2
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Here

https://www2.bc.edu/~reederma/Groups.pdf

on page $112$, a table of the number of groups of order $p^2q$ is given. In the explanations, there is a typo ($\frac{q+5}{5}$ instead of $\frac{q+5}{2}$)

Can someone approve the following formula for $f(p,q)$, the number of groups of order $p^2q$ ?

  • If $p$ does not divide $q-1$ and $q$ does not divide $p^2-1$, then $f(p,q)=2$

    • If $p|q-1$ and $p^2$ does not divide $q-1$, then $f(p,q)=4$

    • If $p^2|q-1$ , then $f(p,q)=5$

    • If $q=2$ , then $f(p,q)=5$

    • If $q>2$ and $q|p+1$ , then $f(p,q)=3$

    • If $q>2$ and $q|p-1$ , then $f(p,q)=\frac{q+9}{2}$

Does anyone know a similar formula for the number of groups of order $p^2q^2$ ?

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  • 1
    $\begingroup$ The case $p=2$, $q=3$ is wrong in both lines 2 and 5. $f(12)=5$. Otherwise it's OK. $\endgroup$ – Derek Holt Dec 8 '15 at 18:28
  • $\begingroup$ @Derek Did you check this for all $p,q$, or did you check the values upto some limit ? $\endgroup$ – Peter Dec 8 '15 at 18:36
  • $\begingroup$ I checked it for all $p$ and $q$. $\endgroup$ – Derek Holt Dec 8 '15 at 18:45
  • $\begingroup$ @Derek Holt Does a formula exists for $p^2q^2$ ? $\endgroup$ – Peter Dec 23 '15 at 9:14

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