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1h
answered property about centralizer of maximal subgroup
1d
awarded  finite-groups
1d
awarded  Nice Answer
1d
answered Subgroups of generalized dihedral groups
Sep
1
awarded  Popular Question
Aug
31
comment If $|H|=112$ then $A_7\cap H \lhd H$?
@dREaM: Ah right, I forgot you wanted $H$ to have order $112$..
Jul
31
comment Composition series and its number determine a group?
Unless I made a mistake again, $A_4$ and the cyclic group of order $12$ both have three distinct composition series.
Jul
31
revised Composition series and its number determine a group?
deleted 97 characters in body
Jul
31
comment Composition series and its number determine a group?
@DerekHolt: Thanks for catching my mistake, you are right. But I do think there is only one reasonable interpretation to this question.
Jul
31
answered Composition series and its number determine a group?
Jul
21
comment Example of a group
Again much later, but.. see here for an example in $A_9$.
Jul
21
accepted What is the smallest $n$ for which the usual “counting sizes of conjugacy classes” proof of simplicity fails for $A_n$?
Jul
21
comment What is the smallest $n$ for which the usual “counting sizes of conjugacy classes” proof of simplicity fails for $A_n$?
Thanks, I think this should do it.
Jul
21
comment What is the smallest $n$ for which the usual “counting sizes of conjugacy classes” proof of simplicity fails for $A_n$?
For shape (5,3) in $A_9$, I count 24192.
Jul
5
comment Maximal closed subgroups in algebraic groups
For $G$ simple you can still find examples, see my answer.
Jul
5
answered Maximal closed subgroups in algebraic groups
Jun
30
awarded  Notable Question
Jun
29
comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative
The paper Groups with preassigned central and central quotient group by Baer should be useful, see corollary 7.2.
Jun
1
comment $|G|=p^nm$ and number of subgroups of order $p^s$
This proof also works if you know $r_1 \equiv 1 \mod{p}$, because $q_i$ is also equal to the number of subgroups of order $p$ in $N_G(Q_i)$. You can prove $r_1 \equiv 1 \mod{p}$ with McKay's proof of Cauchy's theorem
Jun
1
answered $|G|=p^nm$ and number of subgroups of order $p^s$