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Groups of order 2p

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.

Solution

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

        \[\mathcal{G} \cong \mathcal{D}_{2p}\]

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