This question is, in a way, a follow-up of [this earlier question of mine](https://mathoverflow.net/questions/30647/fibered-products-of-cyclic-groups).


**Background**


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](https://mathoverflow.net/questions/30647/fibered-products-of-cyclic-groups/30656#30656) 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:


**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!



Isomorphism type of fibered products of groups
(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".


