Algebra: Group theory... HELPPP

Suppose G is a group of order 2p where p is an odd prime number. Using Lagrange’s theorem, show that G must contain an element of
order p.

Well, I think proof by contradiction would be a good idea but I don't know where to start. Do you have any advice?

Suppose G is a group of order 2p where p is an odd prime number. Using Lagrange’s theorem, show that G must contain an element of
order p.

I honestly have no idea, do you think you could explain it please?

If no elements have order p, then for each $g \in G$ there are three options for the order of $g$. What are those options, and what does each imply?

3 options, are you sure?

How many do you think there are? If it's 2, you might not be considering the order 1, which is still a (pretty trivial) option.

I'm probably not seeing something here, but wouldn't it be What other options are there?

What are the factors of 2p which are not 1,2,p? You've missed the most obvious one :P

For all $g \in G$, we have that $|g| = 1,2,p,2p$, we also have that if $g$ has order $2p$, then $g^2$ has order p.

So if there were no elements with order p, then all elements have order 1 or 2?

My brain is too tired.

Yes, you're correct, but it took a bit of work to rule out $2p$; with less work you can rule out $1$

Does your proof go something like showing that G is abelian and showing that there exists a subgroup with order 4, that is, not a divisor of 2p. Hence p is an element?

Pretty much - I happen to know that if all non-identity elements are of order 2 (that is, if the group is boolean), then the group has order a power of 2. Can't remember if my proof required Cauchy's theorem or not, though. Anyway, that would do.

You don't need Cauchy (which when all is said and done is a non-trivial thereom) to show the existence of a subgroup of order 4. (It's easy enough to exhibit one).

I remember having done this question in the past - I was mainly trying to work out whether I'd done it before or after Cauchy's theorem. Probably before, as you say.