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Questions on group theory which concern finite groups.
12
votes
Accepted
Why a group of order $2^{m}\cdot p^{n}\cdot q^{t}$ is solvable?
Let $G$ be an unsolvable group. Since $G$ is a finite group, then it
has a chief series. Since $G$ is unsolvable it is easy to show that $G$ has series: $1 \unlhd N \lhd H \unlhd G$ such that $H/N$ is …
-1
votes
1
answer
440
views
Order of normalizer Sylow 5-subgroup in Suzuki group
It known the Suzuki group $Sz(q)$, where $q=2^{2n+1}$ is of order $q^2(q^{2}+1)(q-1)$. By $2^{2}$ $=-1$ mod $5$, then $2^{2n}$ $=(-1)^{n}$ mod $5$. So $q=2^{2n+1}$ $=2(-1)^{n}$ mod $5$ and so $q^{2}+1 …