I need a break from actually doing homework, so here's a proof.
Claim: If |G|=8, then G has an element a of order 2.
Pf: A corollary of Lagrange's Theorem states that a^|G|=e for all a in G. Then, a^8=e, implying that the possible orders of a are 1, 2, 4, or 8. By the property of closure, if a is in G, then a^k is in G, where k in any integer. Then, let |a|=8; this implies a^8=e. Then, a^4 is distinct from a^8 and is in G; so, (a^4)^2=e and thus G has an element of order 2. Similarly, let |a|=4; this implies a^4=e. Then, a^2 is distinct from a^2 and is in G; so, (a^2)^2=e and thus G has an element of order 2. If a is of order 2, then, obviously, G has an element of order 2. Finally, if a is of first order, then a=e; however, e can be the only first-order element in any group, so not all 8 elements of G can be e, meaning that there is at least one element of order 2. QED.
That was fun, if a little facile. The interesting thing here is that Lagrange's Theorem states that the order of a subgroup H divides the order of the group G, and, consequentially, the order of any element a in G divides the order of G. This does not imply that G MUST contain elements and subgroups of each divisor of its own order, just that its possible. This problem is oriented towards overcoming the temptation to assume the truth of the converse of Lagrange's Theorem.
Ok. Back to homework. Another proof later, perhaps.