1993 Nov
1993 November
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Prove that there is no non-abelian simple group of order 36.
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Prove that for all \(n\geq 3\), the commutator subgroup of \(S_n\) is \(A_n\).
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a. State, without proof, the Sylow Theorems.\ b. Prove that every group of order 255 is cyclic.
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Let \(G\) be a group and let \(X\) be a set. We say that \(G\) acts on \(X\) (on the left) if there is a multiplication \(G\times X \rightarrow X\) such that \(1x = x\) and \((gh)x = g(hx)\) for all \(g, h\in G\) and \(x\in X\).\ a. Let \(H\) be a subgroup of \(G\). Prove that \(G\) acts on the space of left cosets \(G/H\).\ b. Let \(x\in X\) and let \(H\) be the subgroup of \(G\) fixing \(x\). Define the orbit of \(x\) under \(G\) and prove that there is a one-to-one correspondence between this orbit and the space of left cosets \(G/H\). In what sense can this one-to-one correspondence be called an isomorphism?\ c. Let \(g\in G\) and let \(x' = gx\). Let \(H\) be the subgroup of \(G\) fixing \(x\) and let \(H'\) be the subgroup fixing \(x'\). Prove that \(H\) and \(H'\) are conjugate.
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Prove that a finite group is nilpotent if and only if it is a product of \(p\)-groups.
{\bf Theorem.} [L. Sylow (1872)] Let \(G\) be a finite group with \(| G| = p^m r\), where \(m\) is a non-negative integer and \(r\) is a positive integer such that \(p\) does not divide \(r\). Then \begin{enumerate}[(i)] \item \(G\) has a subgroup of order \(p^m\).
Such a subgroup is called a \emph{Sylow \(p\)-subgroup} of \(G\). \item If \(H\) and \(J\) are Sylow \(p\)-subgroups of \(G\), then \(J\leq g H g^{-1}\) for some \(g\in G\). In particular, the Sylow \(p\)-subgroups of \(G\) form a single conjugacy class. \item Let \(n_p\) denote the number of Sylow \(p\)-subgroups of \(G\) and let \(H\) be any Sylow \(p\)-subgroup of \(G\). Then [ n_p \equiv 1 \pmod p, \qquad n_p \mid [G:H], \quad \text{ and } \quad n_p = |G : N_G(H)|. ] (Note that \([G:H]=r\) so the second condition says that \(n_p\) divides \(r\).) \end{enumerate}