Hey I'm having some trouble with this question
Heres what I've come up with so far.Deduce that any group of order 2p. where p is a prime p>2 Must contain a subgroup of order p
Since G is closed this implies that for every element
where q is a positive integer. Hence each element of g generates a cyclic subgroup of G of order q.
By Lagranges therom q can only be 2 or p.
The trouble I'm having now is proving G cannot just consist of elements of order 2 (or did I misunderstand the question?)
I'm assuming you haven't yet heard of Cauchy's theorem of Sylow theorems, otherwise your question is trivial, so we can begin by checking cases ( and assuming ):
1) if is abelian then:
(i) if is cyclic then solves our problem (use here what you said: the order of any element in the group is exactly one of the following: 1,2, p, 2p)
ii) if is not cyclic: define . Check now that , so solves our problem.
2) If is not abelian then it can't be that all its non-unit elements have order 2 (why? Prove that a group in which all its elements satisty the condition must be abelian), and thus there's an element of order p, and again we're done.