Let $A,B$ be groups, suppose we have an epimorphism $p:A \to B$. Let $\phi \in \operatorname{Aut}(A)$.

Does there exist some $\varphi \in \operatorname{Aut}(B)$ such that the following diagram

\begin{array}{c} A & \xrightarrow{\phi} & A \\ \downarrow{p} & & \downarrow{p} \\ B & \xrightarrow{\varphi} & B \end{array}


My thoughts: Take $b \in B$. Since $p$ is surjective, there is some $a \in A$ such that $p(a) = b$ (the possible non-uniqueness of such an $a$ might be a problem). Then try to define $\varphi(b)$ as $p(\phi(a)) \in B$. If we fix some $a_b \in p^{-1}(b)$ for each $b$ with the properties that $a_{b_1} a_{b_2} = a_{b_1 b_2}$ (**) (not sure we can do this), then the construction might work:

$\bullet$ $\varphi(b_1) \varphi(b_2) = p(\phi(a_{b_1})) p (\phi(a_{b_2})) = p(\phi(a_{b_1} a_{b_2}))$. Since $a_{b_1} a_{b_2} = a_{b_1 b_2}$ $\varphi$ is a homorphism.

$\bullet$ $\varphi$ is maybe injective and surjective (I haven't checked this as I'm not sure of (**)) Also we might need the property that $\phi^{-1}(\{ a_{b} : b \in B\}) , \phi (\{ a_{b} : b \in B\}) \subseteq \{ a_{b} : b \in B\}$.

There seem to be too many properties required for it to be true!!

Edit: The answer is indeed no, as it was pointed out in the answer by user1729, unless $\phi( \ker(A \xrightarrow{p} B)) = \ker(A \xrightarrow{p} B)$.

This example is just to make sure I understand the construction correctly.

Example: Suppose that there is a surjective morphism $p:A_1 \times A_2 \to B$ and that $\phi: A_1 \times A_2 \to A_2\times A_1 (\cong A_1 \times A_2)$ is just a permutation of the factors. The kernel $N$ of $p$ can be embedded as $N_1 \times N_2 \subseteq A_1 \times A_2$. Then $\phi(N) = N$ clearly, and so there is a $\varphi$ making the diagram commute.

(Is the above example correct?)

  • $\begingroup$ What are your own thoughts on the matter? Why would you think such an automorphism exists, and what might be the problem causing its non-existence? $\endgroup$ – Jonas Dahlbæk Aug 18 '14 at 7:49
  • $\begingroup$ @user161825 I've tried to come up with counterexamples but failed (even though my gut feeling is that the statement is not true). On the other hand the statement is "natural" enough that it might be true. Can you give me some hints on how to tackle the question? Thanks! $\endgroup$ – user170297 Aug 18 '14 at 8:06
  • $\begingroup$ The example you give is of a map which works, but my example if of one which does not (so, a counter-example)...I can flesh out my answer if you wish? $\endgroup$ – user1729 Aug 18 '14 at 9:56
  • $\begingroup$ I understand your example, but you can flesh it out anyway if you want! $\endgroup$ – user170297 Aug 18 '14 at 9:58

No. In essence, you are claiming that if $\Phi:A\twoheadrightarrow B$ then $\operatorname{Aut}(A)\rightarrow \operatorname{Aut}(B)$, which does not hold. It does hold if the kernel of $\Phi$ is characteristic, that is, if $\phi(N)=N$ for all $\phi\in\operatorname{Aut}(A)$ (for example, if $\Phi$ was the abelinisation map). However, your proposition does not hold in general.

So, here are some hints to finding a counter-example:


Take $A$ and $B$ such that the kernel of your projection map $p=\Phi$ is not characteristic.

Spoiler 1:

Take $A\cong H\times H$, and take $B=H$, where $H$ is an arbitrary (non-trivial!) group. What should the automorphism be here? What should the projection map be?

Spoiler 2:

Further, take $\phi:(g, h)\mapsto (h, g)$ and take the projection map $p:(g, h)\mapsto g$. So $\phi$ swaps the factors but the projection $p$ just looks at one of them.

  • $\begingroup$ What do you mean by "characteristic"? $\endgroup$ – user170297 Aug 18 '14 at 8:38
  • $\begingroup$ @user170297 Sorry, I will edit that in. I mean that $\phi(N)=N$ for all $\phi\in\operatorname{Aut}(A)$. It is a strictly stronger property than "normal". $\endgroup$ – user1729 Aug 18 '14 at 8:39
  • $\begingroup$ One last question: if we fix just one $\phi \in Aut(A)$, does the property $\phi(N) = N$ imply that there is a $\varphi \in Aut(B)$ making the diagram commute? $\endgroup$ – user170297 Aug 18 '14 at 9:11
  • $\begingroup$ @user Yes, that does work. $\endgroup$ – user1729 Aug 18 '14 at 9:23
  • $\begingroup$ I've accepted it, but I still can't upvote. Will do as soon as I have the right to do it! $\endgroup$ – user170297 Aug 18 '14 at 9:25

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