= Equivalence of group representations
{title2=$E(g)=S^{-1}D(g)S\quad\text{for every }g\in G$}
= Equivalent representation
{synonym}
Two finite-dimensional <group representations> $D,E$ over the same <field> are equivalent if there is one invertible linear map $S$ intertwining all their group actions, equivalently $E(g)=S^{-1}D(g)S$ for every $g$. The same change of basis must work simultaneously for all group elements. The <cyclic property of the trace> implies that <equivalent representations> have identical <characters>: $\operatorname{tr}(S^{-1}D(g)S)=\operatorname{tr}D(g)$. Consequently a difference at a single group element proves inequivalence. This implication does not need the group to be finite, and the reverse implication is not asserted for arbitrary groups or fields.
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