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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.


4h
comment Groups of orders 108, 120, 144, 168, 180, 228 and 240
You mean Isaacs' book?
4h
comment Groups of orders 108, 120, 144, 168, 180, 228 and 240
What have you tried? Which theorems do you know? Sylow's?
6h
revised $N$ a normal subgroup of $G$ with $|G|$ odd and $|N|=5$ satisfies $N \subset Z(G)$
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6h
revised $N$ a normal subgroup of $G$ with $|G|$ odd and $|N|=5$ satisfies $N \subset Z(G)$
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7h
answered $N$ a normal subgroup of $G$ with $|G|$ odd and $|N|=5$ satisfies $N \subset Z(G)$
14h
answered Properties of Group representations, duality and the derived subgroup
23h
revised How to solve $ \frac{\left( \sqrt{2} \right) ^a}{a}=1$
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1d
answered Calculating number of conjugacy classes for a prime power group $G$
Sep
12
comment Number of Sylow 2-subgroups of a special linear group
@user152715 Yes, there maybe an intersection, you have to rule out that as well. Did not want to give away the total solution at once. What I mean with writing down the matrices is reflected by Derek's answer.
Sep
12
comment Number of Sylow 2-subgroups of a special linear group
You noted that there are 4 Sylow $3$-subgroups, which gives 4(3-1)=8 elements of order 3. Now if there are 3 Sylow $2$-subgroups, then they all contain the center of $SL(2,3)$ being $\{±I\}$. Then you would have 3(8-2)=18 2-elements; together with the 8 elements of order 3, this gives you 26 different elements ...
Sep
12
revised Number of Sylow 2-subgroups of a special linear group
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Sep
12
answered Number of Sylow 2-subgroups of a special linear group
Sep
10
revised What is the least $n$ such that it is possible to embed $\operatorname{GL}_2(\mathbb{F}_5)$ into $S_n$?
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Sep
9
comment question on nilpotent group.
A slightly different argument for finite groups, also using the Frattini subgroup and Frattini argument, runs as follows. For this you have to know that if a group $G$ is nilpotent, then $G' \subseteq \Phi(G)$, and if $N \unlhd G$, then $\Phi(N) \subseteq \Phi(G)$. Now let $P \in Syl_p(G)$, then $PN'/N' \in Syl_p(G/N')$, so $PN' \unlhd G$, since $G/N'$ is nilpotent. But $P \in Syl_p(PN')$, so the Frattini argument yields $G=N_G(P)PN'=N_G(P)N'$. But since $N$ is nilpotent we have $N' \subseteq \Phi(N) \subseteq \Phi(G)$, and it follows that in fact $G=N_G(P)$, that is $P$ is normal in $G$.
Sep
9
comment question on nilpotent group.
Same reference as Derek Holt gave
Sep
9
answered question on nilpotent group.
Sep
9
comment Find Pi using integral
@rlartiga Thanks for explaining - and Bob I can see you learned something, well done!
Sep
9
answered Find Pi using integral
Sep
9
comment non-abelian simple group
You are welcome! Please, asd a common habit on StackExchange, if you deem it as the one your were looking for, tick it as such. Thanks.
Sep
9
answered non-abelian simple group