[math-fun] non-square products of squares?
Let some set be closed under an associative multiplication operator. If it commutes then the product of distinct squares is also square: aabb = abab. Contrarily, can you construct a non-commuting system where this is usually false? That is, where non-trivial products of squares are never square? (Trivializers include null/idem-potent elements, perhaps all self-squares a=aa). For example is there a simple model using 2x2 matrices over some Zn?
On Thursday 18 September 2003 12:15 pm, Marc LeBrun wrote:
Let some set be closed under an associative multiplication operator.
If it commutes then the product of distinct squares is also square: aabb = abab.
Contrarily, can you construct a non-commuting system where this is usually false?
That is, where non-trivial products of squares are never square?
(Trivializers include null/idem-potent elements, perhaps all self-squares a=aa).
For example is there a simple model using 2x2 matrices over some Zn?
Actually there is an easy example from computer science. The concatenation operator is associative and non-commutative. and closed over the set of all finite strings of characters. Let a, b, c be strings as in a = "aA", b = "bB" and c = "cC". Where: ab = "aAbB" abc = "aAbBcC" etc. It is clear that (ab)c = a(bc), that is concatenation is associative, but ab is not usually equal to ba, so is concatenation is non-commutative. The product of the squares aabb is also usually not equal to abab. aabb is rarely a square. Regards Otto oto@olympus.net
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Let me ask for the smallest group in which two squares can multiply to a non-square. I'll try to answer this on-line: 1) We can dismiss abelian groups, and also groups of odd order, since in the latter all elements are squares. 2) this leaves D6, D8, Q8, D10, D12, Q12, A4, D14 below order 16. 3) The squares in D(2.odd) are precisely the elements of C(odd), which kills D6, D10, D14. 4) the squares in D8 or Q8 are the elements of their central C2s, which kills them, leaving only D12,Q12,A4. 5) The squares in D12 are its subgroup C3, since elements outside C6 all square to 1. 6) I can see that the squares in A4 are precisely the elements not of order 2, so: A4 HAS THE PROPERTY! 7) Let's think about Q12 = < a,b : a^6 = 1 = b^4, a^b = a^-1, b^2 = a^3 > The square of (a^k)b is a^k.b.a^k.b = a^k.b.b.a^-k = a^3, for all k. On the other hand, the square of a^k is a^2k. So the set of squares is { 1, a^2, a^3, a^4 }, not a group: Q12 HAS THE PROPERTY! Therefore, the smallest groups in which squares can multiply to a non-square are A4 and Q12; indeed, they are the only such groups with order < 16 (and probably the only ones with order < 20, since I don't believe there will be any of order 16). In Q12, the only instances are a^i.a^j = a^(i+j) for (i,j) = (2,3),(4,3),(3,2),(3,4), whereas in A4 there are lots, so A4 is the better example. Let me try to see how many:- The squares are the elements of order 1 or 3, and the ones we multiply must be both of order 3. These 8 elements form the two cosets of the 4-subgroup, and we must multiply ones from distict cosets, since ones from the same coset will have a product of order 3. But finally, they can't be mutual inverses. So the number of products is 8x3 = 24; we have 1 chance in 6 of getting such a pair if we multiply two elements at random. (For Q12 it was 1 chance in 36.) A much harder and more interesting problem is: what's the largest probability we can attain? John Conway PS. The fact that the squares are precisely the elements of odd order also holds in A5 : let's see what probability that gives. We want products of types 3.3=2, 3.5=2, 5.5=2, so to speak, which we can translate into (3 or 5).2 = 3 or 5. For 5.2: (01234) times (12)(34) = (024... works (13)(24) = (032... works (14)(23) = (04)(13) fails so this gives 10 cases per 5-element, for a total of 240. For 3.2: (012) times (01)(34) fails (01)(24) = (142... works (03)(14) = (041... works so this gives 6+6 = 12 cases per 3-element, giving another 240. So we get 480 = 8 times 60, giving probability of 8/60 = 2/15, worse that 1/6 dammit. The group 2A4 might be better, but I'll do that off-line. John Conway
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