Thread: ord(a) = m and a^k=e prove m|k

Ok we have group G and a in G with ord(a) = m and a^k=e prove m|k.

ok so i know that if ord(a) = m in G then m is smalles integer such that a^m=e, we are givin that a^k = e. I have a feeling that we have to use the division algorithm on k so we would get k = nq + r, 0<=r<n, that a^m = a ^(nq+r) then usually in these type of proofs we go ((a^n)^q)a^r and somehow we need to get that this is a multiple of m or a contradiction that nq+r < m but im not sure which way to go.

Originally Posted by ChrisBickle

Ok we have group G and a in G with ord(a) = m and a^k=e prove m|k.

ok so i know that if ord(a) = m in G then m is smalles integer such that a^m=e, we are givin that a^k = e. I have a feeling that we have to use the division algorithm on k so we would get k = nq + r, 0<=r<n, that a^m = a ^(nq+r) then usually in these type of proofs we go ((a^n)^q)a^r and somehow we need to get that this is a multiple of m or a contradiction that nq+r < m but im not sure which way to go.

By the division algorithm we there exists a such that . Thus, . If we see there exists a positive number less than such that contradicting that definition of . Thus, and the conclusion follows.