Let \(G\) be a group (not necessarily finite) that contains a subgroup of index \(n\). Show that \(G\) contains a normal subgroup \(N\) such that \(n\leq[G:N]\leq n!\)
Use part (a) to show that there is no simple group of order 36.
Solution
1. Part (a): Normal core and index bounds:
1.1. Let \(H \le G\) be a subgroup of index \([G : H] = n\).Let \(X = G/H = \{x_1 H, \dots, x_n H\}\) be the set of left cosets of \(H\) in \(G\), with \(|X| = n\).
definition of index.
1.2. The group \(G\) acts on \(X\) by left multiplication: \(g \cdot (x H) = (gx) H\).This action induces a group homomorphism:\[ \rho: G \to S(X) \cong S_n. \]
group action on cosets.
1.3. Define \(N = \ker(\rho) = \bigcap_{g \in G} g H g^{-1} = \operatorname{Core}_G(H)\).\(N\) is a normal subgroup of \(G\) (\(N \trianglelefteq G\)).
kernel of a group homomorphism is normal.
1.4. By the First Isomorphism Theorem, \(G/N \cong \operatorname{im}(\rho) \le S_n\).Thus \([G : N] = |G/N| = |\operatorname{im}(\rho)|\) divides \(|S_n| = n!\), so \([G : N] \le n!\).
Lagrange’s Theorem for subgroups of \(S_n\).
1.5. Furthermore, \(N = \ker(\rho) \subseteq H\) because for any \(g \in N\), \(g \cdot (eH) = eH \implies g \in H\).By multiplicativity of indices:\[ [G : N] = [G : H][H : N] = n [H : N] \ge n. \]Thus \(n \le [G : N] \le n!\), and \([G : N] \mid n!\).
tower law for subgroup indices.
2. Part (b): No simple group of order 36:
2.1. Suppose for contradiction that there exists a simple group \(G\) of order \(|G| = 36 = 2^2 \cdot 3^2\).
assumption for contradiction.
2.2. By Sylow’s Third Theorem, the number \(n_3\) of Sylow 3-subgroups satisfies:\[ n_3 \equiv 1 \pmod 3 \quad \text{and} \quad n_3 \mid 4 \implies n_3 \in \{1, 4\}. \]
Sylow’s Theorem.
2.3. If \(n_3 = 1\), the unique Sylow 3-subgroup \(P\) is normal in \(G\) with \(|P| = 9\).Since \(1 < 9 < 36\), \(P\) is a non-trivial proper normal subgroup, contradicting simplicity.
unique Sylow subgroups are normal.
2.4. If \(n_3 = 4\), let \(P\) be a Sylow 3-subgroup of order 9.Its index in \(G\) is \([G : P] = 36 / 9 = 4\).
Lagrange’s Theorem.
2.5. Apply Part (a) to \(H = P\) with \(n = 4\):There exists a normal subgroup \(N \trianglelefteq G\) such that \(N \subseteq P\) and \([G : N]\) divides \(4! = 24\).
Part (a).
2.6. Since \(G\) is assumed simple, its only normal subgroups are \(N = \{e\}\) and \(N = G\):- If \(N = G\), then \(N \subseteq P \implies G \subseteq P\), which is impossible since \(|P| = 9 < 36\).- If \(N = \{e\}\), then \([G : N] = |G| = 36\). But by 5, \([G : N]\) must divide 24, and 36 does not divide 24.Both cases yield a contradiction.
divisibility in \(\mathbb{Z}\).
3. Conclusion:There is no simple group of order 36. Q.E.D.
1 and 2.