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Math Help - ord(a) = m and a^k=e prove m|k

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    Post 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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ChrisBickle View Post
    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 \ell\in\mathbb{Z} such that k=\ell m+r,\text{ }0\leqslant r<m. Thus, e=a^{m}=a^{\ell m}a^k=a^k. If r\ne 0 we see there exists a positive number r less than k such that a^r=e contradicting that definition of k. Thus, r=0 and the conclusion follows.
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