Homework #5 Solutions (due 10/10/06)
... Yet another way of expressing this is that NG may be regarded as a function on the set of conjugacy classes of subgroups. Now we note that almost all of our subgroups can be identified as either cyclic subgroups or as certain normalizers (or centralizers). Cyclic subgroups are easily divided into c ...
... Yet another way of expressing this is that NG may be regarded as a function on the set of conjugacy classes of subgroups. Now we note that almost all of our subgroups can be identified as either cyclic subgroups or as certain normalizers (or centralizers). Cyclic subgroups are easily divided into c ...
Ring Theory Solutions
... that x ≡ a mod m and x ≡ b mod n. (Hint: Consider the remainders of a, a + m, a + 2m, . . . , a + (n − 1)m on division by n.) Solution: Consider the remainders of a, a + m, a + 2m, . . . , a + (n − 1)m on division by n. We claim no two remainder is same. Suppose if (a + im) mod n = (a + jm) mod n, t ...
... that x ≡ a mod m and x ≡ b mod n. (Hint: Consider the remainders of a, a + m, a + 2m, . . . , a + (n − 1)m on division by n.) Solution: Consider the remainders of a, a + m, a + 2m, . . . , a + (n − 1)m on division by n. We claim no two remainder is same. Suppose if (a + im) mod n = (a + jm) mod n, t ...
