Verified answer

If you prove (2) then (1) follows.

Yes | means is divisible by.

Let the identity of G be e, then the identity of H is h(e), call that i.

Assume a≠e (otherwise trivial).

Say o(a)=n then the powers of a look like

e, a, a^2, a^3, ..., a^(n-1), e, a, a^2, a^3, ..., a^(n-1), e, a, a^2, a^3, ..., a^(n-1), ...

where no term not explicitly written as e is equal to e. So cycles of length n repeating.

Say h(a) = b, then applying the homomorphism h to the above sequence of powers (and using the fact that h is a homomorphism) gives

i, b, b^2, b^3, ..., b^(n-1), i, b, b^2, b^3, ..., b^(n-1), i, b, b^2, b^3, ..., b^(n-1), ...

which proves (1) since it shows o(h(a))=o(b) ≤ n < ∞. It also shows b^n=i.

Suppose o(b) = k, where k<n, then the powers of b cycle with length k, and some integer number of those cycles, say m of them will give b^n=i. So k * m = n, i.e. k divides n, i.e. o(h(a)) | o(a).

Perhaps, if you want to be more precise, use the division algorithm which guarantees that when you divide n by k you get n = mk + r where m, r are integral and unique with 0≤r<k.

Then i = b^n = b^(mk+r) = (b^k)^m * b^r = i^m * b^r = i * b^r = b^r. This says i = b^r where r is less than k, the order of b. This contradicts the minimality of k unless r=0, in which case n=mk, i.e. the order of b=h(a) divides the order of a.