Group Theory
Richard Carr
Consider F=the free group on {a,b,c,d,...,z}. Let E be the equivalence
relation generated by ~ (where x~y iff x and y are English words (having
an entry in the OED, say- possibly obsolete or a variant form) which are
anagrams of each other- so for example, altitude~latitude and
clean~lance). Let Q be the quotient group F/E.
The problem is to show that Q is the free Abelian group on
{a,b,c,d,...,z}. (This problem was given to me several years ago,
together with another one (same F, E was with words that sound the same-
homophones, I guess- and the object was to get that Q was trivial, but
that was much easier.)
Anyway, there are 325 relations to get and I've got 302 of them. Any clues
on the missing ones? Here they are:
cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
Richard Carr
PS It's more of a dictionary puzzle than a mathematical one.
P^2S I didn't use any words with accents- because they wouldn't really be
in F, so I am trying to avoid them.
Hugo van der Sanden
>
> Consider F=the free group on {a,b,c,d,...,z}. Let E be the equivalence
> relation generated by ~ (where x~y iff x and y are English words (having
> an entry in the OED, say- possibly obsolete or a variant form) which are
> anagrams of each other- so for example, altitude~latitude and
> clean~lance). Let Q be the quotient group F/E.
> The problem is to show that Q is the free Abelian group on
> {a,b,c,d,...,z}. (This problem was given to me several years ago,
> together with another one (same F, E was with words that sound the same-
> homophones, I guess- and the object was to get that Q was trivial, but
> that was much easier.)
> Anyway, there are 325 relations to get and I've got 302 of them. Any clues
> on the missing ones? Here they are:
>
> cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
I don't understand enough of the maths to get what is required here. Can
you give some examples of the relations you already have, that might
clarify?
Hugo
Gareth Taylor
Richard Carr <ca...@cpw.math.columbia.edu> wrote:
> Consider F=the free group on {a,b,c,d,...,z}. Let E be the equivalence
> relation generated by ~ (where x~y iff x and y are English words (having
> an entry in the OED, say- possibly obsolete or a variant form) which are
> anagrams of each other- so for example, altitude~latitude and
> clean~lance). Let Q be the quotient group F/E.
> The problem is to show that Q is the free Abelian group on
> {a,b,c,d,...,z}.
Hang on. The OED clause is worrying. With that restriction, then
"clean/lance" and "nclea" are distinct elements in F/E, as "nclea" isn't a
word in the OED, so isn't in the same class as "clean". However, they
_are_ the same element in the free abelian group.
If you drop the "having an entry in the OED" bit and confine yourself to
"x~y iff x is a permutation of y" then it's easy.
Gareth
Gareth Taylor
Gareth Taylor <gta...@chiark.greenend.org.uk> wrote:
...something which someone has added to. A friend of mine sent me the
following:
> Oh! *blinks* _Now_ I see. That weasel word `generated' was important.
>
> The equivalence relation _generated by_ ~. That is, not the equivalence
> relation ~ itself (which is a perfectly sensible equivalence relation),
> but the relation that also contains any other equivalences that can be
> deduced from ~ and group theory. Thus:
>
> cat ~ act (defn of ~)
> => cat = act (if it's true in ~, it must be true in generated
> relation =)
> => ca = ac (= is a relation which preserves group-theoretic
> properties)
>
> _That's_ what he's after. A sequence of proofs like that which
> demonstrate Abelianness pairwise between letters of the alphabet.
Is _that_ what you meant?
Gareth
John Rickard
: Richard Carr wrote:
[...]
: > Anyway, there are 325 relations to get and I've got 302 of them. Any clues
: > on the missing ones? Here they are:
: >
: > cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
:
: I don't understand enough of the maths to get what is required here.
Roughly, for each of those letter pairs, find a pair of single-word
anagrams containing the two letters in which they occur in a different
order in the two words. For example, for the first pair "cj", if
"jolace" and "jenticed" were words then cajole/jolace would work, but
injected/jenticed would not, because the "j" is before the "c" in both
"injected" and "jenticed".
I say "roughly" because Richard Carr's requirements are more stringent
than that: anagrams with the property I've described are necessary,
but not sufficient. However, I think that a solution to the problem
I've described is unlikely not to solve the original problem as well.
--
John Rickard
Richard Carr
:Date: 05 Jul 2000 11:07:35 +0100 (BST)
:From: Gareth Taylor <gta...@chiark.greenend.org.uk>
:Newsgroups: rec.puzzles
:Subject: Re: Group Theory
:
:In article <tEg*ng...@news.chiark.greenend.org.uk>,
Yes- like this. If I already have ar=ra (say, from art=rat) then I can use
that jar=raj and ajar=raja to get
jar=raj=rajaa^{-1}=ajara^{-1}=ajraa^{-1}=ajr and hence ja=aj.
The main problem is coming up with the anagrams.
:
:Gareth
:
Richard Carr
:Date: 05 Jul 2000 13:44:38 +0100 (BST)
:From: John Rickard <John.R...@virata.com>
:Newsgroups: rec.puzzles
:Subject: Re: Group Theory
:
:Hugo van der Sanden <h...@crypt0.demon.co.uk> wrote:
:: Richard Carr wrote:
:[...]
:: > Anyway, there are 325 relations to get and I've got 302 of them. Any clues
:: > on the missing ones? Here they are:
:: >
:: > cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
::
:: I don't understand enough of the maths to get what is required here.
:
:Roughly, for each of those letter pairs, find a pair of single-word
:anagrams containing the two letters in which they occur in a different
:order in the two words. For example, for the first pair "cj", if
:"jolace" and "jenticed" were words then cajole/jolace would work, but
:injected/jenticed would not, because the "j" is before the "c" in both
:"injected" and "jenticed".
:
Right.
:I say "roughly" because Richard Carr's requirements are more stringent
:than that: anagrams with the property I've described are necessary,
:but not sufficient. However, I think that a solution to the problem
:I've described is unlikely not to solve the original problem as well.
:
Indeed, at this stage, with so few relations left, it is more than likely
that this approach would work too- the only problem being
that all the relations involve j, q, x or z so that there is
a chance that finding an anagram may not work simply because it has at
has, for instance, c, f and j in. From caftajack to jaftacack (both of
which I just made up for illustrative purposes) one couldn't conclude
cj=jc of fj=jf because the j has to get past both the f and the c. The
best you could conclude is that jc=cj if and only if fj=jf (which would be
useful in the case of finding some other relation that settled one case or
the other).
:--
:John Rickard
:
Richard Carr
:Date: Wed, 5 Jul 2000 13:04:47 -0700 (PDT)
:From: David Beckman <bec...@its.caltech.edu>
:To: ca...@cpw.math.columbia.edu
:Newsgroups: rec.puzzles
:Subject: Re: Group Theory
:
:In rec.puzzles you write:
:
:
:>Consider F=the free group on {a,b,c,d,...,z}. Let E be the equivalence
:>relation generated by ~ (where x~y iff x and y are English words (having
:>an entry in the OED, say- possibly obsolete or a variant form) which are
:>anagrams of each other- so for example, altitude~latitude and
:>clean~lance). Let Q be the quotient group F/E.
:>The problem is to show that Q is the free Abelian group on
:>{a,b,c,d,...,z}. (This problem was given to me several years ago,
:>together with another one (same F, E was with words that sound the same-
:> homophones, I guess- and the object was to get that Q was trivial, but
:>that was much easier.)
:>Anyway, there are 325 relations to get and I've got 302 of them. Any clues
:>on the missing ones? Here they are:
:>
:>cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
:>
:>Richard Carr
:
:Hi Richard,
:
:I used a word list from some Merriam-Webster dictionary, which has some
:questionable words, and ran it through some anagram programs I had around.
:I checked with OED online and found some of the words. The pairs below
:were in OED as main entries except where noted. (Some were only found in
:a full-text search; some weren't there at all and seem to be variants or
:specialized terminology.) I don't know how much you want to stretch the
:OED words....
Thanks- that was fast. I'm not sure about the status of capitalized words
(I'll have to check with the guy who asked me the puzzle). Also work-box
seems to be hyphenated (whereas words in F are not).
Anyway, angelique and equiangle checked as did philozoic and
zoophilic. Hejazi and jeziah did too (so I'm going to check if proper
words are OK). I found myxofibroma but not fibromyxoma and cajoler but not
jecoral. I couldn't find several of the later words in the OED and don't
have a M-W to check but I'll see if I can get access to one.
I did find both the words for qv- what worries me here is that one is
capitalized (and part of a compound sort of expression) and the other is
accented. Anyway, I thank you for your help and look forward to verifying
all of this.
:
:cj:
:cajoler
:jecoral (jecorary)
:
:fx:
:fibromyxoma (fibroma, myxoma)
:myxofibroma (fibroma, myxoma)
:
:fibromyxosarcoma (fibroma, myxoma, sarcoma)
:myxofibrosarcoma (fibroma, myxoma, sarcoma)
:
:gq:
:angelique
:equiangle
:
:aqueoigneous
:igneoaqueous
:
:hz:
:Hejazi (found in OED full text, main entry Iraqi)
:jeziah
:
:Phytozoa (found in full text)
:Zoophyta (found in full text)
:zoopathy (zoopathology)
:
:philozoic (found in full text)
:zoophilic
:
:philozoist (found in full text)
:zoophilist
:
:Rhizoctonia
:Zonotrichia (found in full text)
:chorization
:
:jz:
:Hejazi (found in full text)
:jeziah
:
:kx:
:boxwork (not found in OED, but a common term for a type of cave formation:
: http://www.goodearthgraphics.com/virtcave/boxwork/boxwork.html)
:workbox
:
:hexakisoctahedron
:octakishexahedron
:
:hexakistetrahedron
:tetrakishexahedron
:
:kz:
:alkalizer (alkalize)
:lazarlike (lazar)
:
:qv:
:eveque (Pont L'Eveque, cheese variety)
:quev'ee (acute accent over second e)
:
:wx:
:boxwork
:workbox
:
:xz:
:azoxine (var. of azoxy? A web search turns up some matches, but not in OED)
:oxazine
:
:
: Dave Beckman (bec...@its.caltech.edu)
:
Hugo van der Sanden
> Hugo van der Sanden <h...@crypt0.demon.co.uk> wrote:
> : Richard Carr wrote:
> [...]
> : > on the missing ones? Here they are:
> : >
> : > cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
> :
>
> Roughly, for each of those letter pairs, find a pair of single-word
> anagrams containing the two letters in which they occur in a different
> order in the two words. For example, for the first pair "cj", if
> "jolace" and "jenticed" were words then cajole/jolace would work, but
> injected/jenticed would not, because the "j" is before the "c" in both
> "injected" and "jenticed".
>
> than that: anagrams with the property I've described are necessary,
> but not sufficient. However, I think that a solution to the problem
> I've described is unlikely not to solve the original problem as well.
Right, along with the extra explained in Richard's reply to Gareth
Taylor I now understand, and was able to confirm that nothing in my own
wordlist will help. :(
Thanks,
Hugo
David Moews
Richard Carr <ca...@cpw.math.columbia.edu> wrote:
|relation generated by ~ (where x~y iff x and y are English words (having
|an entry in the OED, say- possibly obsolete or a variant form) which are
|anagrams of each other- so for example, altitude~latitude and
|{a,b,c,d,...,z}. ...
|Anyway, there are 325 relations to get and I've got 302 of them. Any clues
|on the missing ones? Here they are:
|
|cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
What anagrams make j and v commute?
Conway, Feit, and Sahi have written a paper on this. Here are some anagrams
from their paper, together with some Chris Cole, Chris Long, Murray Pearce,
and I found in 1993:
(ni2 or 3 = merriam-webster's second or third new international dictionary;
9c = merriam-webster's collegiate dictionary, 9th ed.)
fj fijf jiff (oed,9c) [murray pearce]
fq foursquare quarefours (oed) [chris long]
fx expressionful superflexions (ni2)
fx loxify foxily [conway-feit-sahi]
gx exerting genetrix (ni2, ni3 or oed2) [chris cole]
jw jackscrew screwjack [conway-feit-sahi]
jz jiz zij (oed2, funk & wagnall new standard) [chris long]
qv eveque queeve (ni2)
wx boxwood woodbox [conway-feit-sahi]
wz waz zaw (oed2) [chris long]
Also, AZOXINE is in NI2.
Together with the anagrams found by you and David Beckman, these cover all
pairs except for (j,q), (j,x), (k,q), (q,w), (q,x), (q,y), and (v,x).
--
David Moews dmo...@xraysgi.ims.uconn.edu
Richard Carr
:Date: 6 Jul 2000 11:39:50 -0700
:From: David Moews <dmo...@xraysgi.ims.uconn.edu>
:Newsgroups: rec.puzzles
:Subject: Re: Group Theory
:
:In article <Pine.LNX.4.21.000705...@cpw.math.columbia.edu>,
:Richard Carr <ca...@cpw.math.columbia.edu> wrote:
:|Consider F=the free group on {a,b,c,d,...,z}. Let E be the equivalence
:|relation generated by ~ (where x~y iff x and y are English words (having
:|an entry in the OED, say- possibly obsolete or a variant form) which are
:|anagrams of each other- so for example, altitude~latitude and
:|clean~lance)...The problem is to show that [F/E] is the free Abelian group on
:|{a,b,c,d,...,z}. ...
:|Anyway, there are 325 relations to get and I've got 302 of them. Any clues
:|on the missing ones? Here they are:
:|
:|cj,fj,fq,fx,gq,gx,hz,jq,jw,jx,jz,kq,kx,kz,qv,qw,qx,qy,vx,wx,wz,xz.
:
:What anagrams make j and v commute?
javar and vajra
javar: A swelling in the hollow of the pastern of a horse
vajra: In Hinduism and Buddhism, a thunderbolt or mythical weapon esp. one
wielded by the god Indra.
Progress towards the solution of this problem can be found at
http://www.math.columbia.edu/~carr/Matt.html
(Matt told me about the problem.)
I have put up those of David Beckman's relations (which I have
confirmed) together with those I had already found (his are listed
with "found by David Beckman") and as I confirm the other ones (yours
and his) will add those to the list also with appropriate accredation
(e.g. "found by Chris Cole, shown to me by David Moews"). I have also
confirmed that workbox need not be hyphenated as I first thought. I was
thus able to use boxwood=woodbox to get wx=xw and from that kx=kx using
David's workbox=boxwork.
:
:Conway, Feit, and Sahi have written a paper on this. Here are some anagrams
:from their paper, together with some Chris Cole, Chris Long, Murray Pearce,
:and I found in 1993:
:
:(ni2 or 3 = merriam-webster's second or third new international dictionary;
: 9c = merriam-webster's collegiate dictionary, 9th ed.)
:
:fj fijf jiff (oed,9c) [murray pearce]
I was able to verify this (indeed both are in the OED).
:fq foursquare quarefours (oed) [chris long]
four-square is hyphenated, unfortunately. I was also able to confirm that
capitalized words are not allowed, unfortunately.
Actually, I just looked in the Webster's and foursquare is not hyphenated
there (although it is in the OED).
:fx expressionful superflexions (ni2)
:fx loxify foxily [conway-feit-sahi]
:gx exerting genetrix (ni2, ni3 or oed2) [chris cole]
I was also able to confirm this.
:jw jackscrew screwjack [conway-feit-sahi]
Confirmed jackscrew in Webster's but "screw jack" is listed as 2 words.
:jz jiz zij (oed2, funk & wagnall new standard) [chris long]
:qv eveque queeve (ni2)
:wx boxwood woodbox [conway-feit-sahi]
Confirmed.
:wz waz zaw (oed2) [chris long]
:
:Also, AZOXINE is in NI2.
I need to get or find copies of ni2, ni3 and oed2 evidently. : )
I did find lazarlike in Webster's but not alkalizer and also I found
hexakisoctahedron and hexakistetrahedron but not octakishexahedron or
tetrakishexahedron. I will search for them when I get hold of more
dictionaries.
Richard Carr
PS Thanks by the way- I have been on this problem on and off for a few
years (don't have a good way of searching a dictionary, unfortunately so
mainly I just guess words and look them up to see if they exist- it takes
a lot of time that way- I imagine you guys have better methods).
:
:Together with the anagrams found by you and David Beckman, these cover all
:
John Rickard
: (This problem was given to me several years ago, together with
: another one (same F, E was with words that sound the same-
: homophones, I guess- and the object was to get that Q was trivial,
: but that was much easier.)
I've been thinking (not all that hard) about that one; it does indeed
seem to be much easier. I can show that every letter except M and V
is the identity, using only (what I consider to be) common words and,
so far as I know, using homophones that don't depend on the variety of
English one speaks. I can get M in British English, but I don't know
whether my example works in American English.
What did you use for M and V? (For those less mathematically
inclined, find pairs of homophones where M or V occurs in one word but
not the other, or a different number of times in each word.)
Hah -- just thought of a V as I was writing this, though one of the
words is slang.
--
John Rickard <John.R...@virata.com>
Richard Carr
:Date: 07 Jul 2000 18:08:25 +0100 (BST)
:From: John Rickard <j...@255.255.255.192>
:Reply-To: John.R...@virata.com
:Newsgroups: rec.puzzles
:Subject: Re: Group Theory
:
:Richard Carr <ca...@cpw.math.columbia.edu> wrote:
:: (This problem was given to me several years ago, together with
:: another one (same F, E was with words that sound the same-
:: homophones, I guess- and the object was to get that Q was trivial,
:: but that was much easier.)
:
:I've been thinking (not all that hard) about that one; it does indeed
:seem to be much easier. I can show that every letter except M and V
:is the identity, using only (what I consider to be) common words and,
:so far as I know, using homophones that don't depend on the variety of
:English one speaks. I can get M in British English, but I don't know
:whether my example works in American English.
:
:What did you use for M and V? (For those less mathematically
:inclined, find pairs of homophones where M or V occurs in one word but
:not the other, or a different number of times in each word.)
:
It's a few years ago now, but I remember that I used bevy=bevvy for v (both
in OED and have same pronunciation- the phonetic spellings are given
and they matched). I can't recall off hand what I used for m- almost
certainly one word had a double m where the other had just a single m. I
used 26 relations on this one.
:Hah -- just thought of a V as I was writing this, though one of the
: