This is just the set of all which have order . Note, this set can be empty.

Since it means there is an isomorphism . To prove that we need to define a one-to-one correspondence . First, if is empty then has to be empty, because if has order then has order also because isomorphisms preserve orders (and the inverse function is an isomorphism too). Suppose is non-empty. For define . Then is in because isomorphisms preserve orders. Now we need to prove that is one-to-one and onto. It is one-to-one because if . It is onto, because if then and . That means the sets and have equal sizes.(2)

If show that, for all , |G(n)| = |H(n)|.

(3)

I start you off. If then . Pick an such that this is not true. Then this would mean that the isomorphism cannot exist.Deduce that, is not .

Is it true that

Is it true that