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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..
May
5
awarded  Enlightened
May
4
awarded  Nice Answer
Apr
30
awarded  Enlightened
Apr
30
awarded  Nice Answer
Apr
24
comment Is every group of odd order isomorphic to a subgroup of $A_n$ for some $n$?
Actually $S_m$ embeds into $A_{m+2}$ but not in $A_{m+1}$, unless $m = 1$.
Apr
24
comment Conjugacy classes of the nonabelian group of order 21
@RuianChen: Thanks, fixed.
Apr
24
revised Conjugacy classes of the nonabelian group of order 21
edited body
Apr
22
awarded  Nice Answer
Apr
11
comment Given 3 distinct primes {$p,q,r$}, then $|G|=pqr \implies G$ not simple
So you get $\geq (p-1)q$ elements of order $p$, also $\geq (q-1)r$ elements of order $q$, and $\geq (r-1)pq$ elements of order $r$.. putting all this together gives you at least $pqr$ elements, and then you still have the identity
Mar
27
comment Question on proof that maximal normal abelian subgroup is self-centralising in nilpotent groups
$C_G(N)$ is normal if $N$ is normal. Furthermore in the proof $N$ is a central subgroup of $U$ and $U/N$ is cyclic, hence by (*) the subgroup $U$ must be abelian.