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It is easy to show that if $G$ is a group of order $p^2q^2$, where $p,q$ are primes with correspondings Sylow subgroups $P,Q$, that $G$ is a semi-direct product of $P$ and $Q$. Moreover, if $pq\neq 6$, then $$G=Q\rtimes P.$$

Now, I am guessing that this is not the case for $|G|=p^4q^2$, however I don't know any example.

  1. Can anyone give such an example please?
  2. can we esstimate how many such groups are there up to isomorphism?
Ofir Schnabel
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    Are you interested in semi-direct products of Sylow subgroups only? $S_4$ is not a semi-direct product of its Sylow subgroups, so neither is $G=S_4\times C_6$ (so $|G|=2^4\cdot3^2$). It is obviously a direct product though. – Jyrki Lahtonen Jul 01 '15 at 07:55
  • @JyrkiLahtonen , yes I am interested in semi-direct product of Sylow subgroups and therefore, your example is perfect. Can you construct an odd example? – Ofir Schnabel Jul 01 '15 at 07:59
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    For an odd example, note that $C_{13}$ acts in a non-trivial way on $C_3^3$, and $C_3$ acts in a non-trivial way on $C_{13}$, and so, taking the associated semi-direct products, $(C_3^3\rtimes C_{13})\times(C_{13}\rtimes C_3)$ is an example. – Jeremy Rickard Jul 01 '15 at 08:23
  • @JeremyRickard, thanks a lot. Is there a reasonble way to determine for a given $p,q$, how many groups of order $p^4q^2$ are there which are not a semi-direct product of their Sylow subgroup. – Ofir Schnabel Jul 01 '15 at 08:33
  • I don't know, sorry. – Jeremy Rickard Jul 01 '15 at 09:04
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    verret mentioned in a comment to a recent answer by Derek Holt to a similar question the existence of a group of order $144 = 2^4\cdot 3^2$ that cannot be written as semi-direct product of any of its proper subgroups. – j.p. Jul 01 '15 at 09:14
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    Note that, in your case, this property is equivalent to having a normal Sylow subgroup. In particular, by Sylow's theorems, an example must have $p\leq q^2-1$ and $q\leq p^4-1$. – verret Jul 01 '15 at 09:54

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There are two reasons why a group of order $p^a.q^b$ is not a semi-direct product of its Sylow-subgroups. The first case is when the Fitting length is at least three or, said another way, when $G/F(G)$ is not nilpotent with $F(G)$ being the Fitting subgroup. The second case is when $G/F(G)$ is nilpotent with an order multiple of both $p$ and $q$.

In the first case, we must have $(p,q)=(2,3)$, $(p,q)=(3,2)$ or $(p,q)=(3,13)$. There are, according to GAP, 17 suitables groups of order 144 and one group of order 324. GAP is not managing $13689=3^4.13^2$ but, if I'm not mistaken, there are four suitable groups for that order. More precisely, those four groups have a characteristic cyclic subgroup of order 13 with a quotient of order 1053 and of the form $(C_3^3 \rtimes C_{13}) \rtimes C_3$.

In the second case, we have $q=p^2+p+1$ and the group is isomorphic to $(C_p^3 \rtimes C_q) \times (C_q \rtimes C_p)$. According to Bunyakovsky conjecture, this should yield an infinite family of groups.

This answer is based on personnal research, which is not fully completed and not published, so there is risk of errors in it.

Eric Brier
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