# extending homomorphisms to semidirect products

Let $A = F_2$ be the free group on two generators (not sure if this is important.)

Suppose you have a semidirect product $A\rtimes C$ coming from some homomorphism $\beta: C\rightarrow \text{Aut}(A)$.

Let $G$ be a finite group, and let $f_A: A\twoheadrightarrow G$ be a surjection such that $$f_A(^c\cdot) := f_A\circ\beta(c)\equiv f_A\mod\text{Inn}(G)\qquad\text{for all c\in C}$$

then, under these conditions, must there exist:

1. A homomorphism $f : A\rtimes C\twoheadrightarrow G$ such that $f|_A = f_A$, or equivalently...
2. A homomorphism $f_C : C\rightarrow G$ such that we have $$(f_A\circ\beta(c))(a) := f_A(^ca) = f_C(c)f_A(a)f_C(c)^{-1}\qquad\text{for all c\in C, a\in A}$$

Remark: Our assumptions tell us that for every $c\in C$, there is a $g_c\in G$ such that $f_A(^ca) = g_cf_A(a)g_c^{-1}$ for all $a\in A$, so what I'm asking is if it's always possible to pick the $g_c$'s wisely so that the map $c\mapsto g_c$ is a homomorphism.

This question came from me thinking about Teichmuller level structures and how they classify torsors, so in my context $1\rightarrow A\rightarrow A\rtimes C\twoheadrightarrow C\rightarrow 1$ is an exact sequence of fundamental groups associated to some punctured curve (either $\mathbb{P}^1$ minus 3 points or an elliptic curve punctured once).

EDIT: As Dave Witte Morris points out, this is not possible in general. Though, I wonder: Are there conditions we can put on $G$ or $f_A$ to ensure that the statement is true? If $G$ is abelian, then $f_C$ could literally be anything, but that's quite a strong condition. Can we say anything for nonabelian $G$? I'm trying to understand exactly why this fails... Is there some kind of cohomology going on?

• If the center of $G$ is trivial, then $f_C$ certainly exists. In general, you have a homomorphism $\varphi$ from $C$ to $\mathrm{Inn}(G)$, and $G$ is a central extension of $\mathrm{Inn}(G)$, so you also have a cohomology class $\xi \in H^2 \bigl( \mathrm{Inn}(G); Z(G) \bigr)$. I think the obstruction to the existence of $f_C$ is the pullback $\varphi^* \xi \in H^2 \bigl( C; Z(G) \bigr)$. – Dave Witte Morris Sep 22 '15 at 6:40

No. For a counterexample, let $G = \langle i,j \rangle$ be the quaternion group of order $8$, let $f_A$ map the generating set $\{a,b\}$ of $F_2$ to $\{i,j\}$, and let $C = \langle c \rangle$, where $c$ is the automorphism that fixes $a$ and inverts $b$.

We can choose $g_c$ to be $i$, so the hypothesis is satisfied. However, since $c$ has order 2, and the only element of order 2 in $G$ is central, there is no way to pick $g_c$ to get a homomorphism.

• Ah, nice example. I'll accept this answer when I wake up tomorrow, though do you know of any conditions on $G$ or $f_A$ that would make the statement true? (Certainly if $G$ is abelian, then $f_C$ can be anything but is there anything else?) – Will Chen Sep 21 '15 at 7:05