Reputation
15,231
Top tag
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
3 17 37
Newest
 Yearling
Impact
~118k people reached

13h
answered all of subgroups of group
May
19
comment $G/Z(G) \simeq S_3 \implies |G'|=3$
This follows from transfer theory: Lemma If all the Sylow subgroups of $G$ are abelian, then $G' \cap Z(G)=1$. Now, if $S \in Syl_p(G)$ for some $p$ prime, then either $S \subseteq Z(G)$, in which $S$ is abelian. Or, if $S$ is not contained in $Z(G)$, then $SZ(G)/Z(G) \in Syl_p(G/Z(G))$ and hence $SZ(G)/Z(G)$ is cyclic (and has order $2$ or $3$). Put $H=SZ(G)$. Then $Z(G) \subseteq Z(H) \subseteq H$, whence $H/Z(H)$ is cyclic, and it is well-known that this implies $H$ being abelian. But then $S$ must be abelian, and we are done.
May
19
revised Facts on $ \mathbb{C} $-characters
edited body
May
19
revised Facts on $ \mathbb{C} $-characters
edited tags
May
19
answered Facts on $ \mathbb{C} $-characters
May
15
revised What is the difference between Dijkstra's method and dynamic programming when finding the shortest root of a path?
edited body; edited title
May
15
answered Must a non-simple group have a normal Sylow subgroup?
May
12
revised Maximum number of Sylow subgroups
added 179 characters in body
May
12
revised Maximum number of Sylow subgroups
added 632 characters in body
May
12
revised Maximum number of Sylow subgroups
added 632 characters in body
May
12
comment Maximum number of Sylow subgroups
I wonder if there is a (simple) number-theoretic argument, since it concerns particular divisors $n_p$ of the number $|G|$ that are also $\equiv 1$ mod $p$.
May
12
revised Maximum number of Sylow subgroups
added 28 characters in body
May
12
revised Maximum number of Sylow subgroups
added 20 characters in body
May
12
comment Maximum number of Sylow subgroups
@Andreas, Tobias Could be but it is not clear from the post. Let's try to prove it for all $p$ dividing $|G|$ then. Or come up with a counterexample.
May
12
answered Maximum number of Sylow subgroups
May
9
revised Direct product and Sylow subgroups
edited tags
May
7
revised $g$ has order $n$, then $\langle g\rangle=\langle g^2\rangle=\cdots=\langle g^{n-1}\rangle$
added 4 characters in body
May
6
comment Normal group and commutator subgroup question
Yes, even if $N$ is not central. There are nilpotent groups with $\{1\} \lt [G,G'] \subsetneq G'$. So if you take $N=G'$ here, the statement won't be true.
May
6
answered Normal group and commutator subgroup question
May
6
comment Find the coordinates of the point(s) at which the tangent to the curve is parallel to the x-axis.
Hint: can you find the coordinates of the top of the parabola, even without knowing derivatives?