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2d
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.
2d
revised Composition series and its number determine a group?
deleted 97 characters in body
2d
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.
2d
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 $p^s$
May
31
answered The center of a non-Abelian group of order 8
May
27
awarded  Good Answer
May
25
comment If I know the order of every element in a group, do I know the group?
See also math.stackexchange.com/questions/729611/…
May
25
awarded  Nice Answer
May
24
answered An example of a group such that $G \cong G \times G$
May
23
comment Elementary question in Group Theory with less prerequisite
To finish your answer, use the class equation to show that $G$ has nontrivial center, so $G/Z(G)$ has order $5$ or $1$. But then the image of any element in $G/Z(G)$ is trivial (because the image has order $3$ or $1$), so $G/Z(G)$ is trivial and $G$ is abelian. When $G$ is abelian the claim is easy. Of course, with the same amount of work we could just prove Cauchy's theorem..