Are there two nonisomorphic groups with the same representations?
Solution
Yes, if “the same representations” means the same complex character table or the same complex representation ring. A standard example is the dihedral group \(D_8\) of order \(8\) and the quaternion group \(Q_8\).
1. The groups \(D_8\) and \(Q_8\) are not isomorphic.
The quaternion group has a unique element of order \(2\), namely \(-1\). The dihedral group of order \(8\) has five elements of order \(2\) (the half-turn and four reflections). Thus the groups are not isomorphic.
2. Each group has four one-dimensional complex irreducible representations and one two-dimensional irreducible representation.
Both abelianizations are isomorphic to \(C_2\times C_2\), so both have four one-dimensional characters. Each group has five conjugacy classes, hence five irreducible complex characters. The sum-of-squares formula \[ \sum_i (\dim V_i)^2=8 \] forces the remaining irreducible to have dimension \(2\).
3. After a suitable matching of conjugacy classes, their character tables are identical.
In both groups the four linear characters factor through \(C_2\times C_2\). The two-dimensional irreducible character has value \(2\) at the identity, \(-2\) at the unique central nonidentity element in the commutator subgroup, and \(0\) on the remaining three noncentral conjugacy classes. Thus the character tables coincide after relabeling the classes.
Consequently the complex representation rings are isomorphic even though the groups are not.
This does not mean their symmetric tensor categories of representations, together with the forgetful fiber functor, are indistinguishable: Tannakian reconstruction recovers the group from that richer structure. The counterexample concerns coarser representation-theoretic data such as characters or the representation ring.