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1999 Mar

1999 March

  1. There is no simple group of order 328.

  2. Every finite \(p\)-group has a nontrivial center.

  3. Suppose \(G\) is a group and \(H\) is a subgroup of \(G\) and \(N\) is a normal subgroup of \(G\).\ a. If \(G\) is solvable, then \(H\) is solvable.\ b. If \(G\) is solvable, then \(G/N\) is solvable.\ c. If \(N\) and \(G/N\) are solvable, then \(G\) is solvable

  4. Suppose \(H\) and \(K\) are subgroups of the group \(G\). Then \([H, K]\) is the subgroup generated by \(\{hkh^{-1}k^{-1} \mid h\in H, \ k\in K \}\). We recursively define \(G_n\) by \(G_0 = G\) and \(G_{n+1} = [G_n, G]\). A group is nilpotent iff for some \(k\), \(G_k = \{1\}\). Use this definition of nilpotent group in what follows. For the rest of the problem, \(G\) is assumed to be a finite group.\ a. If \(P\) is a Sylow subgroup of \(G\), then \(N(N(P)) = N(P)\), where \(N(P)\) denotes the normalizer of \(P\).\ b. If \(G\) is a nilpotent group and \(H\) is a subgroup of \(G\) not equal to \(G\), then \(N(H) \neq H\).\ c. If \(G\) is nilpotent and \(P\) is a Sylow subgroup of \(G\), then \(P\) is a normal subgroup of \(G\).\ d. If each of the Sylow subgroups of \(G\) are normal in \(G\), then \(G\) is the direct product of its Sylow subgroups.\ e. Every finite, nilpotent group is the direct product of its Sylow subgroups.