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

1. ## 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.

2. 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 $\displaystyle \ell\in\mathbb{Z}$ such that $\displaystyle k=\ell m+r,\text{ }0\leqslant r<m$. Thus, $\displaystyle e=a^{m}=a^{\ell m}a^k=a^k$. If $\displaystyle r\ne 0$ we see there exists a positive number $\displaystyle r$ less than $\displaystyle k$ such that $\displaystyle a^r=e$ contradicting that definition of $\displaystyle k$. Thus, $\displaystyle r=0$ and the conclusion follows.