[math-fun] A "cycle" in N and proving a quaternion-like group is infinite
While we're on the subject of the meaning of infinite paths in N, would anyone like to comment on the validity/invalidity of the following argument? I was looking at the group G defined by the quaternion-like relations ijk = l jkl = i kli = j lij = k If you define two "permutations of N" t = (1)(2 3)(4 5) (6 7).... u = (1 2)(3 4)(5 6)... their product is a sort of infinite-order "cycle" in N tu = (1 3 5 7 9 11 13 15 17 ...... 16 14 12 10 8 6 4 2). Now consider the mapping f f(i) = t f(j) = t f(k)= u f(l) = u The mapping f seems to be a homomorphism from G into the group of permutations of N that respects the 4 group relations I started out with (when you check the relations, the involutions t and u cancel each other out). So in particular the element f(ik) = tu has infinite order, so G is infinite. I'm sure this is nothing new, but I was happy to find it (even more happy if it is valid). Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm
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Thane Plambeck