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January 24

Query on an online document - subgroups of A5

Hi all,

I was just wondering if anyone could explain to me briefly the logic behind a part of this link:

http://docs.google.com/viewer?a=v&q=cache%3Aw82TetVc1DEJ%3Awww.rose-hulman.edu%2Fmathjournal%2Farchives%2F2009%2Fvol10-n1%2Fpaper4%2Fv10n1-4pd.pdf+subgroups+of+A5+order+20&hl=en&gl=uk&sig=AHIEtbSsrQJRFZfAaRS7cE25dmdwlG9dFg

On page 5 of 7, around 2 thirds of the way down the page, it says with regards to A5, 'If there were a subgroup H of order 15 or 20, letting A5 act on the coset space G/H would give a nontrivial homomorphism φ : A5 → S3 or φ : A5 → S4. The kernel of φ in either case would be a proper nontrivial normal subgroup of A5, hence no such H exists.'

Now, how do we know that the homomorphism would necessarily be nontrivial for indexes of 3 or 4, but not for index 5 or 6, say? I can't see where the argument falls down in the step between 3 and 4, other than the fact that we can actually find a group of order 12 in A5? I presume there's a nicer way to show where the argument falls apart, without having to actually exhibit the order-12 group. Why is it that the homom. would necessarily be non-trivial for |A5:H|=3,4 but not 5 or 6?

Many thanks,

Typeships17 (talk) 03:15, 24 January 2010 (UTC)[reply]

It's still nontrivial for 5, but then you just get a homomorphism into S₅, which can be (and is) injective. The problem is that S₃ and S₄ are smaller than A₅, so the homomorphism can't be injective, so the kernel can't be trivial. Since the homomorphism is necessarily nontrivial (because the action on the coset space is clearly transitive), the kernel can't be A₅ either, so it must be a proper nontrivial normal subgroup, which of course doesn't exist. Algebraist 03:21, 24 January 2010 (UTC)[reply]

In general this argument shows that a simple group G of order ~~less~~ more than n! can't have a subgroup of index n. Slightly more thought shows that this holds if G just has order ~~less~~ more than n!/2. Algebraist 03:24, 24 January 2010 (UTC)[reply]

You mean a simple group of order *greater* than n! can't have a subgroup of index n > 1. In fact, a non-alternating simple group with a subgroup of index n (n large enough) has order bounded above by n^(c*log(n)). By comparison, n! is larger than (n/e)^n. JackSchmidt (talk) 07:50, 24 January 2010 (UTC)[reply]

That's nice. Is the proof amenable to brief explanation? Algebraist 15:00, 24 January 2010 (UTC)[reply]

I don't know if the "full" result is easy to explain, but there are some improvements to n!/2 that probably make a lot of sense. They eventually lead to the simple estimate that G cannot have order more than 4^n, which is better than 4^( c * n log(n) ) = n!/2, but worse than 4^( c*log(n)*log(n) ) = n^(c*log(n)) that I mentioned. A result of (Bochert 1889) shows that not only can you divide by 2, you can divide by the larger ((n+1)/2)! instead. Burnside's book apparently has one of the earlier versions, where you divide by a few more primes than just 2. This is mentioned in the paper (Manning 1915) which divides by a whole lot of primes, and might actually make a significant asymptotic difference. Basically it comes from looking at how Sylow subgroups can act; Sylows are small, and group orders are the product of their Sylows. A more advanced version of this is in (Praeger & Saxl 1980) which again uses bounds on the Sylows, splitting the primes into cases like Manning did, but using more modern estimates. They give the 4^n version.

Bochert, Alfred (1889), "Ueber die Zahl der verschiedenen Werthe, die eine Function gegebener Buchstaben durch Vertauschung derselben erlangen kann", Mathematische Annalen, 33 (4): 584–590, doi:10.1007/BF01444035, ISSN 0025-5831, MR 1510562
Manning, W. A. (1915), "On the order of primitive groups. II", Transactions of the American Mathematical Society, 16 (2): 139–147, doi:10.2307/1988714, ISSN 0002-9947, MR 1501006
Praeger, Cheryl E.; Saxl, Jan (1980), "On the orders of primitive permutation groups", The Bulletin of the London Mathematical Society, 12 (4): 303–307, doi:10.1112/blms/12.4.303, ISSN 0024-6093, MR 0576980

I don't remember an easy textbook like treatment of it, but I think results along these lines are discussed in Dixon-Mortimer. I'd check if Burnside's version is interesting and accessible. Let me know if you find an interesting and readable account; "subgroups of simple groups are ridiculously small" is one of my favorite themes to get across. JackSchmidt (talk) 17:27, 24 January 2010 (UTC)[reply]

College question: What material do I need to have a firm grasp on?

I am entering a college algebra course (some people call it pre-calculus) in a few weeks and I would like to know what prior material (Alg 1 & 2) I need to have a firm grasp on in order to make a smooth transition and be successful. I have never been very good at math but I am tremendously hard working. A list of topics or any advice would greatly appreciated. Thank you. —Preceding unsigned comment added by 161.165.196.84 (talk) 04:46, 24 January 2010 (UTC)[reply]

I recently ranted about College Algebra on -- maybe it was the science refdesk? Anyway, here's my perspective, as someone who's taught the course several times.

College Algebra (at least at the institution where I taught it) does not really go very far beyond Algebra I & II; it just does the same material faster. There are a few little wrinkles sometimes introduced, but I think they're fairly worthless (stuff like Dirichlet's Law of Signs or something like that).

My typical student was a lot like you've described yourself — never had a lot of success in math classes but willing to work very hard. This couldn't help but endear the students to me; they were very likable.

But unfortunately they were working hard in the wrong way. They preferred to memorize for an hour rather than think for five minutes. The outcome was usually not all that good.

My advice is not to do it that way. Put some effort into figuring out what's going on. If you don't understand what's going on, getting to where you do should be your priority over memorizing the recipes. Once you understand them, the recipes are pretty much obvious; you no longer need to memorize them.

Now, I should say that I'm not necessarily advising you how to get the best grade here. In the long run, this approach should get you the best grades too, but if this is the last math class you ever intend to take, then it's possible it won't "kick in" in time. So take it for what it's worth.

But if you actually want to get something out of the class, beyond checking off a general-ed requirement, then my way is really the only way. --Trovatore (talk) 05:04, 24 January 2010 (UTC)[reply]

Thank you for the response. I prefer your method due to the fact that it should afford me the greatest degree off success in my future calculus classes. When you say to "think for 5 minutes" what exactly do you mean? Can you explain to me the "right" thought proccess when confronted with a problem? Again, thank you for the help. —Preceding unsigned comment added by 161.165.196.84 (talk) 06:28, 24 January 2010 (UTC)[reply]

Obviously, the thought processes vary from problem to problem. I would suggest going over a similar problem that someone else has solved (I'm sure your lecturer will go over some, make sure you take detailed notes if they aren't provided). You then need to make sure you really understand what they did and why. If you really understand why they solved that problem like that, then you will find it easy to adapt the method to your problem. It is important that you understand the underlying theory, as well as how to solve problems. Make sure you really understand the definitions (working out how the definition works in extreme or trivial situations helps with that) and the statements of any results. Understanding the proofs also helps but, unless they are liable to come up in the exam (check that), you don't need to be able to reproduce them (often that just involves memorising the clever trick that is used - there usually is one - and that isn't useful for anything other than reproducing that proof). --Tango (talk) 07:10, 24 January 2010 (UTC)[reply]

Expanding on what Tango said, there is no single "right way to think", which is what makes it fun. Except of course when you're not getting anywhere; in that case it's what makes it painful. Maybe the most important thing to remember is, if the way you're thinking about it doesn't seem to be leading anywhere, try something else — and don't give up. The problems they give you at this level are all solvable. Sooner or later you'll hit on something.

Then try to remember the clarifying insight for next time.

I don't know how helpful these remarks really are — learning to think is a lifelong endeavor, and it's not realistic that I can help you much at that in a few lines. My encouragement to you is simply to actually do it. Use whatever you've already learned about how your thinking process works, and build on that. It gets more effective the more you do it, and the more you do it in a specific arena. Unfortunately most students, at least in the US, have not been exposed much to the idea that thinking is something that's actually required or even useful in a math class; you need to break out of that mindset. Best wishes for success; it can really be done. --Trovatore (talk) 07:37, 24 January 2010 (UTC)[reply]

I really don't understand US maths education... You're doing a pre-calc course at college? In the UK, if you are going to learn calculus, you do so aged 16/17. --Tango (talk) 07:10, 24 January 2010 (UTC)[reply]

U.S. students definitely have the opprotunity to take higher math courses while still in HS, I just chose not to..... unfortunately, I am now making up for it in college. —Preceding unsigned comment added by 161.165.196.84 (talk) 07:34, 24 January 2010 (UTC)[reply]

I'd answer your question for advice another way. The pre-requisites for such a course are limited. The best thing you could do would be to start on the textbook for the course a little early, and then just try to stay a leap ahead of the instructor.Julzes (talk) 08:30, 24 January 2010 (UTC)[reply]

You are under the misconception that there is one, and only one right way to think mathematically. Mathematics subsumes infinitely many different ways of thinking, and in fact the sorts of ways of thinking in existence will never be known. If otherwise, at one point in time everything about mathematics would be known, and this is of course impossible. It is therefore extremely important to appreciate that you will have to invent new ways of thinking should you pursue mathematics furthur. In fact, whatever you choose to study in the future, you will have to discover your own answers; neither your lecturer, nor even the most intelligent person in the world (not that one exists) can give you an answer to everything.

In any case, given the level of mathematics you are studying at present, you are not expected to invent new ways of thinking; ways unknown to all professional mathematicians. Nonetheless, it does help to develop your own ways of thinking; only you know those ways of thinking that result in maximal productivity for you. As a specific example, I would note that even after many years of experience in mathematical thinking, I occassionally find that I am thinking "too hard"; by this I imply thinking in a manner which gives one many original intuitions useful elsewhere but not effective in the context of the idea he/she is investigating. Thus, the best piece of advice I can offer you with regards to mathematical thinking is: "Think simply but abstractly" (you will learn the intended meaning of this "phrase" with time).

On another note, there are many ways in which you can "master" your precalculus course. Firstly, it is always useful to at least attempt to solve worked examples given in your book before (thoroughly) examining their solutions. This way, even if your attempts at the solutions result in failure, you will have some idea about the worked example making it easier to comprehend later on. Secondly, do not work too hard; if you are confident that you have a firm grasp of the material given in one chapter (that is, that you can solve a variety of textbook exercises without necessarily being able to solve them all), move on to the next chapter. Then, if time permits, you can return to solve any remaining exercises in previous chapters before the "final exam". By maintaining your workload at a minimum, you will remain interested in the course, and this is perhaps the most important factor in achieving good grades.

Lastly, specifically in response to your question, you should be comfortable around solving basic algebraic equations, factoring low degree polynomials, and perhaps, if possible, visualizing the graphs of basic polynomials. It is also useful to be familiar with trigonometry (trigonometric functions), but that very much depends on the content of your course. To summarize, do not be concerned if you cannot stick exactly to my advice; with time you will appreciate different ways of learning, and the advice I have given you now will become clear. Hope this helps. --PS T 12:38, 24 January 2010 (UTC)[reply]

A big thank you to all who replied!!!! your information is as helpful as it is inspirational. Finally, does anyone know of any good online resources (preferably FREE) that provide math instruction and/ or tests to keep the skills I have aquried fresh? Thanks again! —Preceding unsigned comment added by 161.165.196.84 (talk) 11:53, 26 January 2010 (UTC)[reply]

Factoring Polynomials

Is there a general method for substituting a polynomial (in one variable) into another so as to get a factorable result? In particular, I'm interested in whether the assumption of Schinzel's hypothesis H allows the claim that for any p the ratio of the largest to smallest prime factors of n^p+1 may be made arbitrarily close to 1. For p=2 and p=3, it's not difficult to get a desired factorization into irreducible polynomials (which can be prime, by the hypothesis). Whether something of the same kind can be done with larger p is what I'm after specifically, but of course I'd like to know the general answer if there is one. I'm not quite sure where to look it up.Julzes (talk) 04:55, 24 January 2010 (UTC)[reply]

Just in the simplest case after the ones mentioned, I guess I'd like quartic Q(x) such that P(Q(x)), P(x)=x^4-x^3+x^2-x+1, factors into four quartics.Julzes (talk) 05:02, 24 January 2010 (UTC)[reply]

Actually, Q wouldn't need to be quartic, and I wouldn't generally expect Q to have the same or a lower degree than P. The only thing needed is to get the same number of factors as the degree of P and for the leading terms--coefficient and degree--of the factors to be equal.Julzes (talk) 08:46, 24 January 2010 (UTC)[reply]

Complex icosidodecahedra

Are these polyhedra?

Small complex icosidodecahedron (Cid)
Great complex icosidodecahedron (Gacid)

4 T C 05:42, 24 January 2010 (UTC)[reply]

Yes. The clue is in the name - if it ends with "hedron" then it is a polyhedron (you don't get monohedrons - I suppose it would be another name for a solid polygon, but you never hear them called that). A polyhedron is a 3D shape made up of lots of flat faces (poly=many, hedron=faces). --Tango (talk) 07:14, 24 January 2010 (UTC)[reply]

Wouldn't a solid polygon be a dihedron as it has a top face and a bottom face? ;-) But anyway, since I made the articles from Mathworld, I still don't get where the faces are in these polyhedra. 4 T C 07:23, 24 January 2010 (UTC)[reply]

Yes, it could be a dihedron too. It depends on your exact definition - dealing with degenerate cases always tests just how good your definition is! --Tango (talk) 08:04, 24 January 2010 (UTC)[reply]

The first one just looks like an ordinary icosahedron to me; I don't see where the "dodeca" part comes in. --Trovatore (talk) 07:28, 24 January 2010 (UTC)[reply]

After some checking on MathWorld, it seems like the first one is a compound of an icosahedron and a great dodecahedron. That could be where the "dodeca" part comes in. 4 T C 07:37, 24 January 2010 (UTC)[reply]

I don't see a dodecahedron there, do you? Of course it's the dual of the icosahedron, so I suppose you can visualize a dodecahedron connecting the centers of the faces, but that's a bit strained — there's no obvious dodecahedron in the pic. --Trovatore (talk) 07:43, 24 January 2010 (UTC)[reply]

A big problem, yes. Actually, I meant great dodecahedron not a Platonic dodecahedron, but still we have the same problem: you can only see the icosahedron. Maybe someone could edit the picture to make this clearer. 4 T C 07:51, 24 January 2010 (UTC)[reply]

And in case you were wondering, I don't see a dodecahedron either. 4 T C 07:52, 24 January 2010 (UTC)[reply]

The shape shown in the picture is just an icosahedron. Presumably, if one could examine the actual 3-D model, then one would be able to see the more complex polyhedron with the dodecahedron inside. The picture needs clarification. Dbfirs 07:57, 24 January 2010 (UTC)[reply]

(undent) Or perhaps doing some rerendering. Actually, what really would help is to draw the icosahedron in wireframe, thus exposing the (solid) great dodecahedron inside. 4 T C 08:00, 24 January 2010 (UTC)[reply]

And the problem holds for the second one too. It's supposed to be a compound of a great icosahedron and a small stellated dodecahedron, but the picture is just a small stellated dodecahedron. The "wireframe solution" should also work here. 4 T C 08:02, 24 January 2010 (UTC)[reply]

Oops, no it doesn't. The "wireframe solution" would give the same problem in reverse - you would now see the other polyhedron in the compound, and just that. Has anyone got any ideas to resolve these issues? 4 T C 08:04, 24 January 2010 (UTC)[reply]

Something seems very sinister about polyhedra with coincident edges and vertices. Can they really be considered polyhedra after all? 4 T C 08:05, 24 January 2010 (UTC)[reply]

They're definitely not what I've ever thought of as polyhedra. The polyhedron page laments that there's no clear definition. I think I would at least require a polyhedron to be a compact C⁰ 2-manifold. I was going to add "injectively immersed in 3-space" but I suppose a triangulation of the Klein bottle ought to count as a polyhedron. --Trovatore (talk) 08:11, 24 January 2010 (UTC)[reply]

Could you please explain what a "compact C⁰ 2-manifold" is? I don't know much about manifolds... 4 T C 08:12, 24 January 2010 (UTC)[reply]

It means that at every point, locally, it has to look like Euclidean 2-space. So for example having three faces joining at the same edge is ruled out, because any neighborhood of a point on the edge would have a flap coming off it. --Trovatore (talk) 08:15, 24 January 2010 (UTC)[reply]

Incidentally, how can a manifold not be C⁰? Doesn't C⁰ just mean it is a topological space? --Tango (talk) 08:24, 24 January 2010 (UTC)[reply]

Right. The point of specifying C⁰ is to clarify that I'm not interested in the differential structure. --Trovatore (talk) 08:26, 24 January 2010 (UTC)[reply]

These two polyhedroids (since you don't think they're polyhedra anymore), have four faces at an edge - are they still compact C⁰ 2-manifolds? 4 T C 08:31, 24 January 2010 (UTC)[reply]

No, unless I'm missing something. --Trovatore (talk) 08:42, 24 January 2010 (UTC)[reply]

(undent) Thought so, but can't figure out why. 4 T C 08:44, 24 January 2010 (UTC)[reply]

No. Basically, you should be able to smoothly deform a polyhedron into a sphere (or, I suppose, a torus, Klein bottle, etc., but normally it's a sphere). You need to be able to deform in so, at any given point, it is nice and flat around it. There is no way of making a point where more than two faces meet flat. --Tango (talk) 08:49, 24 January 2010 (UTC)[reply]

OK, thanks for the great explanation. I see now - such "polyhedroids" have different Euler characteristics depend on how you consider the edges. 4 T C 09:06, 24 January 2010 (UTC)[reply]

The edges and vertices are fine. It's the faces that trouble me - I've never seen a polyhedron where the faces intersect each other, other than at an edge. I would consider the intersection of two faces to be an edge, by definition, really. Your articles describe them as "generalised polyhedra", which seems reasonable to me (I haven't come across the term before, but it makes sense). An article, generalised polyhedron, would be good, though. It seems that the definition of a polyhedron is generalised by allowing more than two faces to meet at an edge and allowing faces the intersect away from edges. --Tango (talk) 08:20, 24 January 2010 (UTC)[reply]

Then take a look at the dodecadodecahedron - some points and lines where the faces intersect are not true vertices and edges. 4 T C 08:28, 24 January 2010 (UTC)[reply]

Oh, yes, there are some edges meeting each other away from vertices, as well. I haven't noticed that. --Tango (talk) 08:49, 24 January 2010 (UTC)[reply]

How about rendering the outer polyhedron with 50% transparency? --Tango (talk) 08:20, 24 January 2010 (UTC)[reply]

Now that's a great idea! How about, then, chopping up the faces of the outer polyhedron until you can only see thin rectangles running along the edges? 4 T C 08:28, 24 January 2010 (UTC)[reply]

Most 3D graphics programs can do transparency themselves, the faces would just look slightly see-through. --Tango (talk) 08:49, 24 January 2010 (UTC)[reply]

(undent) But apart from these sticky issues, there are still more: how do you define the vertex figure for such a polyhedron? The Euler characteristic shows that they're compounds - one element in both of them has chi = -6 and the other has chi = 2. You can figure out that both polyhedra (cid and gacid) have chi = -4 and -6 + 2 = -4. But that only holds if you don't merge coincident edges and vertices. 4 T C 08:36, 24 January 2010 (UTC)[reply]

Oh dear. There's yet another problem: do these polyhedra have duals? Let's just call them the small complex icosidodecacron and the great complex icosidodecacron like the uniform polyhedra. And - well - if they had duals, what would they look like? 4 T C 08:41, 24 January 2010 (UTC)[reply]

Without having given it much thought, I would think the dual of a compound would be the compound of the duals. That will most likely be just as messy as the originals, but if we accept the original as existing, we should accept the dual as existing too. --Tango (talk) 08:51, 24 January 2010 (UTC)[reply]

By your definition, the dual of cid would be a compound of a small stellated dodecahedron with its stellation core - a dodecahedron - in the centre. Darn, coincident edges and vertices again. Then the dual of gacid would be a compound of a great dodecahedron with a great stellated dodecahedron outside it, completely swallowing it up - the duals are just as messy as the originals. 4 T C 08:54, 24 January 2010 (UTC)[reply]

Well, cid's dual looks like gacid, and gacid's dual looks like a great stellated dodecahedron. (Not sure if I should tell all of you this, but I generated gacid as cid's dual here, as you couldn't see the inside. Looks like Stella uses Tango's definition.) 4 T C 08:56, 24 January 2010 (UTC)[reply]

Who's Stella? I didn't define anything, I just naively applied the standard definition of the dual of a polyhedron - the definition still works in this weird world we're discussing, it just gets an equally weird result. --Tango (talk) 08:59, 24 January 2010 (UTC)[reply]

This is Stella. ;-) Sorry for not being clear about that. 4 T C 09:04, 24 January 2010 (UTC)[reply]

(undent) For completeness, here are some freshly baked dual images. However they suffer from the same problems!

4 T C 13:05, 24 January 2010 (UTC)[reply]

And could they be considered uniform? They satisfy the requirements, but something seems strange. 4 T C 08:45, 24 January 2010 (UTC)[reply]

MathWorld says yes - but I'm not sure. 4 T C 09:20, 24 January 2010 (UTC)[reply]

There really is something very sinister about these two polyhedra. I wonder if there are any more like them? 4 T C 08:45, 24 January 2010 (UTC)[reply]

Well, their duals of course. There are a lot - say, a dodecahedron with an interior great stellated dodecahedron, or a compound of a cube, an octahedron, and a cuboctahedron. 4 T C 09:20, 24 January 2010 (UTC)[reply]

Now I'm not even sure they're compounds, as no compound has such coincident edges and vertices. 4 T C 08:51, 24 January 2010 (UTC)[reply]

I don't think it really matters - you get the same problems with other compounds, so excluding this case form being a compound doesn't seem to gain much. --Tango (talk) 08:55, 24 January 2010 (UTC)[reply]

I meant excluding cases where all vertices and edges coincide. This excludes these two but leaves all the other uniform compounds intact. 4 T C 09:02, 24 January 2010 (UTC)[reply]

Whoever wrote Polyhedral compound seems to think you can have polyhedra that aren't 2-manifolds. Those pictures certainly aren't of 2-manifolds, and the first line says they are polyhedra. --Tango (talk) 08:55, 24 January 2010 (UTC)[reply]

While you can have polyhedra with chi > 2 that aren't compounds (like the great disnub dirhombidodecahedron), something is wrong with the article in my opinion. Oh well, that's what you get for having no proper definition of a polyhedron. 4 T C 08:58, 24 January 2010 (UTC)[reply]

These things are weird. Weird, weird, weird. 4 T C 09:10, 24 January 2010 (UTC)[reply]

So weird we'd better shove them under the rug! 4 T C 09:21, 24 January 2010 (UTC)[reply]

An interesting debate as to Euler characteristics and what consitutes a polyhedron is to be found in Lakatos' book Proofs and refutations. Tinfoilcat (talk) 12:52, 24 January 2010 (UTC)[reply]

Very interesting - there's even an image of a great stellated dodecahedron on the cover. (That is, if it isn't a great complex icosidodecacron.) 4 T C 12:57, 24 January 2010 (UTC)[reply]

Until Stella (software) was mentioned, in my naivety I assumed that the pics were of physical models, possibly copper-foil stained glass nicely soldered and with spherical bumps of gilt solder at the vertices - now that would be craftsmanship.→86.132.233.255 (talk) 13:39, 24 January 2010 (UTC)[reply]

If only I could really do that... ;-) 4 T C 08:12, 25 January 2010 (UTC)[reply]