Odd prime $p$ implies positive divisors of $2p$ are $1,2,p,$ and $2p$ $1,2,p,$ and $2p$ are indeed divisors of $2p$. I want to show these are the only positive divisors. Is there a more elegant or concise way to prove this besides the proof I have below?

Suppose that positive $a \in \left([3,2p-1] \cap \mathbb{N}\right) \setminus\{p\}$ divides $2p$. So $ak$=$2p$ for $k \in \mathbb{Z}$, and clearly $2 \leq k \leq p$. Since $ak=2p$ is even, at least one of $a$ or $k$ must be even.
If $k$ is even, then $a\frac{k}{2}=aj=p$ for integer $1 \leq j \leq \frac{p}{2}<p$, so $j | p$, so $j=1$, so $a=p$ which is a contradiction.
Similarly, if $a$ is even then $k | p$, so $k=p$. But then $a=2$.
So there are no other positive divisors besides $1,2,p$, and $2p$.

The motivation for this is to show that if a group $G$ has order $2p$ for odd prime $p$, then nonabelian $G$ is isomorphic to $D_{2p}$. The proof begins with "the possible orders for nonidentity elements of $G$ are $2,p,$ and $2p$," which I am trying to prove with Lagrange's Theorem. If there is an alternative way to justify this statement using group theory, then I would appreciate seeing that as well.
 A: If $p$ is an odd prime number, then by the fundamental theorem of number theory, $2\times p$ is the unique primary decomposition of $2p$. Once you express a positive integer $n$ as it's unique primary decomposition, say $p_1^{a_1}\dots p_k^{a_k}$, then all the positive factors will be of the form $p_1^{b_1}\dots p_k^{b_k}$ where $0\leq b_i\leq a_i$ for each $i$. With this observation you should be able to answer your question.
A: 
By the fundamental theorem of Arithmetic the only prime factors of $2p$ are $2$ and $p$ and so every factor must be a combination of $2$ and $p$ of which $1,2,p$ and $2p$ are the only options.

That is more than sufficient and more than anyone can reasonable expect to require proof.
....
But if you want to smash an ant with a sledgehammer:
The fundamental Theorem of Arithmetic say each number has a unique prime factorization of $n = \prod p_i^{a_i}$.  So if any factor $m; m|n$ can only have $\{p_i\}$ as prime factors and only to the powers less than or equal to $a_i$.  
So all if $m|n$ then $m$ must be of the for $\prod p_i^{b_i}$ where $0\le b_i \le a_i$ and so there are $\prod (a_i+1)$ such factors.
So the factors of $2p$ are all of the form $2^b p^c$ where $b = 0,1$ and $p = 0,1$.  There are four of these numbers and they are $2^0p^0 =1; 2^1p^0 = 2; 2^0p^1 = p; $ and $2^1p^1 = 2p$.
That's it.
That is a result that it is reasonable to expect every reader is either familiar with or can justify on their own.
....
But frankly it is enough to say:
"The only factor of $2p $ are $1,2,p$ and $2p$" and assume that is completely self-evident.
And it is.
A: By the Fundamental Theorem of Arithmetic, the only primes that divide $2p$ are $2$ and $p$. Note that $2p$ is squarefree and has only two prime factors.
A: The other answerers are of course right to point you to the Fundamental Theorem of Arithmetic as it will come in handy at many stages of your life. Still I wanted to say two things about your own proof.
1) It is correct and quite elegant. As one of the other answerers points out: using the Fundamental Theorem is a bit of a sledgehammer.
2) The key step in your proof is the lemma 'since $ak$ is even at least one of $a, k$ is even'. I wanted to point out to you that this has a very nice generalization, called Euclid's lemma. It states:

Let $q$ be any prime number. Then whenever $ak$ is divisible by $q$ we necessarily have that at least one of $a, k$ is divisible by $q$. 

So your lemma is the case $q = 2$. Knowing this, we see that you could also make a copy of your proof but with $p$ in the role of 2 and 2 in the role of $p$ although it would be less intuitively appealing.
Euclid's lemma implies the uniqueness part of the Fundamental Theorem (and is mostly proved first) but of course if you have a different proof of the Fundamental Theorem, Euclid's lemma readily follows from it.   
A: You can do this fairly simply with Euclid's Lemma, that if $q$ is a prime number and $q\mid ab$, then either $q\mid a$ or $q\mid b$.  
If $d$ were a divisor of $2p$ other than $1$, $2$, $p$, or $2p$, then $d$, since it's not equal to $1$, must be divisible by some prime. So let $q$ be a prime divisor of $d$. But now $q\mid d\mid2p$ implies, by Euclid's Lemma, that either $q\mid2$ or $q\mid p$, which implies either $q=2$ or $q=p$. The latter is not possible, since if we write $d=qk$ and let $q=p$, we cannot have $k=1$ or $2$, since $d\not=p$ or $2p$, nor can we have $k\gt2$ since a divisor cannot be larger than the number it divides. So $q=2$. Thus $d$ can only be a power of $2$, and since $d\not=2$, it must be divisible by $4$. But since $p$ is an odd prime, we have $p=2n+1$ for some $n$, in which case $2p=4n+2$, which is not divisible by $4$. The contradiction tells us that there is no divisor of $2p$ other than $1$, $2$, $p$, and $2p$.
Remark: Laying out the logic of this was a little more complicated than I initially anticipated.  It might be possible to compress the argument, but I don't offhand see any suitably slick way to do so. Maybe someone else does.
