Prove that a group of order n>2 cannot have a subgroup of order n-1?

Answer 1

Let G be a group with three or more elements. Let e be the identity. G - { e } is obviously not a subgroup, so consider instead G - { a } for some a != e. G has at least three elements so there is also a b with b != e and b != a.

Let x be the element of G such that bx = a, i.e., x = b^-1 a. If x is in G - { a }, then we have b in G - { a } and x in G - { a ] and their product bx = a not in G - { a }, so G - { a } is not closed under the group operation and so is not a subgroup of G. So x cannot be in G - { a }, which means x = a. But then bx = ba = a implies b = e, contradicting the assumption that b != e.

Go G - { any one element } either lacks the identity element or is not closed under the group operation if |G| > =3, and so no group G or order |G| >= 3 can have a subgroup omitting just one element.

The proof works for infinite G, too.

Dan

Answer 2

Lagrange's Theorem says the order of a subgroup must divide the order of the group. If n>2, then n-1 does not divide n.