Confusion regarding matrix representation properties

Question

I am studying about matrix representation of finite groups. If the group is defined as \begin{equation} G=\{e,a,b,c,.....\} \end{equation} then the matrix representation is defined by the collection of non-singular matrices \begin{equation} T=\left\{T(e),T(a),T(b),T(c),.....\right\} \end{equation} The representation has the property that \begin{equation} T(a)T(b)=T(ab) \tag{1} \end{equation} I am trying to prove $(3)$ using the fact that if $\{\phi_i\}$ be a set of orthonormal basis vectors of the vector space where the operators of $G$ act, then operation of an element $a\in G$ on a basis vector would be given by: \begin{equation} a\phi_i=\sum_j T_{ij}(a) \phi_j \tag{2} \end{equation} Using $(2)$, I can write for $a,b \in G$, \begin{align} (ab)\phi_i &=a\left(\sum_j T_{ij}(b) \phi_j\right)\\ &=\sum_j T_{ij}(b) a\phi_j\\ &=\sum_{k}\left(\sum_j T_{ij}(b) T_{jk}(a)\right)\phi_k \tag{3} \end{align} Also, as $ab \in G$ for $a,b \in G$, we can write \begin{equation} (ab)\phi_i=\sum_k T_{ik}(ab)\phi_k \tag{4} \end{equation} Comparing $(3)$ and $(4)$, I am getting \begin{equation} T_{ik}(ab)=\sum_j T_{ij}(b) T_{jk}(a) \end{equation} However, in order to prove $(1)$, I had to show instead that \begin{equation} T_{ik}(ab)=\sum_j T_{ij}(a) T_{jk}(b) \end{equation} I am not getting where did I go wrong. Can anyone please help me in this derivation?

Answer 1

The standard definition of the matrix $T(a)$ representing the linear action of $a$ relative to a (not necessarily orthonormal) basis $\lbrace\phi_i\rbrace$ is $$ a\phi_i = \sum_j T_{ji}(a)\phi_j. $$ Note this differs from your expression (2) in the order of the indices. Using this should give the result you seek.