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 Post subject: Groups of order $2p$Posted: Thu May 18, 2017 9:12 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Let $\mathcal{G}$ be a group, $p$ a prime number and $|\mathcal{G}|=2p$. Prove that either $\mathcal{G}$ is cyclic or $\mathcal{G} \cong \mathcal{D}_{2p}$, where $\mathcal{D}_{2p}$ is the dihedral group of order $2p$.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Groups of order $2p$Posted: Thu May 18, 2017 11:24 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
By Cauchy's theorem for the group $G$ of order $2p$, $p$ prime, we have that there is an element $a\in G$ of order $p$. But then every element $a^{k}\,,\;k\in\{1,2,\ldots,p-1\}$ of the subgroup $\langle{a}\rangle$ has order $p$.
By Sylow's 3rd theorem we have that the number of the subgroups of $G$ of order $p$ is of the form $1+mp$, for some $m\in{\mathbb{N}}$. But then
$$({p-1})\,(1+mp)+1\leqslant 2p\quad\Rightarrow\quad m=0\,.$$ So, there is just one subgroup of order $p$, or, in other words, the only elements of $G$ of order $p$ are the (non-zero) elements of $\langle{a}\rangle$. The other $p$ elements of $G$ must have order either $2p$, either $2$.

• If $G$ has an element of order $2p$ then $G$ is cyclic.
• If $G$ has not an element of order $2p$, then $G$ has $p-1$ elements of order $p$ and $p$ elements of order $2$ (and of cource a zero element of order $1$).

Remains to show that the last case makes $G$ isomorphic to the dihedral group $${\cal{D}}_{2p}=\left\langle{\rho,\,\tau \ | \ \rho^{p}=\tau^2={\rm{id}}, \ \tau\rho\tau=\rho^{-1}}\right\rangle\,.$$
• For $p=2$, we have $G\cong {\cal{D}}_{4}$.
• For $p>2$, it is suffice to show that for the element $a$ -which is a generator of the subgroup $\langle{a}\rangle$- and for an element $b\in G$ of order $2$ holds $bab=a^{-1}$.
Because $\big[G:\langle{a}\rangle\big]=2$, the subgroup $\langle{a}\rangle$ is normal. So, $b^{-1}ab\in\langle{a}\rangle$, or equivalently, there exists $k\in\{2,3,\ldots,p-1\}$ such that $b^{-1}ab=a^k$ [The case $k=0$ leads to $p=1$ and the case $k=1$ leads to $p=2$] and because $b^{-1}=b$ we have:\begin{align*}
\end{align*} But, then \begin{align*}
&\stackrel{p \,{\text{prime}}}{=\!=\!\Longrightarrow}\quad p\,|\,k-1 \; \vee\; p\,|\,k+1\,.
\end{align*} If $p\,|\,k-1$, then \begin{align*}
\end{align*} Contradiction. So must $p\,|\,k+1$, or equivalently \begin{align*}
\end{align*} and $G$ is isomorphic to the dihedral group ${\cal{D}}_{2p}$.

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Grigorios Kostakos

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 Post subject: Re: Groups of order $2p$Posted: Thu May 18, 2017 11:49 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
In the same spirit as above.

It is clear for $p=2$. So we will assume that $p$ is an odd prime. Choose $a,b \in \mathcal{G}$ with $\circ(a)=2 , \circ(b)=p$. Let $H=\langle a \rangle, \ K = \langle b \rangle$. Since every subgroup of index $2$ is normal, $K$ is a normal subgroup of $\mathcal{G}$ and thus
$aba^{-1}=b^j$

for some integer $j$. Note that, since $p$ is odd, $H \cap K = \{1\}$ and hence

$\mathcal{G}=HK = \langle a,b \rangle$

Now

$b^{j^2} = (aba^{-1})^j = ab^j a^{-1}=a(aba^{-1})a^{-1}=b$

because $a^2=1$. Thus $b^{j^2 - 1}=1$ and hence $p \mid j^2 - 1$ because $\circ(b)=p$. Therefore either $p \mid j - 1$ or $p \mid j+1$. So we will consider two cases:

• $p \mid j - 1$. In this case
$aba^{-1}=b^j = b$

and so $ab=ba$. Thus $\mathcal{G}$ is abelian and hence $\circ(ab)=2p$ because $p$ is odd. So $\mathcal{G}$ is cyclic in this case.

• $p \mid j+1$. In this case
$aba^{-1}=b^j = b^{-1}$

and so

$\mathcal{G} \cong \mathcal{D}_{2p}$

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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