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Let $|G|=p^2q$ with following conditions:

  1. Sylow-$p$ subgroup is normal and is $\langle x,y\rangle \cong \mathbb{Z}_p\times \mathbb{Z}_p$.

  2. Sylow-$q$ subgroup is $\langle z\rangle$, and is not normal.

  3. No subgroup of order $p$ is normal in $G$.

Question: How many isomorphism types of such groups are?

For $p=2$ and $q=3$, we know that there is such a group isomorphic to $A_4$.

While giving classification of groups of order $p^2q$ in a different way, I stuck at this case.

Groups
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    Hmm. The conjugation action of $z$ on $\Bbb{Z}p^2$ is a linear transformation of that vector space. Item 3 tells us that it must not have eigenvalues in $\Bbb{Z}_p$. Therefore the eigenvalues are a conjugate pair of elements of order $q$ of the field $\Bbb{F}{p^2}$. Therefore $q\mid p^2-1=(p-1)(p+1)$, and the previous consideration forces $q\mid p+1$. Using a different conjugate pair of eigenvalues amounts to replacing $z$ with its power, so we can fix a conjugate pair of roots of unity of order $q$. This specifies the linear transformation up to conjugacy in $GL_2(\Bbb{F}_{p^2})$. – Jyrki Lahtonen Sep 05 '15 at 08:30
  • (cont'd) But the transformation is in $GL_2(\Bbb{F}_p)$, and hence (a known but non-trivial step here) it is determined up to conjugacy in $GL_2(\Bbb{F}_p)$. Did I just argue that the number is "one", if $q\mid p+1$ and "zero" otherwise? Needs checking... – Jyrki Lahtonen Sep 05 '15 at 08:33
  • any hint towards non-trivial step? I know that conjugate transformations will give isomorphic semi-direct product; but is converse true here? (I will not worry about "one", "zero" now. Thanks for some clarifications!) – Groups Sep 05 '15 at 08:36
  • Groups, what I was thinking was that if $\omega$ and $\overline{\omega}=\omega^p$ are a conjugate pair of elements of order $q$ in $\Bbb{F}_{p^2}$, then all other such pairs are of the form $(\omega^k,\omega^{pk})$ for some $k, 0<k<q$. And if $z$ has eigenvalues $(\omega,\omega^p)$ then $z^k$ has eigenvalues $(\omega^k,\omega^{pk})$. – Jyrki Lahtonen Sep 05 '15 at 08:36
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    Ok. Got it. The elements of order $q$ in GL(2,p) with no eigenvector are conjugate to elements of order $q$ in $GL(2,p^2)$ with eigenvalues in $\mathbb{F}{p^2}\setminus \mathbb{F}_p$, which are (Galois) conjugate in pair and these eigenvalues are elements of order $q$ in $ \mathbb{F}{p^2} \setminus {0}$, which form cyclic group (of prime order); hence powers of each other. – Groups Sep 05 '15 at 08:42
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    The non-trivial fact is the following: Let $K/F$ be a field extension. If two matrices in $M_n(F)$ are conjugate in $M_n(K)$, they are conjugate already in $M_n(F)$. Searching for a link.... here. See also Marc van Leeuwen's answer in the same thread. – Jyrki Lahtonen Sep 05 '15 at 08:43
  • A first thing to count here (being an upper bound for the number) is the number of irreducible $2$-dimensional representations of the cyclic group of order $q$ over the field of $p$ elements (which is given in http://math.stackexchange.com/questions/75350/irreducible-representations-of-a-cyclic-group-of-order-p-over-a-field-of-q-eleme). It is not quite clear when such will give isomorphic groups when starting with distinct reps. – Tobias Kildetoft Sep 05 '15 at 10:22

1 Answers1

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By Sylow theorem, there is only $1$ Sylow $p$-subgroup which is $\Bbb{Z}_p\times \Bbb{Z}_p$.

And by Sylow theorem again, the number of Sylow $q$-subgroup $n_q|1,\:p,\:p^2$, and $n_q\equiv1\mod q$.

  1. If $n_q=1$, then there is only one Sylow $q$-subgroup which is normal. So this is impossible.

  2. If $n_q=p$, then there are $p$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup. So there are at most $p(q-1)+p^2$ elements. But since $$p^2q-(p(q-1)+p^2)=p(q-1)(p-1)>0$$ total elements are less than $p^2q$, which is impossible.

So only possibility is $n_q=p^2$, and there are $p^2$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup of $\Bbb{Z}_p\times \Bbb{Z}_p$ that is normal. Since there are more than one Sylow $p$-subgroup, $G$ is non-abelian. We prove that it is the only isomorphism type of $G$.

Suppose $\:H=\Bbb{Z}_p\times \Bbb{Z}_p=\{h:h^p=1,\:h\in H\}$ and $K=\Bbb{Z}_q=\{k:k^q=1\}$. Clearly $K$ is cyclic. Since $H\vartriangleleft G, \:kHk^{-1}=H, \:\forall k\in K$. Any conjugate mapping $\{khk^{−1}=h_1,h,h_1∈H, k\in K\}$ can be made isomorphic by inner automorphism as $\{\phi:\:\phi(x)=g^{-1}xg,\:x\in G,\:g\in G, \: g\ne 1, \:g\notin Z(G)\}$ for the same center.

Since isomorphism keeps center of group invariant, the number of isomorphism types depends on the number of center of $G$. Since $Z(G)\vartriangleleft G$ and $|Z(G)|\mid|G|$, $|Z(G)|=1,q,p,p^2$. By given condition, $|Z(G)|\ne p$. If $|Z(G)|=p^2$, then $|G/Z(G)|=q$. So $G/Z(G)$ is cyclic and $G$ is abelian, which is contradiction. $|Z(G)|=q$ is also impossible for none of Sylow $q$-subgroups is normal. So $|Z(G)|=1$.

So there is only one isomorphism type for $G$ with $p^2$ Sylow $q$-subgroup and $1$ Sylow $p$-subgroup of $\Bbb{Z}_p\times \Bbb{Z}_p$.

Eugene Zhang
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  • This does not answer the question, which was about the number of isomorphism classes of such groups, not about the possible number of Sylow subgroups. – Tobias Kildetoft Sep 09 '15 at 06:49
  • No, this still does not answer the question. – Tobias Kildetoft Sep 12 '15 at 05:20
  • The question asks for the number of such groups. I am not even quite sure what this answer concludes. – Tobias Kildetoft Sep 12 '15 at 05:27
  • Possibly, but that is not what the answer says, and I see no arguments in the answer to support the claim. – Tobias Kildetoft Sep 12 '15 at 05:43
  • Yes, I do understand the basics of group theory, and your answer does not show that there is only one such group. – Tobias Kildetoft Sep 12 '15 at 06:15
  • That last part makes no sense. There are a lot of isimorphisms, but that is besides the point because we are not counting isomorphisms but rather isomorphism classes. You seem to be under the impression that the properties you mention at the end uniquely determine the group which I see no reason to suspect without further arguments. – Tobias Kildetoft Sep 12 '15 at 06:25
  • You now seem to argue that any two groups with same Sylow subgroups and trivial centers are isomorphic. This seems clearly false. – Tobias Kildetoft Sep 13 '15 at 06:24
  • I did not actually look for a counter example as the claim is simply too good to be true. And in any case, you have certainly not proved the claim. – Tobias Kildetoft Sep 13 '15 at 08:02