Let \(G\) be a finite group with \(|G| = 150\). Prove that for at least one \(k \in \{5, 10, 15, 25\}\) there exists a normal subgroup \(N \trianglelefteq G\) of order \(k\).
Solution
1. A Sylow \(5\)-subgroup \(P\) has order \(25\) and index \(6\).
The factorization is \(150=2\cdot3\cdot5^2\). Sylow’s existence theorem [@DF04] supplies \(P\leq G\) of order \(5^2=25\). The index is \([G:P]=150/25=6\).
2. The action on the six left cosets gives a normal subgroup\(N\lhd G\) contained in \(P\).
Left multiplication defines a homomorphism \[ \rho:G\longrightarrow\operatorname{Sym}(G/P)\cong S_6, \qquad \rho(g)(aP)=gaP. \] It is well-defined because multiplication preserves equality of left cosets, and the inverse permutation is induced by \(g^{-1}\). Put \(N=\ker\rho\), which is normal in \(G\). An element of \(N\) fixes the coset \(P\), so it lies in its stabilizer \(P\). Thus \(N\subseteq P\) and \[ |N|\in\{1,5,25\}. \]
3. The kernel \(N\) is nontrivial, so \(|N|\in\{5,25\}\).
If \(N=1\), the homomorphism \(\rho\) would embed \(G\) in \(S_6\). Lagrange’s theorem would then imply \[ 150\mid |S_6|=6!=720=2^4\cdot3^2\cdot5. \] This is impossible because \(25\mid150\) but \(25\nmid720\). Therefore step 2 leaves \(|N|=5\) or \(25\), both members of \(\{5,10,15,25\}\).