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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.

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For the first couple you can look at A000001 to see if you can find a pattern. – Sam DeHority Jan 25 '13 at 20:22
What's "the $p$-Group $P$" in this context? – user123454321 Jan 25 '13 at 20:45
I mean Sylow p-subgroup. It's simple to prove a subgroup of order $p^2$ is normal and abelian. – user59671 Jan 25 '13 at 20:48
up vote 4 down vote accepted

Let $\,G\,$ be a group of order $\,2p^2\,\,\,,\,p\neq 2\,$ a prime.

First, there are $\,2\,$ abelians groups up to isomorphism of order $\,2p^2\,$ , namely

$$A_1:=C_2\times C_{p^2}\cong C_{2p^2}\;\;,\;\;A_2:=C_2\times C_p\times C_p\cong C_{2p}\times C_p$$

Next, we have to get into non-trivial semidirect products. As there's always one unique Sylow $\,p-$subgroup of order $\,p^2\,$ , say $\,P\,$ , which is then normal in $\,G\,$, we get:


$$\;\;\;\;\;\;\;P=C_{p^2}:=\langle x\rangle\,\Longrightarrow \operatorname{Aut}(P)\cong C_{p(p-1)}\Longrightarrow|\operatorname{Aut}(P)|=p(p-1)$$

In this case, $\,C_2:=\langle c\rangle\,$ acts on $\,P\,$ by inversion, or what is the same: there exists a homomorphism $\,f:C_2\to\operatorname{Aut}(P)\,\,,\,f(x):=\psi\,$ , where $\,\psi(x):=x^{-1}\,$ . We can write shortly this action by automorphisms of $\,C_2\,$ on $\,P\,$ as $\,(x^i)^c:=x^{-i}\,$ . We thus get the semidirect product

$$P\rtimes_\psi C_2\,$$

Do note that the above is the only possibility as there's one single element of order two in $\,\operatorname{Aut}(P)\,$ (why?)


$$\;\;\;\;\;\;\;\;\;\;\;P=C_p\times C_p=\langle a\rangle\times\langle b\rangle\Longrightarrow\operatorname{Aut}(P)\cong GL_2(\Bbb Z/p\Bbb Z)\Longrightarrow$$ $$ |\operatorname{Aut}(P)|=(p^2-1)(p^2-p)=p(p-1)(p^2-1)$$

This time we can choose inversion on the first generator $\,a\,$ or on the second one $\,b\,$: the resulting semidirect products are isomorphic to each other but, of course, each different (i.e., non isomorphic) from the first one (1) above. You may want to read page 95 (version "ereaders) of this excellent source.


But there's one more option: inversion of both generators! I'll leave it to you to show this additional possibility gives a semidirect product non-isomorphic with the first one in (2) (of course, also non-isomorphic with the one in (1)), for example by showing that in this latter case the group is centerless (hint: you may want to check dihedral groups....), whereas in the first case above (in (2)) we have non-trivial center...

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Great. But I have to check a few propositions, e.g. $(2)$. thanks. – user59671 Jan 25 '13 at 22:06

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.

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more complicated. – user59671 Jan 25 '13 at 23:14
Which part/argument is more complicated ? – tj_ Jan 25 '13 at 23:25
More sophisticated. I know little about extensions and eigenvectors in Group Theory. But it may be simpler for someone else. – user59671 Jan 25 '13 at 23:51
Of course it's your opinion and there is nothing bad with it. But what I'm surprised about is that you feel in my answer exactly those items more complicated where I have given precise arguments while DonAntonio just states results. In detail: ... – tj_ Jan 26 '13 at 12:55
... (1) $G$ is the semi-direct product of a Sylow p-subgroup and a Sylow 2-subgroup. But why ? DonAntonio just says "Next, we have to get into non-trivial semidirect products". For my taste one should give some argument in order to explain that there are no other possibilities than semidirect products. For example the Quaternion group of order 8 has a normal subgroup of order 4 without being a semidirect product. – tj_ Jan 26 '13 at 12:56

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