Groups of order 2p

Posted: January 4, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields

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 :/

  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.

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