8 votes
Accepted

Center of group of order 3773

Let $P$ be a $7$-Sylow subgroup, $Q$ is an $11$-Sylow subgroup. Then $P,Q$ are normal in $G$, $P\cap Q=\{e\}$ and by comparing cardinalities, $G=PQ$. It is well known that in this case we have: $G\...
Mark's user avatar
  • 40.1k
4 votes

Center of group of order 3773

If $G$ is abelian, then you are done. So, let's suppose $G$ nonabelian. From $n_{11}=1$ follows that the Sylow $11$-subgroup is normal and, as such, it is union of conjugacy classes. Now, the ...
citadel's user avatar
  • 3,035
3 votes
Accepted

GAP orthogonal groups: Specifying the invariant bilinear form

This is a feature of the forms package. The problem is that the matrix you give is not equivalent over $GF(7)$ to the matrix used internally (but its negative is). ...
ahulpke's user avatar
  • 18.5k
3 votes
Accepted

Prove the index of a proper subgroup of a simple group of order 17971200 is at least 14.

Let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$, $X_G:=\{H\le G\mid H\ne G\}$. Lagrange's theorem already puts a limitation on the order of the elements of $...
citadel's user avatar
  • 3,035
1 vote

Group of order $p^{\alpha}q$ is not simple.

Since the Sylow $p$-subgroups are all pairwise conjugate (Sylow II), their intersection is the normal core in $G$ of any of them. So, once you know this intersection is not trivial, you get a ...
citadel's user avatar
  • 3,035

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