Hmm, so if a is the identity in G, then it has order 1. Thus f(a) will give the identity in H, and then the order of f(a) is simply 1 again, so this meaning o(a)=o(f(ah))??
Is that what you're getting at?
...And then you would just prove it for arbitrary a and b of H, such that o(a+b)=o(f(a)+f(b))? Etc, basically applying the order to the ring axioms with order being n?...
^not sure if that top part makes sense lol
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