Recall that if H is normal in G, then we can form the quotient group G/H. The elements of the group G/H are of the form aH where a is an element of G. The operation of the group is defined via
(aH)(bH) = abH.
Moreover, the identity in G/H is eH = H since (aH)(eH) = aH. By our standard theorem about order of an element, we know that |aH| = k if and only (aH)^k = H and k is the smallest such positive exponent. But notice that (aH)^k = a^kH. So, in order to compute the order of aH, we need to find the smallest positive exponent k such that a^kH = H, but a^kH = H exactly when a^k is in H. The upshot is that to find the order of aH, we need the smallest positive k such that a^k is in H.
dcernst
commented
5 years ago
I should outline something like this:
Recall that if H is normal in G, then we can form the quotient group G/H. The elements of the group G/H are of the form aH where a is an element of G. The operation of the group is defined via
(aH)(bH) = abH.
Moreover, the identity in G/H is eH = H since (aH)(eH) = aH. By our standard theorem about order of an element, we know that |aH| = k if and only (aH)^k = H and k is the smallest such positive exponent. But notice that (aH)^k = a^kH. So, in order to compute the order of aH, we need to find the smallest positive exponent k such that a^kH = H, but a^kH = H exactly when a^k is in H. The upshot is that to find the order of aH, we need the smallest positive k such that a^k is in H.