Looking for a simple proof that groups of order $2p$ are up to isomorphism $\Bbb{Z}_{2p}$ and $D_p$ for prime $p>2$.

I'm looking for a simple proof that up to isomorphism every group of order $$2p$$ ($$p$$ prime greater than two) is either $$\mathbb{Z}_{2p}$$ or $$D_{p}$$ (the Dihedral group of order $$2p$$).

I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff.

Thanks a lot!

• your result is not true if you take p=2.So you should mention that p is an odd prime. Commented Jul 15, 2015 at 9:38
• Even for $p=2$ it is true, with the convention that $D_2=C_2\times C_2$ is the Klein four group. Commented Mar 24, 2016 at 15:04

Since we are allowed to use Sylow, we can assume $G$ is generated by $x,y$ with $x^p=y^2=1$, where $\langle x \rangle \lhd G$, so $y^{-1}xy = x^t$ for some $t$ with $1 \le t \le p-1$. Then $x = y^{-2}xy^2 = x^{t^2}$, so $p$ divides $t^2-1 = (t-1)(t+1)$, hence $p$ divides $t-1$ or $t+1$ and the only possibilities are $t=1$ or $p-1$, giving the cyclic and dihedral groups.

• Hi Derek, how is "$x = y^{-2}xy^2 = x^{t^2}$" true? Commented Dec 7, 2014 at 7:40
• @jstn $x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^t y=\underbrace{(y^{-1}xy)\cdot (y^{-1}xy)\cdots(y^{-1}xy)}_{\text{tea time}}=(x^t)^t=x^{t^2}$. Commented Mar 21, 2016 at 9:34
• Hello Derek!! I am reading your answer and I have some questions... We have that $|G|=2p$, so there are $2$-Sylow and $p$-Sylow in $G$, right? $$P\in \text{Syl}_2(G) , \ |P|=2 \\ Q\in \text{Syl}_p(G) , \ |Q|=p$$ Why do we have that $G$ is generated by $x,y$ ? Commented Mar 24, 2016 at 15:46
• Because $x$ has order $p$ and $y$ has order $2$ so the subgroup they generate has order divisible by $G$, and so it must equal $G$. Commented Mar 24, 2016 at 16:54
• Note that $xy$ has order $2p$ so the group has to be cyclic. Commented Mar 25, 2016 at 7:04

Since the $2$-Sylow subgroup is cyclic, the group has a normal $2$-complement (corollary to Burnside's transfer theorem), which means that the $p$-Sylow subgroup is normal (or just use that any subgroup of index $2$ is normal). Thus, the group is a semidirect product of a cyclic group of order $p$ and one of order $2$. Since the Automorphism group of the cyclic group of order $p$ has a unique subgroup of order $2$, this means that there can only be one non-trivial such semidirect product, and since $D_p$ is such a semidirect product, it must be it. If the semidirect product is trivial, we of course get the cyclic group of order $2p$.

• You really don't need Burnside's transfer Theorem! A Sylow $p$-subgroup has index 2 and so must be normal. Commented Jan 31, 2013 at 13:53
• @DerekHolt I know (I also mentioned that simpler version). But as the proof can actually be done about as elegantly by elementary means, but the OP explicitly allowed advanced results, I could not resist using this approach (which can of course be used in much more general settings). Commented Jan 31, 2013 at 13:55
• For $p=2$ the trivial semidirect product is not the cyclic group, but it is $D_2$ in that case. But otherwise a nice proof. Commented Jan 31, 2013 at 14:27
• @MarcvanLeeuwen right, I assume $p\neq 2$ as otherwise the result is trivial. Commented Jan 31, 2013 at 14:36

If the group is Abelian then by the Classification of Finitely generated abelian groups we know that $$G=\mathbb{Z}_{2p}$$ is the only possibility (if $$p$$ is an odd prime).

If it is non Abelian:

By Cauchy's theorem one gets that there exits an element of order $$p$$ and order $$2$$. Call them $$x$$ and $$y$$ respectively, then all possible elements of the group are as follows:

$$\{1,x,x^2,...,x^{p-1},y,yx,yx^2,...,yx^{p-1},xy,x^2y,...x^{p-1}y\}$$ since the order of the group is $$2p$$, we get that $$yx=x^{j}y$$ for some $$2\leq j\leq p-1$$. ($$j=1$$ will force the group to be Abelian, so it not possible).

Then $$yx^2=(yx)x=x^j(yx)=(x^{2j\bmod{p}})y\implies yx^k=(x^{kj\bmod{p}}y)$$ by induction.

Then $$y(yx)=x=y(x^jy)=(yx^j)y=x^{j^2 \bmod{p}}yy=x^{j^2 \bmod{p}}\implies x^{j^2 -1\bmod p}=1\implies j=\pm1\bmod p$$. Since the group is not Abelian we get that $$yx=x^{p-1}y$$ is the only reasonable relation possible, now consider the homorphism induced by $$y\mapsto s$$ and $$r=x\mapsto r$$. This is clearly an isomorphism to $$D_{p}$$.

Therefore only groups of order $$2p$$ are $$\mathbb{Z_{2p}}$$ or $$D_{p}$$.

• If $p=2$ you get a second abelian group. Commented Jun 14, 2021 at 20:02
• @MarkBennet Yes, your observation is right. I was unconsciously assuming that $p$ was an odd prime. Commented Jun 16, 2021 at 7:16

By Sylow-1 we can say that $$\mathbb{Z}/(2)\simeq H, $$\mathbb{Z}/(p)\simeq K, by Lagrange $$H\cap K=\{e\}$$ and it is easy ro see that $$G=HK$$.

And since $$(G:K)=2$$, $$K\triangleleft G$$.

So $$G=K \rtimes_f H$$ for some $$f:H\to \operatorname{Aut}(K)$$. It is well-known that $$\operatorname{Aut}(\mathbb{Z}/(p))\simeq \mathbb{Z}/(p-1)$$. So there are 2 such morphisms since $$\gcd(2;p-1)=2$$.

So, there are 1 or 2 groups of order $$2p$$, and we can name 2 of them: $$\mathbb{Z}/(2p)$$ and symmetries of a regular $$n$$-gon.

For $$p=2$$ our group has 4 elements, so it is either cyclic or the Klein-4-group. So let $$p > 2$$.

By Cauchy's theorem you know there is a cyclic subgroup of order $$p$$ and $$2$$, call this $$H \leq G$$. Let $$a$$ be a generator of $$H$$ and let $$b$$ be an element of order $$2$$. Then $$a,b$$ generates $$G$$, and thus we only need to consider how $$a$$ acts on $$b$$.

$$H$$ is normal since $$|G : H| = 2$$, so $$bab^{-1} \in H \implies bab^{-1} = a^j,\ 0 \leq j < p$$ And $$b$$ has order $$2$$, so $$a = b^2 ab^{-2} = b (bab^{-1})b^{-1} = b a^j b^{-1} = \overbrace{(b a b^{-1}) \cdots (bab^{-1})}^{j \text{ times}} = a^{j^2}$$

Thus $$j^2 \equiv 1 \mod p \implies j \in \{1, p-1\}$$.

If $$j = 1$$, then $$G$$ is Abelian, so $$G \simeq C_{2p}$$ (The cyclic group of order $$2p$$)

If $$j = p-1$$, it is clear that the mapping of $$a$$ to a rotation and $$b$$ to a mirroring is an isomorphism to $$D_p$$. (Since rotating then mirroring is the same as mirroring then rotating the other direction)

By Cauchy's theorem, $$G$$ has an element $$x$$ of order $$p$$ and an element $$y$$ of order $$2$$. If $$\text{o}(xy)=2p$$ then the group is cyclic, and if $$\text{o}(xy)=1,2$$ then the group is dihedral because we would have the presentation $$x^p=y^2=(xy)^2=1$$. So suppose that $$\text{o}(xy)=p$$. Clearly, $$xy\not\in \langle x\rangle$$, since that would imply $$y\in \langle x \rangle$$. Thus, $$G$$ has at least $$p$$ elements of order $$p$$, namely $$xy,x,x^2,\ldots,x^{p-1}$$. However, $$p\leq |\{x\in G:\text{o}(x)=p\}|\leq 2p-2=|G\backslash \langle y\rangle|,$$ but none of the numbers $$p,p+1,\ldots,2p-2$$ are $$\equiv -1\pmod{p}$$. This contradicts Cauchy's theorem on the number of elements of order $$p$$.

You can say we have an element of order p or 2p because if all of the elements has order at most two we can say the group is abelian. Then if you have an element with order 2p it's solved. If you have an element with order p. Then you can say we have an element with order 2 because the number of elements with order two is odd in groups with even order. After finding this two elements the remaining part is the same as "Derek"s solution. The point is you can prove it easily without using Cauchy or Sylow theorem.

Let $$|G| = 2p$$, and it can be written as $$G = \{e, a, \ldots, a^{p-1}, ba, \ldots, ba^{p-1}, ab, \ldots, a^{p-1}b\}$$. However, this appears to be of order $$3p$$, so there must be an equivalence $$ba = a^{n}b$$ for some $$n \leq p-1$$.

We observe that if we have an element $$b \in G$$ of order 2, there exists an automorphism of $$H \cong \mathbb{Z}_{p}$$ defined by $$\alpha(a) = bab^{-1}$$ such that $$\alpha^2 = \text{id}$$. To understand why this is the case, consider this geometrically where $$a$$ and $$b$$ represent either rotations or reflections of a regular $$2p$$-sided polygon. This leads to $$\alpha^2(a) = a^{n^2} = \text{id} = a \implies n^2 = 1$$, which has two solutions in the ring $$\mathbb{Z}_{p}$$, namely $$n = \pm 1$$. If $$n = 1$$, $$G$$ is abelian and isomorphic to $$\mathbb{Z}_p \times \mathbb{Z}_2$$. Otherwise, we obtain precisely $$bab^{-1} = a^{-1}$$ (which also has geometric significance), making it isomorphic to $$D_{2p}$$.

Use Cauchy Theorem

Cauchy's theorem — Let $$G$$ be a finite group and $$p$$ be a prime. If $$p$$ divides the order of $$G$$, then $$G$$ has an element of order $$p$$.

then you have an element $$x\in G$$ of order $$2$$ and another element $$y\in G$$ of order $$p$$. Now you have to show that $$xy$$ is of order $$2p$$
using commutativity we get:

$$(xy)^2 = y^2$$, and hence $$ord(xy) \not| 2$$

$$(xy)^p = x$$ , therefore $$ord(xy) \not| p$$

and $$(xy)^{2p} = y^{2p} = e$$, then $$ord(xy) | 2p$$

hence $$1 <2
an then $$ord(xy) = 2p$$ because it doesn't equal to any divisor different of $$2p$$.

• You have shown that, if $G$ is Abelian of order $2p$, then $G$ is cyclic. But this is just "half" of what is claimed in the OP.
– user1007416
Commented Jan 28, 2022 at 10:10
• @fitzcarraldo true, thanks. Commented Jan 29, 2022 at 14:55