- #1
SiddharthM
- 176
- 0
Ok, well a corollary to Lagrange's theorem is that every group of prime order, call it G, must be cyclic. Consider the cyclic subgroup of G generated by a (a not equal to e), the order of the subgroup must divide the order of p, since the only number less than or equal to p that divides p is p, the cyclic subgroup of G is G and ANY element of G that is not the identity generates G.
Now, looking at the finite group of {1,...,11} whose multiplication is ab reduced mod12.
Take 3, by the corollary above, 3 must generate the entire group because 11 is a prime number.
3*3=9
3*3*3=3 (because 27 = 3mod12)
3*3*3*3= 9 (because 81 =9mod12)
(3*3*3*3)*3=9*3 = 3
and so on and so forth, so 3 does NOT generate the group {1,...,11}.
Clearly Lagrange's theorem is not false and neither is it's corollary above. So I must be missing something pretty simple and i feel retarded because i can't see it.
Now, looking at the finite group of {1,...,11} whose multiplication is ab reduced mod12.
Take 3, by the corollary above, 3 must generate the entire group because 11 is a prime number.
3*3=9
3*3*3=3 (because 27 = 3mod12)
3*3*3*3= 9 (because 81 =9mod12)
(3*3*3*3)*3=9*3 = 3
and so on and so forth, so 3 does NOT generate the group {1,...,11}.
Clearly Lagrange's theorem is not false and neither is it's corollary above. So I must be missing something pretty simple and i feel retarded because i can't see it.