Discussion
For which of the following values of \(n\) is every group of order \(n\) abelian ?
Answer: B
Every group of order 121 is abelian.
\(121 = 11^{2}\), and every group of order \(p^{2}\) is abelian: its centre is non-trivial (a \(p\)-group has non-trivial centre), and if \(G/Z(G)\) were cyclic G would be abelian, so \(Z(G) = G\). Such a group is \(\mathbf{Z}_{121}\) or \(\mathbf{Z}_{11} \times \mathbf{Z}_{11}\).
Each other order has a non-abelian group: 100 has \(D_{50}\), the dihedral group of order 100; 24 has \(S_{4}\); and 120 has \(S_{5}\).
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