Prove that every proper subgroup of a group of order 6 is cycli
Let G be a group of order 6 and H be a proper subgroup of G. Since H is a proper subgroup, its order must be a divisor of 6 and must be greater than 1.
Case 1: |H| = 2
If |H| = 2, then H must contain the identity element and one other element. Since every group of order 2 is cyclic, H is cyclic.
Case 2: |H| = 3
If |H| = 3, then H must contain the identity element and two other elements. Since 3 is a prime number, any group of order 3 is cyclic. Therefore, H is cyclic.
Case 3: |H| = 6
If |H| = 6, then H = G and is not a proper subgroup.
Since all proper subgroups of G of order 2 and 3 are cyclic, we can conclude that every proper subgroup of a group of order 6 is cyclic.