TITLE: Finding all groups of order $2p^2$
QUESTION [8 upvotes]: How can one find all groups of order $2p^2$ up to isomorphism, where p is an odd prime.
I know there's 5 groups of order 18. The p-Group $P$ is normal and abelian.

REPLY [3 votes]: There are two cases: 
1) $p=2$. It's well-known that the Dihedral group and the Quaternion group are the only non-abelian groups of order 8. Hence the groups are 
$$C_8,C_4 \times C_2, C_2 \times C_2 \times C_2, D_8, Q_8$$
2) $p>2$. Let $G$ be a group of order $2p^2$ and let $P$ be a Sylow $p$-subgroup. Since $P$ has index 2, $P$ is normal in $G$. Moreover, $P$ has order $p^2$, whence $P=C_{p^2}$ or $P=C_p \times C_p$. In particular, $P$ is abelian and there is an extension 
$$1 \to P \to G \to C_2 \to 1$$
This extension splits (this follows from $H^2(C_2,P)=0$ or can be shown with some more work directly). Thus $G=P \ltimes C_2$ is a semi-direct product ($P$ normal). 
2.1) Direct products are $C_{p^2} \times C_2,\;C_p \times C_p \times C_2$. 
2.2) In order to find the semi-direct products that aren't direct products we have to determine the elements $\sigma$ of order 2 in $Aut(P)$. 
2.2.1) $P=C_{p^2}$. Here $Aut(P)=C_p \times C_{p-1}$ is cyclic. Hence there is exactly one element $\sigma$ of order 2. $G$ has the presentation 
$$G=\langle x,y\mid x^{p^2}=y^2=1, yxy^{-1}=x^{-1}\rangle$$
2.2.2) $P=C_p\times C_p$. First, let's assume that $\sigma$ has an eigenvector. Since the char. polynomial has degree 2, $\sigma$ has 2 eigenvectors and is thus (up to conjugacy) represented by the matrices 
$$\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\qquad \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$$
The corresponding presentations of $G$ are 
$$G=\langle x,y,z\mid x^p=y^p=z^2=1, xy=yx, zx=xz, zyz^{-1}=y^{-1}\rangle$$
$$G=\langle x,y,z\mid x^p=y^p=z^2=1, xy=yx, zxz^{-1}=x^{-1}, zyz^{-1}=y^{-1}\rangle$$
Finally suppose $x \in C_p \times C_p$ is no eigenvector of $\sigma$. Then $\sigma$ is represented with respect to the basis $\{x,\sigma(x)\}$ by  $=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$.  Hence $\sigma$ has the eigenvalues $\pm 1$ and this case is already covered.