Wikipedia:Reference desk/Archives/Mathematics/2010 July 14

= July 14 =

Free stuff
What's the free ring over an abelian group? Is it this: given an abelian group A, take the free ring generated by A then quotient it by the ideal generated by all the i(a)+i(b)-i(a!b), where i is insertion of generators and ! is the addition in A?

Also I was reading about Freyd's adjoint functor theorem. If you run through the whole proof it seems to be somewhat constructive. My book (MacLane's 2nd edition) on page 125 proves the forgetful functor from compact hausdorff spaces to SET has a left adjoint using adjoint functor theorem; he finds a small solution set explicitly then applies the theorem. Did this suggest how to construct an explicit stone cech compactification? Similarly for Grp. Does the adjoint functor theorem suggest ways to construct free objects explicitly? Because free groups are so much more complicated than free abelian groups (you could put the abelian relation on the free group but why do that when you can use set of all cofinally 0 function from X to the integers) I wonder how they managed to get free groups in the first place. Money is tight (talk) 04:17, 14 July 2010 (UTC)
 * Regarding the first question, I don't specifically know that terminology, but I think it would be the group ring of Z over A. Using your phrasing I guess that would be the free ring generated by A modded out by the ideal generated by all the i(a)*i(b) - i(a!b). Rckrone (talk) 06:11, 14 July 2010 (UTC)
 * I think what you described is the tensor algebra over A with A considered as a Z-module. (Can anyone confirm if that is correct?) Rckrone (talk) 06:26, 14 July 2010 (UTC)
 * Sorry for the confusion I was in a rush. What I wanted was the left adjoint to the forgetful Rng->Ab, sending each ring to its underlying addition abelian group. I've never heard of this construction before. Also my 2nd question is more clearly stated as: How do people explicitly construct free objects for the first time? Is it raw ingenuity or is there some algorithm in doing so? Money is tight (talk) 09:20, 14 July 2010 (UTC)
 * For algebraic categories like the examples you mentioned there is a uniform "algorithm". If U and V are varieties such that the signature σ of U includes the signature τ of V and V includes all reducts of algebras from U, then the forgetful functor U → V has a left adjoint F: V → U defined as follows: F(A) is the quotient of the free U-algebra (which can itself be constructed as the quotient of an absolutely free σ-algebra, e.g. the term algebra, over the congruence generated by the defining identities of U) in generators A over the congruence ~ generated by the atomic diagram of A (i.e., f(a1,...,ak) ~ a for each k-ary operation f ∈ τ and each a1,...,ak,a ∈ A such that a = fA(a1,...,ak)). (Incidentally, the coverage of universal algebra on Wikipedia is extremely poor.)—Emil J. 12:57, 14 July 2010 (UTC)
 * Oh, and the answer to your first question is yes, your construction is in fact a special case of the general construction above (more precisely, the general construction would also put i(−a) + i(a) into the ideal, but that's easily seen to be redundant).—Emil J. 13:22, 14 July 2010 (UTC)

Thanks! I've never had any exposure to universal algebra so I don't exactly know what you mean. I had a look at the online pdf version "A course in universal algebra" page 73, and I think this is how: Say we want the free group over a set X (with signiture multiplication binary inverse unary and 1 constant). Take the term algebra T(X), which is like the free magma except it's got terms with inverse and 1. Next take the intersection E of all congruences on T(X) such that the quotient is actually a group (since T(X) is the "absolutely free algebra" you described, it has no relations on it, not even associativity), and intersection of congrugence is a congruence. The quotient under E is a group too, the free group on X.

I really have no idea what's going on lol. I'll take a proper look some other time.

Another question: this "uniform algorithm" is not the standard way to construct free groups. Is it because this construction is so much more difficult to use practically than the standard construction using equivalence class of words? The free abelian group can be constructed from the free group by using further identifications, but why do that when we can use all confinally 0 functions to the integers. So sure, this algorithm proves every forgeful functor on some type of algebras has a left adjoint, but it might as well just be a pure existence because of the insane complexity in its construction. Money is tight (talk) 05:19, 15 July 2010 (UTC)
 * The congruence E can also be described "from below": t E s iff there exist n ≥ 0 and terms t0 = t, t1, ..., tn = s such that for each i < n, ti = ui(vi) and ti+1 = ui(wi) for some terms ui, vi, wi, where either vi ≈ wi or wi ≈ vi is an instance of one of the defining identities of groups (for example, we may have vi = 1 and wi = r−1·r for some term r). The standard group-specific construction is actually equivalent to this: another description of E is that t E s iff (the flattened forms of) t and s are equivalent as words. In this way it gives more information, as it provides an explicit description of representants of the equivalence classes. This is useful for various group-theoretic considerations, but I don't think it is any easier than the general construction if you just need to prove that the free group is a free group. Of course, in the case of abelian groups the explicit description of free objects is so simple that it beats any generic construction.—Emil J. 12:28, 15 July 2010 (UTC)

Polish notation
What's the point of Polish notation? Not only is it harder to read, it's more difficult to work with algebraicly. For example, compare the normal version ab+cd to its Polish notation equivalent, + * a b * c d. --138.110.206.101 (talk) 16:45, 14 July 2010 (UTC)


 * It removes ambiguity. How do you KNOW that ab+cd=(ab)+(cd) and not ab+cd=a(b+c)d?  You are using an assumption that multiplication precedes addition.  Polish notation never requires assumptions or parenthesis. --  k a i n a w &trade; 16:59, 14 July 2010 (UTC)
 * It isn't really an assumption is it, when everyone is using previously agreed-upon rules? There isn't really any ambiguity to ab+cd is there?   misli  h  20:07, 14 July 2010 (UTC)
 * It's very easy to evaluate without ever rereading any of the input (which is helpful if the input is large). You just note each operation as you come to it and then evaluate it with the next one or two values (depending on the operator).  One or both of those might itself be the result of an operator, so you proceed recursively.  In your example you would say "Add... multiply a and b..." (at which point you have the two things + and ab to remember; you need not remember a and b separately or that you multiplied them) "...multiply c and d" (at which point you obtain cd and then immediately the overall result).  In practice, RPN is used for this purpose because then only numbers (and never operators) have to be temporarily remembered (and the stack that remembers them can often then be smaller).  --Tardis (talk) 18:59, 14 July 2010 (UTC)
 * So Polish or reverse Polish would be better for determining the value of a given expression, but isn't it worse for algebraic manipulation? --138.110.206.101 (talk) 19:01, 14 July 2010 (UTC)
 * Not really. You are just assuming it would be because you aren't used to seeing it.  If you only learned Polish notation, you would look at all the precedence rules and parenthesis used in inline notation and think it was a confusing mess. --  k a i n a w &trade; 19:04, 14 July 2010 (UTC)
 * How would you, for example, solve x 2 ^ x 1 + 2 ^ + x - 1 - 0 = without converting to regular notation? It's a lot harder. --138.110.206.101 (talk) 19:09, 14 July 2010 (UTC)


 * You don't include = in the notation. Keep it as two separate expressions, one being x 2 ^ x 1 + 2 ^ + x - 1 - and the other being just 0.  Anything you do to one you do to the other.  ie: To get rid of "1 -", you use "1 +" and get x 2 ^ x 1 + 2 ^ + x - and 0 1 +.  Since there are no parenthesis or order of operations, we had no issues with deciding what we could and could not do.  The next thing to work on is the trailing x -.  But, it is apparent that you will hit a point at which you want to have one side simply 0 and the other side will contain a squared x.  You will have to simplify.  How do you simplify?  You memorize patterns.  You've probably memorized them as inline patterns, but you could have memorized them as Polish notation patterns just as well. --  k a i n a w &trade; 19:25, 14 July 2010 (UTC)


 * English is like the standard arithmetic expressions if you consider verbs as operators. Latin is reverse polish as in Jim the ball hit and Irish is polish with the equivalent of hit jim the ball. I gues sin them polish or reverse polish might appear a bit more natural. Dmcq (talk) 20:54, 14 July 2010 (UTC)
 * In Hebrew, all three are correct. And the subject doesn't even have to precede the object, giving a total of 6 variations! (Though I'm unsure of the "Hit the ball Jim" variant). -- Meni Rosenfeld (talk) 09:50, 15 July 2010 (UTC)
 * Verb Object Subject doesn't seem to be all that common, and Object Verb Subject even less so - it says Klingon uses that order because it is so strange sounding :) Dmcq (talk) 11:15, 15 July 2010 (UTC)
 * To avoid potential misunderstanding, all six variations are possible in Polish as well, though the default word order is SVO as in English. Polish notation has nothing to do with Polish language, it's named after members of the Polish logical school who invented it.—Emil J. 11:41, 15 July 2010 (UTC)

Rubik's cube group
How many conjugacy classes has the Rubik's cube group? --84.61.131.18 (talk) 20:05, 14 July 2010 (UTC)

How large is the largest order of any element in the Rubik's cube group? --84.61.131.18 (talk) 20:41, 14 July 2010 (UTC)
 * We have some info on the group in Rubik's cube group. The answer to your questions is not there, but some of the stuff may be useful. I suppose it shouldn't be hard to compute conjugacy classes and maximal orders using some computer algebra system.—Emil J. 15:38, 15 July 2010 (UTC)
 * Off the top of my head there are elements of order 8.3.5.7=840. A complete listing of conjugacy classes would be rather lengthy, I'm guessing there are at least a thousand. Conjugacy class of Wreath products of the symmetric groups are fairly easy to work with so a computer algebra system may not be necessary.--RDBury (talk) 16:50, 15 July 2010 (UTC)


 * According to the Magma computer algebra system there are 81120 conjugacy classes and the largest element order is 1260. The group has elements of the following orders: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45, 48, 55, 56, 60, 63, 66, 70, 72, 77, 80, 84, 90, 99, 105, 110, 112, 120, 126, 132, 140, 144, 154, 165, 168, 180, 198, 210, 231, 240, 252, 280, 315, 330, 336, 360, 420, 462, 495, 504, 630, 720, 840, 990, 1260.  Most orders are made up of many classes of several sizes.  Ignoring class sizes, and just sorting by order, one gets the following table:


 * By hand calculations of the orders/sizes of all conjugacy classes are likely much too difficult due to the length, but David Joyner's Adventures in Group Theory (chapter 11, I believe) works out the elements of order 2 in quite some detail by hand. I believe Joyner also includes the by hand calculation of the maximum order, though it might be in one of his other books. JackSchmidt (talk) 17:25, 15 July 2010 (UTC)

How many sporadic subgroups has the Rubik's cube group? --84.61.131.18 (talk) 21:53, 15 July 2010 (UTC)

Which sporadic groups can be subgroups of the Rubik's cube group? --84.61.131.18 (talk) 16:24, 16 July 2010 (UTC)
 * M11 and M12 are subgroups of A12, which is a subgroup of the cube group. Every simple subgroup is isomorphic to a section of some simple composition factor, in other words, is contained as a section of A12 (or A8, but A8 ≤ A12, so this adds nothing new).  The only sporadic simple groups contained in the cube group are M11 and M12. JackSchmidt (talk) 02:19, 17 July 2010 (UTC)

Are there any sporadic groups which have the Rubik's cube group as a subgroup? --84.61.131.18 (talk) 16:24, 16 July 2010 (UTC)
 * No. The only possibility not ruled out by Lagrange's theorem is the monster group, but the cube group is not contained in any of the monster's maximal subgroups, again by Lagrange's theorem. JackSchmidt (talk) 02:19, 17 July 2010 (UTC)

Help with a definite integral
I am trying to solve this integral,


 * $$f(x)=\frac{1}{2\pi} \int_0^{2 \pi} d\theta \, e^{i x \theta} \, \theta \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x \in \mathbb{Z} \!$$

which, I believe, has the solution


 * $$f(x)=\begin{cases}

\pi, & \mbox{if } x = 0 \\ \frac{1}{i x}, & \mbox{if } x \neq 0 \end{cases}\!$$.

However I am having a hard time getting this answer. Using integration by parts, I get


 * $$ \begin{align} f(x)&= \frac{ \theta e^{i x \theta}}{2 \pi \, i \, x} \Bigg|_{\theta=0}^{\theta=2\pi} -\frac{1}{2\pi}\int_0^{2 \pi} d\theta \, \frac{e^{i x \theta}}{i x}  \\

&= \frac{1}{i x} \left( 1-\frac{1}{2\pi}\int_0^{2 \pi} d\theta \, e^{i x \theta} \right)   \\ &= \frac{1}{i x} \left( 1-\delta_{x,0} \right) \end{align} $$.

This gives the correct answer when x is any integer other than zero, but is infinite when $$x=0$$. I could get the correct result for $$x=0$$ just by plugging that in at the beginning. Is there something wrong with my derivation above? Is there a more elegant way to arrive at the correct answer? Thanks   misli  h  20:03, 14 July 2010 (UTC)
 * $$\forall x\;\frac{de^{ix\theta}}{d\theta}=ixe^{ix\theta}$$, but $$\int e^{ix\theta}\,\text{d}\theta\not\equiv \frac{e^{ix\theta}}{ix}+C$$ because of the $$x=0$$ possibility, in which case you have $$\int\text{d}\theta=\theta+C$$, of course. If you write $$\int e^{ix\theta}\,\text{d}\theta=\frac{e^{ix\theta}-1}{ix}+C$$ (which is just using a different C), then the limit works out properly, so you might try that as an intuitive approach, but it's still technically undefined.  So when you do integration by parts you have to split off the $$x=0$$ case just as if you were dividing by x on both sides of an equation.  Then of course you're allowed to assume $$x\neq0$$ later in the other branch, so you would avoid bringing in the Kronecker delta.  --Tardis (talk) 23:36, 14 July 2010 (UTC)
 * Thanks for clearing that up.  misli  h  21:14, 15 July 2010 (UTC)

Denesting Radicals
Hello. Why does $$ \sqrt{a^2 - b^2c} $$ have to be an integer if some nested radical, $$\sqrt{a+b \sqrt{c}\ }$$, can be denested into a sum of surds, $$\sqrt{d}+\sqrt{e}$$? If I rewrite $$ \sqrt{a^2 - b^2c} $$ in terms of $$d$$ and $$e$$, I get $$d - e$$. Thanks in advance. --Mayfare (talk) 22:17, 14 July 2010 (UTC)

If
 * $$ \sqrt{a + b\sqrt{c}} = \sqrt{d} + \sqrt{e}, \, $$

then squaring both sides yields
 * $$ a + b\sqrt{c} = d + e + 2\sqrt{de}. \, $$

Now if a, b, d, e are rational and &radic;c is not, then we have
 * $$ a = d + e \text{ and } b\sqrt{c} = 2\sqrt{de}. \, $$

Therefore
 * $$ a - b\sqrt{c} = d + e - 2\sqrt{de} = ( \sqrt{d} - \sqrt{e} )^2 \, $$

and if everything in site is nonnegative, we can then say
 * $$ \sqrt{a - b\sqrt{c}} = \sqrt{d} - \sqrt{e}. \, $$

Finally we have these two products:

\begin{align} \sqrt{a + b\sqrt{c}} \sqrt{a - b\sqrt{c}} = \sqrt{ a^2 - b^2 c} \\[12pt] (\sqrt{d} + \sqrt{e})(\sqrt{d} - \sqrt{e}) = d - e \end{align} $$ Therefore
 * $$ \sqrt{ a^2 - b^2 c} = d - e. \, $$

So if d &minus; e is an integer then we're done. Michael Hardy (talk) 23:50, 14 July 2010 (UTC)