5

Let $\pi:\ \operatorname{GL}(2,k)\ \longrightarrow\ \operatorname{PGL}(2,k)$ be the canonical homomorphism, and pick some finite subgroup $G\subset \operatorname{PGL}(2,k)$. Then we have an exact sequence

$$1\ \longrightarrow\ \{\alpha I\mid \alpha \in k\}\ \longrightarrow\ \pi^{-1}(G)\ \longrightarrow\ G\ \longrightarrow\ 1.$$

Specific question: Let $k$ be algebraically closed of characteristic zero and let $G$ be some subgroup of $\operatorname{PGL}(2,k)$ isomorphic to $A_5$. Does this sequence split in this case? (It seems to me it shouldn't, but I don't have a proof.)

General question: How would you go about determining, for any particular case of $k$ and $G$, if this sequence splits?

Servaes
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Ben Blum-Smith
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  • Do you mean ${\rm SL}(2,k) \to {\rm PSL}(2,k)$ or ${\rm GL}(2,k) \to {\rm PGL}(2,k)$? – Derek Holt May 09 '14 at 08:37
  • @DerekHolt - I mean $GL(2,k)\rightarrow PGL(2,k)$, but since I had algebraically closed $k$ in mind, we have $PGL(2,k) = PSL(2,k)$ in that case. I've fixed the question for the general case. – Ben Blum-Smith May 11 '14 at 15:08
  • I wonder if $\pi^{-1}(A_5)$ is always isomorphic to the binary icosahedral group (and not to $S_5$)... – Grigory M May 17 '14 at 21:14

2 Answers2

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The sequence does not split: Suppose toward a contradiction that there does exist a splitting $s$. Then there exists a subgroup of $\operatorname{GL}(2,k)$ isomorphic to $A_5$, which has a presentation $$A_5=\langle a,b\vert\ a^2=b^3=(ab)^5=1\rangle.$$ In particular there exist $A,B\in\operatorname{GL}(2,k)$ satisfying $$A^2=B^3=(AB)^5=I,$$ and hence $\det(A)$, $\det(B)$ and $\det(AB)$ are second, third and fifth roots of unity, respectively. Because the determinant is multiplicative, it follows that $\det(A)=\det(B)=\det(AB)=1$.

Note that $A$ is a root of its characteristic polynomial, which is given by $$p_A(X)=X^2-\operatorname{Tr}(A)\cdot X+\det(A)\cdot I.$$ As $\det(A)=1$ and $A^2=I$ it follows that $-\operatorname{Tr}(A)\cdot A+2\cdot I=0$, and hence $$A=\frac{2}{\operatorname{Tr}(A)}\cdot I,$$ which shows that $\pi(A)=I\in\operatorname{PGL}(2,k)$. Therefore $(\pi\circ s)(g)=I$ for any $g\in G$ of order $2$, contradicting the fact that $s$ is a splitting of the sequence.

Note that this works for any $k$ with $\operatorname{char}k\neq2$.

Servaes
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    It works for any $k$ where 2 is invertible. The map splits in fields of characteristic 2 (containing a 3rd root of unity; A5 doesn't exist in PGL(2,k) otherwise). – Jack Schmidt May 19 '14 at 15:56
  • Indeed $2$ must be invertible for my argument to hold, I'll edit. However your final statement is false. For example $A_5\cong\operatorname{PSL}(2,\Bbb{F}_5)\subset\operatorname{PGL}(2,\Bbb{F}_5)$, and in $\operatorname{PGL}(2,\Bbb{C})$ the vertices of an icosahedron are stabilised by a subgroup isomorphic to $A_5\times C_2$. – Servaes May 19 '14 at 22:24
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    Sorry, my last statement was unclear. It included the characteristic 2 hypothesis. – Jack Schmidt May 19 '14 at 22:33
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    For instance, neither PGL(2,2) nor PGL(2,8) contains a subgroup isomorphic to A5. – Jack Schmidt May 19 '14 at 22:34
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For the "specific question", inspired by Servaes' answer, I note that knowledge of $A_5$'s character table (for algebraically closed, characteristic zero $k$) allows us to rule out a splitting of the sequence, simply on the grounds that $GL(2,k)$ doesn't have a subgroup isomorphic to $A_5$ because $A_5$ does not have a 2-dimensional faithful representation.

$$\begin{array}{c | c c c c c} &\text{id}&\text{dbl transp}&\text{3-cycle}&\text{5-cycle}_A&\text{5-cycle}_{A^2} \\ \hline \text{Triv} &1&1&1&1&1\\ \text{3-dim}_A&3&-1&0&\varphi&\hat\varphi\\\text{3-dim}_B&3&-1&0&\hat\varphi&\varphi\\\text{4-dim}&4&0&1&-1&-1\\\text{5-dim}&5&1&-1&0&0\end{array}$$

where $\varphi,\hat\varphi$ are the roots of $x^2-x-1$. The smallest non-trivial representation of $A_5$ is dimension $3$, and any faithful representation of $A_5$ must be a direct sum containing at least one nontrivial representation. (In fact, since $A_5$ is simple, all nontrivial representations are faithful; but 2 dimensions just isn't enough...)

I am still very interested in your thoughts about the "general question".

Ben Blum-Smith
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    2 is a very small number, and the possible groups are so limited one can actually list them all. Then character tables of all of them are computed once and for all. This is actually done in some areas of chemistry and physics. (3 is also “small”). To handle other characteristics, you just use Brauer character tables. To check that a group has a PGL rep, you use Schur covers. For non-algebraically closed you use Schur indexes. – Jack Schmidt May 20 '14 at 16:04
  • http://math.stackexchange.com/questions/42904/finite-subgroups-of-pgl3-c -- this has a reference to the PGL2 question. The list of finite subgroups of GL2 is oddly fuzzier than the one for PGL2, imho. But for lots of people the lists are fine. – Jack Schmidt May 20 '14 at 16:07