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March 24

Pronunciation

How does one pronounce the name of Don Zagier? I couldn't figure out from his article whether his mother tongue is German or English. The name would be pronounced [ˈtsaːɡiːɐ̯] in German I guess, but I have no idea how would it go in English if the name is English after all. — Emil J. 11:27, 24 March 2009 (UTC)[reply]

I don't know the answer to your question but I suggest that you post in on the Language desk where it is more appropriate. Cheers--Shahab (talk) 12:41, 24 March 2009 (UTC)[reply]

I'm well aware of the Language RD, but since answering the question requires factual information about a mathematician, I assumed I'd have better luck at the Math desk. — Emil J. 14:40, 24 March 2009 (UTC)[reply]

His mother tongue is German, and the pronouciation you suggest is indeed the right one. That said, he mostly works in France, so he's quite used to hear his name pronouced in all manners of funny ways. Bikasuishin (talk) 12:54, 24 March 2009 (UTC)[reply]

Thanks! — Emil J. 14:40, 24 March 2009 (UTC)[reply]

I overlapped with him at college (in the U.S.), and at least at that time he pronounced his name with a Z, a long A and a hard G: ZAY-ghee-ur. (But that may have been for the benefit of Americans.) Before college he was living in California.Daqu (talk) 19:18, 20 June 2010 (UTC)[reply]

Sylow 2-subgroups of classical groups

Previous stuff

hello reference desk! Right, I'm trying to count the size of a general group of elements $\langle p,\rho \rangle \cap P=\langle p,\rho (p),\rho ^{2}(p),\ldots ,\rho ^{2^{n}-2}(p)\rangle$ , where p is an element of P of order 2 and $\rho$ is a coprime automorphism of order (2ⁿ-1) for some n. I'm also assuming that the group is abelian... just a bit of background!

I think that this means the order of the group is the sum $\Sigma _{k=0}^{2^{n}-1}(2^{n}-1:k)$ (where the thing being summed is the appropriate binomial coefficient), which should by my calculations equal $2^{2^{n}-1}$ . The problem is, I have a feeling the order of the group is meant to be $2^{2n}$ and I'm 99% sure I've made a mistake (say in the limit of the sum, for example) - especially since this answer means that if n gets large enough, the subgroup will be massively larger that the parent group (which is of size $2^{3n}$ . Any ideas? SetaLyas (talk) 15:58, 24 March 2009 (UTC)[reply]

Your binomial-coefficient-summing is correct, so the error must lie elsewhere. How did you come by that sum? Algebraist 16:04, 24 March 2009 (UTC)[reply]

Your group is a Z₂-vector space, spanned by 2ⁿ-1 elements. Why are you assuming that those elements are linearly independent? Algebraist 16:11, 24 March 2009 (UTC)[reply]

Isn't it linearly independent because the order of

\rho

is coprime to the order of the group? SetaLyas (talk) 16:28, 24 March 2009 (UTC)[reply]

That doesn't follow. For example, let P be the (additive group of the) field of eight elements, realized as an extension of Z₂ by an element x such that x³+x+1=0. Let p be x and let

\rho

be the squaring map, an automorphism of order 3. Then

\langle p,\rho \rangle =\langle x,x^{2},x^{4}\rangle

, but these generators are linearly dependent: x⁴+x²+x=0. The group generated has only 4 elements in this case. Algebraist 16:39, 24 March 2009 (UTC)[reply]

Oh. Damn! So the problem lies with the faulty assumption that

\langle \rho \rangle

acts fixed-point-freely on p, and so given the assumptions I made the answer was correct? Can you think of any conditions I'd be missing to make this an elementary abelian group of order

2^{2n}

? SetaLyas (talk) 16:57, 24 March 2009 (UTC)[reply]

Nothing springs to mind. What is the actual case you're interested in? Algebraist 17:31, 24 March 2009 (UTC)[reply]

It's from a paper from Landrock and Solomon, don't know if it was ever published. The lemma I'm trying to prove is: Let P be a group of order

2^{3n}

, class 2 and exponent 4 admitting an automorphism

\rho

of order

2^{n}-1

such that (i) Z is elementary abelian of order

2^{n}

such that

\rho

acts transitively on Z\{e} (ii) P/Z is elementary abelian of order

2^{2n}

(iii) P/Z is the direct sum of two irreducible

\rho

-modules, each of which is isomorphic (as such a module) to Z. THEN P is isomorphic to the Sylow 2-subgroup of PSU(3,2ⁿ) or PSL(3,2ⁿ). In the paper, he deduces that such a P is generated by the subgroup I was asking about, and a similar one with "ph" instead of "p" (where p inverts h). I was thinking that I might count the orders of both these subgroups, and see what was gonna be left, and try to work from there to prove the lemma... does that help at all? SetaLyas (talk) 17:43, 24 March 2009 (UTC)[reply]

What does 'class 2' mean here, and what do you mean by 'p inverts h' (the obvious thing to mean is that p is the inverse of h, but this makes no sense)? Also, what is p? Is it just a general non-central element of order 2? Algebraist 17:48, 24 March 2009 (UTC)[reply]

btw, I take it we're talking about A characterization of the Sylow 2-subgroups of PSU (3, 2 n) and PSL (3, 2 n), P Landrock, R Solomon - 1975 - Aarhus Universitet, Matematisk Institut? Algebraist 17:52, 24 March 2009 (UTC)[reply]

yeah, that paper - though the copy I have doesn't look like it's come from a journal (I may be wrong!). By "class" I mean "nilpotency class" and by "p inverts h" I assume that if you conjugate h by p then you get the inverse of h i.e. $[p,h]=h^{2}$ . And yes, p is just a general non-central element (though not necessarily of order 2, it could be of order 4, I was trying to do things bit-by-bit) SetaLyas (talk) 18:10, 24 March 2009 (UTC)[reply]

(ec) Bah, misleading heading. I've been bored for days with no ref desk group theory. Class 2 either means that the derived subgroup is contained in the center, or that even more strongly the Frattini subgroup (of the p-group) is contained in the socle of the center (so that both p'th powers and commutators are central and have order dividing p). I think if these are sylows of PSU(3,2^n), then it could be either (that is, I think squares and commutators are central of order dividing 2). The "ph" thing is a little unclear without context, but I think h is a general element, and the conjugation h^p is h^-1. This should be a little similar to dihedral 2-groups where for a specific non-central involution p, the others are either conjugate to p or of the form ph for some element h inverted by p. JackSchmidt (talk) 18:14, 24 March 2009 (UTC)[reply]

Do you have an electronic copy, btw? OCLC 255552770 lists only one copy and neither Math reviews or Zentralblatt has heard of it. (Neither had I, but it sounds cool). JackSchmidt (talk) 18:23, 24 March 2009 (UTC)[reply]

One copy? I can't find any. Algebraist 18:31, 24 March 2009 (UTC)[reply]

If SetaLeyas does not have an easily transmittable copy, I believe that I know how to (slowly) request a copy from Uni Hannover. There is sort of a gathering of master's thesis type work at many of the math departments in the German universities, and they send them on request. I am making a laundry list of theses, and will see if this one can be requested similarly. It strikes me as similar to:

Collins, Michael J. (1972), "The characterisation of the unitary groups U3(2^n) by their Sylow 2-subgroups", The Bulletin of the London Mathematical Society, 4: 49–53, doi:10.1112/blms/4.1.49, ISSN 0024-6093, MR 0340441

so it could have been expository, so not widely circulated. BTW SetaLyas, page 50 of the Collins paper has a nice list of properties. JackSchmidt (talk) 18:38, 24 March 2009 (UTC)[reply]

More chatter

The sketch (probably for algebraist; if it is confusing to anyone, please ignore it) is that the sylow 2-subgroup of U3(2^n) is basically quaternion, but a very strange sort of quaternion group. Instead of "i" "j" and "k" having coefficients +-1 (like a field of size 2), they get coefficients from a field of size 2^n, or from that module V with P/Z = V + V and Z = V. "p" is like the quaternion "i", and "ph" is like the quaternion "j", and P/Z = Vi + Vj, where Vi is the ρ-module generated by i and Vj the ρ-module generated by j. I never found a sane way of expressing this, but the module idea probably will work to unify my intuition and sane exposition. JackSchmidt (talk) 18:57, 24 March 2009 (UTC)[reply]

Yes Jack, what you've said is (as usual) pretty spot on! I don't have an e-copy sadly, but as it's only 4 A4 pages I could always scan it in for you this weekend if you were that bothered. Anywho... the bit I'm stuck at is the final stage of the proof, basically... once we've shown that this general 2-group is generated by these two

\rho

-modules, how can we show it must necessarily be one of these two Sylow subgroups? The proof's penultimate line is, after showing that the commutators uniquely determine most the elements, "thus there exists at most one such group of a given order with Vi (using your notation) elementary abelian, and at most one with Vj homocyclic of exponent 4". I can't see how they get to this point... after you've shown that all commutators of elements between these modules are of a certain form (

\rho ^{i}(h^{2})

for some i</math>), how do you get to the final bit? SetaLyas (talk) 19:22, 24 March 2009 (UTC) (oh, and thanks for the link to that Collins paper... the useful bits I'd already managed to work out/prove myself, but it's good to know I wasn't mistaken!)[reply]

Ok, so the trick is that we need to identify the field in this abstract group "P". Algebraist's example is actually the key: "V" is the field as additive group, and "ρ" is the multiplicative generator of the field. The trick is just going to be when we put the three copies of the field together, do we do it like the dihedral group or like the quaternion group. In the former case we get the Sylow subgroup of PSL3 and in the latter PSU3. Here the field is the one of size 2^n, the "reals" of the unitary group not the "complexes". I'll see if I can actually produce something like a proof, but that is the idea I think. It is just the classification of groups of order 8 (with centers of order 2) again, but over the field with 2^n elements instead of over the field with 2 elements. JackSchmidt (talk) 19:14, 24 March 2009 (UTC)[reply]

Looks like you got there first. Don't worry about knocking an exact proof up, I was just looking for some general direction - thank you so much! If I need help again, you know I'll shout ^_^ SetaLyas (talk) 19:22, 24 March 2009 (UTC)[reply]

No problem. Yeah, I've now gotten it to flow exactly like the classification of non-abelian groups of order p^3. We know that P = < x,y,z : x^p = { z or 1 }, y^p = { z or 1 }, z^p = 1, [x,y] = z > (where x,y,z are parameterized by field elements, like we have elements x(a) for all a in some field K, and x(a)*x(b) = x(a+b)). So there are four possibilities: (z,z), (z,1) =~ (1,z), and (1,1). The case (z,z) is Q8/Syl2(PSU3); the case (1,1) is by definition D8/Syl2(PSL3). The case (1,z) and (z,1) are obviously equivalent, but I think at this point it becomes delicate. If K has odd characteristic and prime order, then (1,z) gives the same group as (z,z). However if K has even characteristic and prime order (K=2), then (1,z) gives the same group as (1,1). This is also exactly the case you mention above, <y,z> = <y> is homocyclic of exponent 4 (that is, the Z[ρ]-module extension of V and V is the unique nonsplit one). Neat stuff.

I'd definitely like a copy of the paper. A stronger understanding of these low rank groups is very helpful to me. My wikipedia email should work or my academic address is easy to find (I'll reply when I get it, so you know you didn't just mail some random guy). JackSchmidt (talk) 19:41, 24 March 2009 (UTC)[reply]

I'll try to get on scanning it in this weekend ^_^ so would I be right in saying that, in your group presentation, I should be taking

x\mapsto p,y\mapsto ph,z\mapsto h^{2},p\mapsto 2

? If so, then both cases (1,z) and (z,1) yield a contradiction (as they'd make the groups abelian). Also, what do you mean when you say that the group elements are parametrised by field elements, and how should I be transferring from P/Z to P (do the extra

2^{n}

elements just come from multiplying by central elements?) SetaLyas (talk) 22:46, 24 March 2009 (UTC)[reply]

Thanks! Yes, I think your mapping is right. Might be some small detail wrong in my presentation or your identification, but as far as getting the direction, that is it. And yes, the extra factor of 2^n elements are the products by central elements.

(1,z): When n=1 (groups of order 8), then I think (1,z) and (z,1) are ok (not abelian). At least <x,y,z: x^2=1, y^2=z, z^2=1, yx=xyz, zx=xz, zy=yz > is dihedral of order 8. I haven't had a chance to handle the general case.

Parameterise: I am using a shorthand. The group presentation I have in mind is very redundant, but nice: The generators are the 3*2^n symbols x(a), y(b), z(c) where a,b,c are in some field K (of order 2^n). The relations are:

x(a)*x(b) = x(a+b), y(a)*y(b) = y(a+b), z(a)*z(b) = z(a+b).
x(a)^p is either z(a) or 1, y(a)^p is either z(a) or 1
z(a)^p = 1
y(b)*x(a) = x(a)*y(b)*z(ab)
z(c)*x(a) = x(a)*z(c)
z(c)*y(b) = y(b)*z(c)

This is called the Steinberg presentation of the maximal unipotent subgroup of PSL3 (or PSU3) over an arbitrary field if I choose (1,1) (or (z,z)). I'm still trying to make the field identification very explicit and hopefully working out the general field case. JackSchmidt (talk) 04:14, 25 March 2009 (UTC)[reply]

I haven't really thought this through much, but could it be that, say, $x(a)=\rho ^{i}(p),~x(a)x(b)=\rho ^{i+j}(p)$ ? I'm still a bit confused by your notion of parametrising group elements by this field... (I tried looking up stuff about the Steinberg presentation this morning in our library, but found nothing useful, sadly) 79.73.251.120 (talk) 14:33, 25 March 2009 (UTC)[reply]

That is how I want it to work, but I'm having a heck of a time making it make sense in odd or zero characteristic (and it should be universal), and I'm worried about some little details I assumed were easy. The problem is basically defining "z" and trying to prove y(b)*x(a) = x(a)*y(b)*z(ab) in the abstract group P (in SU3 or SL3 it is easy).

The x(a)*x(b) thing is actually trickier. x(a) = ρⁱ(p) where a = ζⁱ in K, so we actually need Zech logarithms to write a + b = ζ^k for some k = k(i,j). Even worse is when a = −b, then we are trying to solve ζ^k = 0, oops. I can define x,y,z naturally so that [ y(a), x(a) ] = z(a^2) in the abstract group P, but I've had no luck showing more generally that [ y(b), x(a) ] = z(ab). Also my "natural" definition of z(a) only specifies z(a) for squares a (because it is roughly equivalent to defining z(a^2) = [ y(a), x(a) ]), and in characteristic not two (or imperfect char 2), squares might not be the whole field.

The Steinberg presentation is covered in most books on finite simple groups of Lie type. For the maximal unipotent subgroup it is also just called "Chevalley's commutator formula". Steinberg's Lectures on Chevalley Groups and Carter's Simple Groups of Lie Type are the books I used as I learned it. I found it very difficult to learn, and required very patient instruction. JackSchmidt (talk) 16:53, 25 March 2009 (UTC)[reply]

ALGORITHM carl leffler

I seek information about Carl Leffler, a mathematics professor circa 1930-1931, who created the Bingo cards now in use. The mathematical tool used was algorithm. He did this "manually", this being before computers

He is said to have been affiliated with Columbia University in New York City

All pertinent items desired: Professional papers, anecdotes, mathematics institutional affiliations. In short, anything that will permit a fuller image of one who, singly, successfully completed work requiring enormous mental ability.

I am the Bingo caller at a Salvation Army Independent Living facility in New York City. The information requested is for a Bingo research project

Thanks Center39 (talk) 19:15, 24 March 2009 (UTC)[reply]

You might try contacting the Columbia archives department [1]. You can also try calling random Lefflers in the phonebook in case you discover a relative. By the way, everyone uses algorithms, that does not not narrow it down. McKay (talk) 23:40, 24 March 2009 (UTC)[reply]