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$1,2,p,$ 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.

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    $\begingroup$ Use the Fundamental Theorem of Arithmetic. $\endgroup$ – Shaun May 26 at 16:36
  • $\begingroup$ @Shaun clearly $2 \cdot p$ is a prime factorization of $2p$, and the prime factorization is unique. How does this show there cannot be nonprime divisors of $2p$ besides itself and $1$? $\endgroup$ – jskattt797 May 26 at 16:41
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    $\begingroup$ Because $2p$ is squarefree (and there is only two primes). $\endgroup$ – Shaun May 26 at 16:42
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    $\begingroup$ It's enough to state that the only two prime factors of $2p$ are $2,p$ so (by fundamental th) the only factors are combinations of $2$ and $p$ so they are $1,2,p$ and $2p$. Honestly that sentence is probably too long and saying too much.... I think that if you simply state the only factors of $2p$ are $1,2,p$ and $2p$ that ought to be utterly self-evident and obvious. If anyone asks way state its immediate but the fundamental th. $\endgroup$ – fleablood May 26 at 20:28
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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.

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    $\begingroup$ And we have $2p=2^1p^1$, so the set of all positive divisors is $$\{ 2^{a_1}p^{a_2} \mid a_1,a_2 \in \{0,1\} \}=\{1,2,p,2p \}$$. $\endgroup$ – jskattt797 May 26 at 17:17
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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.

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  • $\begingroup$ Yes, $m|n$ iff $m=\prod p_i^{b_i}$. And each factor $\prod p_i^{b_i}$ maps bijectively to a tuple $(b_1, \dots , b_k)$ for $0 \leq b_i \leq a_i$. So there are exactly $\prod(a_i + 1)$ factors. So an alternative proof is to notice that for $2p = 2^1 p^1$ there are exactly $2 \cdot 2=4$ factors, so there can be no others besides $1,2,p,$ and $2p$. $\endgroup$ – jskattt797 May 27 at 3:19
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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.

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  • $\begingroup$ Nice, so suppose there is another nonprime positive divisor $j=p_1 p_2 \dots \neq 1$ (at least two primes in $j$'s factorization because $j > 1$ is not prime) such that $j \neq 2p$, then $jk=2p$ for integer $k$. Notice $k \neq 1$, so $k > 1$ and $k = p_3 \dots$. But then $2p$ has at least 3 primes in its factorization. Why do we need squarefree? $\endgroup$ – jskattt797 May 26 at 16:58
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    $\begingroup$ To rule out $2^2=4$ and $p^2$. That's all. $\endgroup$ – Shaun May 26 at 17:13
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    $\begingroup$ Your argument seems valid though, @jskattt797. $\endgroup$ – Shaun May 26 at 17:14
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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.

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  • $\begingroup$ You're right the argument is exactly the same, perhaps even simpler using the bounds on $k$: $ak=2p \implies p|k$ or $p|a$. If $p|k$ then $k=p$ and $a=2$. If $p|a$ then $k|2$ so $k=2$ and $a=p$. Both cases contradict the choice of $a$. $\endgroup$ – jskattt797 May 27 at 3:49
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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.

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  • $\begingroup$ $q|d|2p \implies q|2$ or $q|p$ i.e. prime $q=2$ or $q=p$. And our choice of $d$ cannot be divisible by $p$ as you noted so $q=2$, so $d=2k | 2p$ for some positive integer $k$, so $k | p$, so $k=1$ ($d=2$) or $k=p$ ($d=2p$), both of which contradict our choice of $d$. $\endgroup$ – jskattt797 May 27 at 4:08
  • $\begingroup$ But doesn't "$d$, since it's not equal to $1$, must be divisible by some prime" rely on the fundamental theorem of arithmetic? $\endgroup$ – jskattt797 May 27 at 4:11
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    $\begingroup$ @jskattt797, nice alternative. Regarding the "$d$ must be divisible by some prime," no, that's more basic than the fundamental theorem. Or if you like, it's the "easy" part of the fundamental theorem, which says that every integer greater than $1$ can be written as a product of primes (the "easy" part) in exactly one way (the "hard" part). $\endgroup$ – Barry Cipra May 27 at 8:02

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