Prove that every proper subgroup of a group of order 6 is cyclic.
Let G be a group of order 6 and let H be a proper subgroup of G.
Since H is a proper subgroup, its order must divide the order of G by Lagrange's Theorem. Thus, the order of H can be 1, 2, 3, or 6.
If the order of H is 1, then H is the trivial subgroup {e}, which is trivially cyclic.
If the order of H is 2, then H must be isomorphic to the group Z_2, which is cyclic.
If the order of H is 3, then H must be a cyclic group of order 3, since the only group of order 3 is isomorphic to Z_3.
If the order of H is 6, then H = G, which would not be a proper subgroup.
Therefore, every proper subgroup of a group of order 6 is cyclic.