Discussion
Every finite group is isomorphic to a subgroup of :
Answer: C
Every finite group is isomorphic to a subgroup of \(S_{n}\) for some \(n\).
This is Cayley's theorem: a group G acts on itself by left multiplication, \(g \mapsto (x \mapsto gx)\), and this gives an injective homomorphism from G into the symmetric group on the \(|G|\) elements of G. So G is isomorphic to a subgroup of \(S_{n}\) with \(n = |G|\).
The other options are abelian groups, and their subgroups are abelian, so a non-abelian group such as \(S_{3}\) cannot embed in a cyclic group, in \(\mathbf{Z}_{n}\) or in \((\mathbf{R}, +)\).
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