Groups of order 2p

Posted: January 4, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $G$ be a group, $p$ a prime number and $|G|=2p.$ Prove that either $G$ is cyclic or $G \cong D_{2p},$ where $D_{2p}$ is the dihedral group of order $2p.$

Solution. It is clear for $p=2.$ So we will assume that $p$ is an odd prime. Choose $a,b \in G$ with $o(a)=2, \ o(b)=p.$ Let $H=\langle a \rangle, \ K = \langle b \rangle.$ Since every subgroup of index 2 is normal, $K$ is a normal subgroup of $G$ and thus $aba^{-1}=b^j$ for some integer $j.$ Note that, since $p$ is odd, $H \cap K = \{1\}$ and hence $G=HK = \langle a,b \rangle.$ 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 $o(b)=p.$ Therefore either $p \mid j - 1$ or $p \mid j+1.$ So we will consider two cases:

Case 1 : $p \mid j - 1.$ In this case $aba^{-1}=b^j = b$ and so $ab=ba.$ Thus $G$ is abelian and hence $o(ab)=2p,$ because $p$ is odd. So $G$ is cyclic in this case.

Case 2 : $p \mid j+1.$ In this case $aba^{-1}=b^j = b^{-1}$ and so

$G = \langle a,b : \ a^2=b^p=1, \ aba^{-1}=b^{-1} \rangle \cong D_{2p}.$ $\Box$

1. Tizian says:

Hi, thanks for sharing your proof 🙂 I just don’t understand how you can conclude that G= HK. Is there a crucial Theorem I miss ? Doesn’t look obvious to me

• Yaghoub Sharifi says:

$|HK|=\frac{|H| \cdot |K|}{|H \cap K|}.$

2. B Nolan says:

Hi, thanks so much for this proof. In case 2, we say that G has the given presentation but how do we know that there are no other relations which a and b must satisfy?

• Yaghoub Sharifi says:

Let $u,v$ be the generators of $D_{2p}$ which satisfy the relations $u^2=v^p=1, \ uvu^{-1}=v^{-1}$. Since $a,b$ satisfy the same realtions, there exists a group homomorphism $f: D_{2p} \to G$ sending $u,v$ to $a,b$ respectively. Obviously $f$ is surjective because $a,b$ generate $G.$ Thus $D_{2p}/\ker f \cong G$. But $|G|=|D_{2p}|=2p$ and so $|\ker f|=1,$ i.e. $f$ is injective as well.