Theo Johnson-Freyd Jul 14 '10 at 17:44
(I do not understand the comments here -- did I miss something?) Thanks -- this is a very nice example. – José Figueroa-O'Farrill Jul 15 '10 at 0:17
Jose - Originally there was another comment on which I was commenting, but then it was deleted. – Steve D Jul 20 '10 at 18:45

Theo's example works for question (2) as well. Let $p$ be a prime (it doesn't have to be $3$). Let $\alpha:Z_{p^2}\times Z_p\to Z_p\times Z_p$ be the projection map and $\beta=\alpha$. Then the fibre product is isomorphic to $Z_{p^2}\times Z_p\times Z_p$.

Now let $\tau:(x,y)\to(y,x)$ be an automorphism of $Z_p\times Z_p$. Then the twisted fibre product is isomorphic to the fibre product of $\alpha$ and the projecion map $\beta':Z_p\times Z_{p^2}\to Z_p\times Z_p$ and this is isomorphic to $Z_{p^2}\times Z_{p^2}$. So these fibre products are not isomorphic.

This doesn't even need $p$ prime, $p>1$ will suffice :-)

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+1. Thanks! I wish I could accept both your answers! – José Figueroa-O'Farrill Jul 15 '10 at 17:15
+1. I just like the number "3". – Theo Johnson-Freyd Jul 20 '10 at 22:52

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