Re: [math-fun] non-square products of squares?
John Conway writes:
Let me ask for the smallest group in which two squares can multiply to a non-square....
It's nice to see this. I was going to look with Gap, and your analysis helped convince me.
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).
I found two such groups of order 16, but I haven't figured out what their names are. A presentation for the first is <a,b : a^2 = b^4 = (b a b)^2 = (b^-1 a b a)^2 = 1> . Its squares are 1, b^2, and (a b)^2, and the product of the last two is not a square. The other group may be presented <a,b : a^4 = b^4 = a b a^-1 b = 1>. Its squares are 1, a^2, b^2, and again the product of the two non-identity squares is not a square.
A much harder and more interesting problem is: what's the largest probability we can attain?
The largest among groups up to order 255 is 5/6, in a 192-element group. Gap calls it SmallGroup([192,1022]), but again I don't know a name for it. The smallest presentation I've found is <a,b,c : a^3 = b^4 = c^4 = [b,c] = [a,b^2] = [a,c^2] = (a c)^3 = (a b)^3 = [c,a^-1 b^-1 a] = 1> where [x,y] means x y x^-1 y^-1. This is edged out by probability 27/32, in SmallGroup([384,572]), which I haven't looked at in detail. Dan
On Mon, 22 Sep 2003, Dan Hoey wrote:
John Conway writes:
Let me ask for the smallest group in which two squares can multiply to a non-square....
It's nice to see this. I was going to look with Gap, and your analysis helped convince me.
I found two such groups of order 16, but I haven't figured out what their names are. A presentation for the first is
<a,b : a^2 = b^4 = (b a b)^2 = (b^-1 a b a)^2 = 1> .
Its squares are 1, b^2, and (a b)^2, and the product of the last two is not a square.
I don't immediately recognise this group in that presentation. I'll play with it offline to see which one it is.
The other group may be presented <a,b : a^4 = b^4 = a b a^-1 b = 1>.
Its squares are 1, a^2, b^2, and again the product of the two non-identity squares is not a square.
This group is called 4:4 in my system. [In general, A:B denotes the split extension of a group of structure A by one of structure B, and it is usually assumed that it's not the direct product, since we'd call that A x B.]
A much harder and more interesting problem is: what's the largest probability we can attain?
The largest among groups up to order 255 is 5/6, in a 192-element group. Gap calls it SmallGroup([192,1022]), but again I don't know a name for it. The smallest presentation I've found is
<a,b,c : a^3 = b^4 = c^4 = [b,c] = [a,b^2] = [a,c^2] = (a c)^3 = (a b)^3 = [c,a^-1 b^-1 a] = 1>
where [x,y] means x y x^-1 y^-1.
This is edged out by probability 27/32, in SmallGroup([384,572]), which I haven't looked at in detail.
Exactly what are these the probabilities of? I was using the probability that a given triple a,b,ab should be of form square,square,nonsquare. Are yours the conditional probability that ab should be a non-square GIVEN that a,b are squares? [I mention that there is a third natural probability around, namely the probability that a^2.b^2 should be a nonsquare.] Regards, JHC
There's a digital clock at my bedside that shows hours minutes and seconds, and which I usually look at the moment I wake up. A few days ago, I happened to wake up at a palindromic time, and out of interest, waited until the next palindromic time, and noticed the interval between them. Last night exactly the same thing happened, but the interval was exactly four times as long as in the previous case. What did the clock read when I woke up last night? John Conway
On Tuesday 23 September 2003 08:23 am, John Conway wrote:
There's a digital clock at my bedside that shows hours minutes and seconds, and which I usually look at the moment I wake up. A few days ago, I happened to wake up at a palindromic time, and out of interest, waited until the next palindromic time, and noticed the interval between them.
Last night exactly the same thing happened, but the interval was exactly four times as long as in the previous case. What did the clock read when I woke up last night?
John Conway
12:21 to 1:01 = 40minutes 1:01 - 111 = 10 111 - 121 = 10 etc. Regards Otto otto@olympus.net
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Dear Otto, You didn't read carefully enough - it said hours, minutes AQND seconds - JHC On Tue, 23 Sep 2003, otto2hoh wrote:
On Tuesday 23 September 2003 08:23 am, John Conway wrote:
There's a digital clock at my bedside that shows hours minutes and seconds, ...
Last night exactly the same thing happened, but the interval was exactly four times as long as in the previous case. What did the clock read when I woke up last night?
12:21 to 1:01 = 40minutes
1:01 - 111 = 10 111 - 121 = 10 etc.
Regards Otto otto@olympus.net
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For a related problem, last year's US Puzzle Championship included the following problems from Nob Yoshigahara: What are the shortest and longest time spans from one palindromic time to the next? This assumes a 24-hour clock without leading zeros. Nick
On Tuesday 23 September 2003 08:23 am, John Conway wrote:
There's a digital clock at my bedside that shows hours minutes and seconds, and which I usually look at the moment I wake up. A few days ago, I happened to wake up at a palindromic time, and out of interest, waited until the next palindromic time, and noticed the interval between them.
Last night exactly the same thing happened, but the interval was exactly four times as long as in the previous case. What did the clock read when I woke up last night?
John Conway
On Tue, 23 Sep 2003, Nick Baxter wrote:
For a related problem, last year's US Puzzle Championship included the following problems from Nob Yoshigahara: What are the shortest and longest time spans from one palindromic time to the next? This assumes a 24-hour clock without leading zeros.
I think my question is neater, because to justify your answer you have to enumerate all the possible intervals. I should have said that it assumes a 12 hour clock. JHC
280 secs. It also assumes a clock without markings between the hours and the minutes and the minutes and the seconds. Regards Otto On Tuesday 23 September 2003 02:05 pm, John Conway wrote:
On Tue, 23 Sep 2003, Nick Baxter wrote:
For a related problem, last year's US Puzzle Championship included the following problems from Nob Yoshigahara: What are the shortest and longest time spans from one palindromic time to the next? This assumes a 24-hour clock without leading zeros.
I think my question is neater, because to justify your answer you have to enumerate all the possible intervals.
I should have said that it assumes a 12 hour clock. JHC
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What are the shortest and longest time spans from one palindromic time to the next? This assumes a 24-hour clock without leading zeros.
I think my question is neater, because to justify your answer you have to enumerate all the possible intervals.
I should have said that it assumes a 12 hour clock. JHC
Here is another related problem, I am watching palindromic numbers go by in local time on my on my computers 12-hour clock without leading zeros. I count 281 seconds between two palindromes. How long will it be before I count 281 seconds between two number again? Regards Otto
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On Tuesday 23 September 2003 08:23 am, John Conway wrote:
There's a digital clock at my bedside that shows hours minutes and seconds, and which I usually look at the moment I wake up. A few days ago, I happened to wake up at a palindromic time, and out of interest, waited until the next palindromic time, and noticed the interval between them.
Last night exactly the same thing happened, but the interval was exactly four times as long as in the previous case. What did the clock read when I woke up last night?
I should learn to read more carefully. 12:55:21 Regards Otto
John Conway
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On Tue, 23 Sep 2003, John Conway wrote:
On Mon, 22 Sep 2003, Dan Hoey wrote:
I found two such groups of order 16, but I haven't figured out what their names are. A presentation for the first is
<a,b : a^2 = b^4 = (b a b)^2 = (b^-1 a b a)^2 = 1> .
Its squares are 1, b^2, and (a b)^2, and the product of the last two is not a square.
I don't immediately recognise this group in that presentation. I'll play with it offline to see which one it is.
I've now done so. As I expected, it's what I think of as "the last group" of order 16, which Coxeter calls (4,4|2,2) (except that I may have the wrong style of bracket or punctuation): 1 = a^4 = b^4 = (ab)^2 = (a^-1.b)^2. [This is more symmetrical than the presentation you give - my a can be taken as your ab.] The elements can be written a^i.b^j, and exactly half of them (those for which i+j is odd) have order 4. They are of two "sexes" according as i is odd and j even or i even and j odd, and any two elements of opposite sexes generate the group in the above presentation. The square of an element of order 4 is a^2 or b^2, depending only on its sex, so of course it's an example. [I remark that a^2 and b^2 generate the center.]
The other group may be presented <a,b : a^4 = b^4 = a b a^-1 b = 1>.
Its squares are 1, a^2, b^2, and again the product of the two non-identity squares is not a square.
This group is called 4:4 in my system. [In general, A:B denotes the split extension of a group of structure A by one of structure B, and it is usually assumed that it's not the direct product, since we'd call that A x B.]
This group is much easier to think about, since it's that split extension. The square of a^i.b^j is a^(2i) if j is even, and b^2 if j is odd, so superficially it looks very like the other (in the new presentation) in respect of this behavior, though this time there's no symmetry between a and b. When I guessed there wouldn't be examples of order 16, I planned to check (4,4|2,2) just in case, because it's the most interesting one, but forgot. [I "learned" the groups of order 16 by heart long ago, though now I may be a bit shaky. The standard reference for them is a table in the back of Coxeter and Moser, but they describe them in ways that don't tell you the structure, so it's better to use ATLAS-type notations that do.] John Conway
On Tue, 23 Sep 2003, John Conway wrote:
[I "learned" the groups of order 16 by heart long ago, though now I may be a bit shaky. The standard reference for them is a table in the back of Coxeter and Moser, but they describe them in ways that don't tell you the structure, so it's better to use ATLAS-type notations that do.]
Since writing that a few minutes ago, I've reconstructed them. They are the abelian ones: 16, 8x2, 4x4, 4x2^2, 2^4, the dihedral and quaternionic ones D16, Q16, 2xD8, 2xQ8, 2D8=2Q8, the split extensions 8 :^3 2, 8 :^5 2, 4 :^3 4, and the last group (4,4|2,2). In general D2n = < a,b | 1 = a^n = b^2, a^b = a^-1 >, Q4n = < a,b | 1 = a^2n, b^2 = a^n, a^b = a^-1 >, while 2D4n = 2Q4n is obtained from either of these by adjoining a new central square root of a^n. Finally m :^k n = < a,b | 1 = a^m = b^n, a^b = a^k > and (m,n|p,q) = < a,b | 1 = a^m = b^n = (ab)^p = (ab^-1)^q >. [I'm a little bit scared that maybe I've missed one out and two of the above are isomorphic, but I don't think so.] JHC
I said:
[I'm a little bit scared that maybe I've missed one out and two of the above are isomorphic, but I don't think so.]
I've now checked their abstract distinctness, so know they're OK, since I do remember the number 14. They are mostly distinguished by giving their centers and order spectrum, thus: number elts group center given order 16 16 1+1+2+4+8 8x2 8x2 1+3+4+8 4x4 4x4 1+3+12 4x2^2 4x2^2 1+7+8 2^4 2^4 1+15 D16 2 1+9+2+4 Q16 2 1+1+10+4 2xD8 2^2 1+11+4 2xQ8 2^2 1+3+12 2Q8 4 1+7+8 8 :3 2 2 1+5+6+4 8 :5 2 2^2 1+3+4+8 4 :3 4 2^2 1+3+12 4,4|2,2 2^2 1+7+8 The only candidate for an isomorphism that these invariants leave is between 2xQ8 and 4 :^3 4, which is prohibited by the unique factorisation theorem for direct products, since it's easy to see that the latter of these two isn't a non-trivial direct product. The argument is fairly robust, and in any case agrees with my memory, so these are indeed the groups. However, if anyone finds any errors in the above little table, I'd like to hear about them. JHC
participants (4)
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Dan Hoey -
John Conway -
Nick Baxter -
otto2hoh