Let \(G\) be a finite group of order \(n\) and \(H<G\) a proper subgroup of order \(m\). Assume \[ (n/m)!<2n. \] Prove that \(G\) has a proper nontrivial normal subgroup.
Solution
The statement is false as printed. The exceptional case is \[ G\cong C_2, \qquad H=\{e\}. \] Indeed, then \(n=2\), \(m=1\), and \[ (n/m)!=2!<4=2n, \] but \(C_2\) has no proper nontrivial subgroup.
The following argument shows that this is the only exception.
1. Put\[ k\coloneqq[G:H]=\frac nm. \]The action of \(G\) on the \(k\) left cosets of \(H\) gives a homomorphism\[ \varphi:G\longrightarrow S_k \]whose kernel \(K\) is a proper normal subgroup of \(G\).
The kernel of any homomorphism is normal. Since \(H\) is proper, \(k\ge2\), and the coset action is transitive on at least two points. Hence the action is not trivial, so \(K\ne G\). Thus \(K\lhd G\) and \(K\) is proper.
2. If \(K\ne\{e\}\), then \(G\) has the required proper nontrivial normalsubgroup.
This is immediate from step 1.
3. Suppose \(K=\{e\}\). Then\[ G\cong S_k. \]
If \(K\) is trivial, then \(\varphi\) is injective, so \[ n=|G|=|\varphi(G)| \] divides \(|S_k|=k!\). Therefore \[ [S_k:\varphi(G)] = \frac{k!}{n}. \] The hypothesis \(k!<2n\) gives \[ 1\le\frac{k!}{n}<2. \] The index is an integer, so it equals \(1\). Hence \(\varphi(G)=S_k\) and \(G\cong S_k\).
4. Under the hypothesis of step 3, if \(k\ge3\), then \(G\) has aproper nontrivial normal subgroup.
For \(k\ge3\), the alternating group \[ A_k \] is a proper nontrivial normal subgroup of \(S_k\). Transporting it through the isomorphism in step 3 gives a proper nontrivial normal subgroup of \(G\).
5. Under the hypothesis of step 3, if \(k=2\), then\[ G\cong C_2 \qquad\text{and}\qquad H=\{e\}. \]
Step 3 gives \(G\cong S_2\cong C_2\), so \(n=2\). Since \[ k=\frac nm=2, \] one has \(m=1\), hence \(H=\{e\}\).
6. Consequently the sharp conclusion under the printed hypotheses is:\[ \boxed{ \text{either }(G,H)\cong(C_2,\{e\}), \text{ or }G\text{ has a proper nontrivial normal subgroup}.} \]
2. If it istrivial, step 3 identifies \(G\) with \(S_k\); step 4 settles \(k\ge3\),while step 5 identifies the unique remaining case \(k=2\) with thedisplayed exception.
7. Q.E.D.
The opening counterexample disproves the statement as printed, and step 6 proves the corrected assertion with the unique exceptional case.