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I am a former assistant-professor of mathematics at the Univ. of Amsterdam, with specialization in group theory and representation theory of groups. I work for IBM, though not in research.


1h
answered How is number of conjugacy class related to the order of a group?
15h
revised find $\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$
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15h
answered find $\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$
16h
comment If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
@Timbuc, the theorem referred to is a generalization of the following. Let $P \in Syl_p(G)$ be abelian, then $|G' \cap Z(G)|$ is not divisible by $p$. In other words: $P \cap G' \cap Z(G)=1$. The proof can be found in many textbooks and most of the time is proved using transfer. Transfer proofs can often be replaced by arguments using the determinant of a character, see the proof Theorem (5.6) in I.M. Isaacs book Character Theory of Finite Groups. The proof given there can be used to prove a general and beautiful result for $P$ also being non-abelian: $P \cap G' \cap Z(G) \subseteq P'$.
17h
comment If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
@Ben Blum - Why do you want to rule out that case? $P$ must be non-abelian, since $|P'|=2$ ... so it cannot equal the center.
17h
revised If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
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17h
comment If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
Let $N=\{1,n\}$, $n \neq 1$ and $g \in G$. Since $N$ is normal, $g^{-1}ng \in N$. Then $g^{-1}ng=1$ or $=n$. But if it would be $=1$, then $n=1$, a contradiction. So $g$ centralizes $n$, whence $N \subseteq Z(G)$.
17h
comment If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
@Timbuc - if $H \leq K$, then $H' \leq K'$ for subgroups $H$ and $K$ is general.
17h
comment If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
Which part do you like to be explained?
18h
answered If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$
1d
revised Explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$
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1d
answered Explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$
1d
revised Groups of Order $n$
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1d
answered Groups of Order $n$
1d
revised Show that $G' = \bigcap_{C\subseteq N \triangleleft G} N$
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1d
answered Isomorphisms based on Conjugacy Classes
Nov
18
revised Show that a certain normal subgroup or a product is abelian
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Nov
18
comment A commutator relation
@Mesel, no thanks and good question! +1 from me.
Nov
18
revised A commutator relation
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Nov
18
answered A commutator relation