Timeline for Normal subgroups of projective special linear group over a ring
Current License: CC BY-SA 3.0
17 events
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| Aug 26, 2011 at 12:54 | comment | added | Frieder Ladisch | @bobgiraffe: for $G= SL_2(\mathbb{Z})$, $G/[G,G]$ is cyclic of order 12. The group $SL_2(\mathbb{Z}/m\mathbb{Z})$ is generated by two elements of order $m$, so commutator factor group of this group is cyclic of order dividing $gcd(m,12)$. If you know, for example, that $SL_2(\mathbb{Z}/4\mathbb{Z})$ has factor group $\mathbb{Z}/4\mathbb{Z}$, then you have an epimorphism $H=SL_2(\mathbb{Z}/2^n \mathbb{Z}) \to SL_2(\mathbb{Z}/4\mathbb{Z}) \to \mathbb{Z}/4\mathbb{Z}$ for $n\geq 2$, which implies $|H/[H,H]|\geq 4$. See math.uconn.edu/~kconrad/blurbs (the note on $SL_2(\mathbb{Z})$). | |
| Aug 25, 2011 at 16:54 | comment | added | Adam Harris | @Felix and Geoff: Also, do you have a reference for showing that $SL_ 2(\mathbb{Z}/p^n\mathbb{Z})$ is perfect for $p\geq5$ please? Or any reference where these groups over $\mathbb{Z} / m \mathbb{Z}$ for $m$ not prime is discussed. | |
| Aug 25, 2011 at 15:02 | comment | added | Adam Harris | I've been trying to reduce the problem to something simpler by the following: I know it is true for $SL_2(\mathbb{Z} / 4 \mathbb{Z})$ and the reduction homomorphism $\phi$ from $SL_2(\mathbb{Z} / p^n \mathbb{Z})$ to $SL_2(\mathbb{Z} / 4 \mathbb{Z})$ is surjective and goes through commutators i.e. $\phi[g,h] = [\phi(g), \phi(h)]$. Now the problem reduces to various things (e.g. I think it would do if $ker(\phi) \subseteq [G,G]$.), but all seem difficult since I can't really get hold of what the commutators look like. | |
| Aug 24, 2011 at 23:28 | comment | added | Geoff Robinson | I was going to suggest that you try GAP (not that I am an expert), but decided against it, since you said you were not a group theorist. I think you are interpreting the output correctly. | |
| Aug 24, 2011 at 22:48 | comment | added | Adam Harris | I think it's $\mathbb{Z}/4\mathbb{Z}$. I tested it on GAP and get the following: gap> IsCyclic(FactorGroup(SL (2,Integers mod 2^3),CommutatorSubgroup(SL(2,Integers mod 2^3),SL(2,Integers mod 2^3)))); true I'm new to GAP so I hope I'm not misinterpreting this? It also looks like the coset reps are upper unitriangular with top right entry 0,1,2,-1. – bobgiraffe 0 secs ago | |
| Aug 24, 2011 at 15:12 | comment | added | Adam Harris | @Felix and Geoff: Thank you. How about when $p \leq 5$? Looking at the case $p=2$, it looks like if we let $SL_2(\mathbb{Z} / 2^n \mathbb{Z}) = G$ (with $n \geq 2$) then $G / [G,G] \cong \mathbb{Z} / 4 \mathbb{Z}$. I'm not a group theorist and am struggling to prove this is true, but it seems like it should be known. Could you please offer any thoughts or a reference for this classical theory? | |
| Aug 2, 2011 at 11:47 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
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| Aug 2, 2011 at 11:38 | comment | added | Frieder Ladisch | well, in my last comment, $H\times Z$ (not $H$) is the unique maximal normal subgroup. | |
| Aug 2, 2011 at 11:35 | comment | added | Geoff Robinson | @Felix: That's my recollection too, and it certainly does what's necessary, thanks. I hadn't tried to reconstruct the proof. I was going in the direction of thinking that for each $i$, $H_i/H_{i+1}$ is the direct sum of a trivial module and a $3$-dimensional irreducible module for ${\rm SL}(2,p)$, and by my previous remark, in $G/H_{i+1}$, every normal subgroup must contain $H_i/H_{i+1}$. | |
| Aug 2, 2011 at 11:20 | comment | added | Frieder Ladisch | @bobgiraffe: For $p>3$, $G={\rm SL}(2, \mathbb{Z}/p^n\mathbb{Z})$ is perfect, if I remember correctly. (I think this can be shown in a similar way as showing that ${\rm SL}(2, \mathbb{F})$ is perfect for fields with more than 3 elements.) Since the other composition factors of $G$ are abelian, it follows that $H$ is the unique maximal normal subgroup of $G$. | |
| Aug 1, 2011 at 22:04 | comment | added | Geoff Robinson | I promised to get back to this. A start is that when $n$ is at least $2$, all elements of order $p$ in $G$ actually lie in $H$ ( the idea is that a solution of $x^p \equiv I$ (mod p^2) with $x \not \equiv I$ (mod p) lifts to an element of order $p$ in ${\rm SL}(2,\mathbb{Z}_{p})$, then its mininum polynomial contradicts the irreducibility of $x^p + \ldots + x +1$. | |
| Aug 1, 2011 at 13:54 | vote | accept | Adam Harris | ||
| Aug 1, 2011 at 13:01 | comment | added | Adam Harris | Thanks Geoff. Is it known whether there are any normal subgroups which aren't contained in H? | |
| Aug 1, 2011 at 11:08 | comment | added | Geoff Robinson | Yes, thanks, I started to write the chain the other way round. Now corrected, I think | |
| Aug 1, 2011 at 11:07 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
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| Aug 1, 2011 at 10:57 | comment | added | Frieder Ladisch | I guess it should be: "... $H_i$ is the set of elements congruent to the identity $\mod p^i$", and $H= H_1$ in the displayed equation? Alternatively, $H_i$ is the kernel of $ {\rm SL}(2, \mathbb{Z}/p^n \mathbb{Z}) \to {\rm SL} (2,\mathbb{Z}/p^i \mathbb{Z})$. | |
| Aug 1, 2011 at 8:17 | history | answered | Geoff Robinson | CC BY-SA 3.0 |