1. ## Deduce Subgroup property

Hey I'm having some trouble with this question
Deduce that any group of order 2p. where p is a prime p>2 Must contain a subgroup of order p
Heres what I've come up with so far.

Since G is closed this implies that for every element $g \in G$

$g^q = I$

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?)

2. Originally Posted by thelostchild
Hey I'm having some trouble with this question

Heres what I've come up with so far.

Since G is closed this implies that for every element $g \in G$

$g^q = I$

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?)
If you apply the Cauchy's theorem, the group of order 2p ( p: a prime number) has an element of order 2 and an element of order p. Find the relations of the elements of this group. Then you will see that a group of order 2p is either a cyclic group or a dihedral group. In either case, this group contains a subgroup of order p.

3. Originally Posted by TheArtofSymmetry
If you apply the Cauchy's theorem, the group of order 2p ( p: a prime number) has an element of order 2 and an element of order p. Find the relations of the elements of this group. Then you will see that a group of order 2p is either a cyclic group or a dihedral group. In either case, this group contains a subgroup of order p.
Yes there is a theorem that states a Group of order 2p is isomorphic to either
$Z_{2p}$ or $D_p$ . if its $Z_{2p}$then it's cyclic and the answer is immediate. if it's $D_p$ then it contains a subgroup of order .... it's immediate also.

4. Originally Posted by thelostchild
Hey I'm having some trouble with this question

Heres what I've come up with so far.

Since G is closed this implies that for every element $g \in G$

$g^q = I$

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 $p\neq 2$ ):

1) if $G$ is abelian then:

(i) if $G=$ 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 $G$ is not cyclic: define $G_2:=\{x\in G\;;\;x^2=1\,,\,x\neq 1\}$. Check now that $y:=\prod_{x\in G_2}x\notin G_2\cup\{1\}$ , so $$ solves our problem.

2) If $G$ 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 $g^2=1$ must be abelian), and thus there's an element of order p, and again we're done.

Tonio