TITLE: Looking for a BAMS text about the group with commutation relations defined using meaningful words
QUESTION [5 upvotes]: What I definitely remember is that I saw a description of the following in the Bulletin of the American Mathematical Society, sometime in eighties (or maybe nineties?)
One considers the group generated by letters of the Latin alphabet, with the following relations: $W(a,b,...,y,z)=W'(a,b,...,y,z)$ for any words that are permutations of each other and moreover each of them is a meaningful word in English. For example, $name=mane=mean=amen$ in this group.
The question was whether this group is commutative.
Does anybody remember it? Or maybe knows the answer?
Slightly later
Neil Strickland points out that it is Monthly (1986) rather than BAMS.
So formally he answered the question, except that there is another thing I did remember incorrectly: the question there was what is the center of this group!
So, what is the center of this group?

REPLY [5 votes]: I checked that in /usr/dict/words there are a handful of words in which q appears without a following u, but none of them are free of repetition and have any anagrams.  This means that we have a surjective homomorphism to the free group $\langle q,u\rangle$ sending all other letters to the identity, so $q$ and $u$ do not commute, as Jeremy suggested.
[UPDATE] My original answer assumed that we were only using words with no repeated letters, in which case Jeremy's comment would not be applicable.  I now see that this restriction is not in fact in the original problem.  Nonetheless, I have now run a brute-force check including words with repeated letters and reached the same conclusion.  If you take any word that possesses anagrams and map it to $\langle q,u\rangle$ you get either a power of $u$ or the element $qu$.