I recently came across a doubt I've already partially exposed in this post without getting a solution. So I'd like to isolate my final concern from the textbook I mentioned there. The real doubt is:

Is true that in a group $G$, if $P$ is a Sylow $p$-subgroup and $N$ is the normalizer of $P$, then the only subgroups of order $|N|$ are the normalizers of the Sylow p-subgroups?

I've checked within $A_5$ and that seems true in that case:

  1. there $6$ 5-sylow, with their related $6$ normalizers of order $10$, which are exhausting all subgroups of order $10$ in $A_5$
  2. there are $10$ 3-sylow, with their related $10$ normalizers of order 6, which are exhausting all subgroups of order $6$ in $A_5$
  3. there are $5$ 2-sylow of order $4$, which their related $5$ normalizers of order $12$, which are exhausting all subgroups of order $12$ in $A_5$

My doubt is this holds true for any generic group $G$ or just for $A_5$.

Thanks in advance.


1 Answer 1


No, a counterexample is $\mathtt{SmallGroup}(324,160)$, which has the structure $3^3\!:\!A_4$. It has a unique minimal normal subgroup $K$ of order $3^3$ and $G/K \cong A_4$.

The normalizer $N$ of a Sylow $2$-subgroup has order $12$ with $N \cong A_4$, but there is another conjugacy class of subgroups of order $12$, which do not have a normal Sylow $2$-subgroup.

  • $\begingroup$ Thanks @Derek I had the same feeling but I wasn't able to demonstrate It. So now It could be interesting to find the rule for which groups this holds true, if a general rule exists $\endgroup$ Aug 16 at 9:07
  • $\begingroup$ I was inspecting some candidates for the other subgroups of order 12. The dihedral D12 would have the same Sylow 2-subgroups as $A_4$, i.e. Klein four groups, without having them normal in D12. I will double check my Guess with SAGE. $\endgroup$ Aug 16 at 13:10
  • 1
    $\begingroup$ The other subgroup is isomorphic to $D_{12} \cong C_2 \times D_{6}$. It has a normal subgroup of order $3$ that is contained in the minimal normal subgroup of $G$. $\endgroup$
    – Derek Holt
    Aug 16 at 17:15
  • $\begingroup$ thank @DerekHolt for confirming my Guess. I'm new to the SmallGroup library and I'm using It from SAGE. I thought It would have been Easy to create SmallGroup(324, 160) and the recalling the usual order() subgroups() etc, but It seems doesn't work and probably I have to build the group by hands. I'm a professional Math programmer and I cannot understand why to have a library which output cannot be used as normal SAGE groups. $\endgroup$ Aug 16 at 17:47
  • $\begingroup$ I see now. Actually using a gap library returns objects meant to be used by other GAP commands, not SAGE ones. (Like ConjugacyClassesSubgroups for instance). So opening a GAP console from SAGE and working in olain GAP environment will di the trick. $\endgroup$ Aug 16 at 18:16

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