# Equality in commutator subgroup

I will start by apologizing if this questions seems twisted.

I am reading the paper Cohomology theory of groups with a single defining relation (Lyndon, $1950$) and the question comes from page $659$ Corollary $7.1$.

Basically let $R_{t}=Q_t^{q_t}$ be elements in a free group $F$. It is stated that if: $$\prod T_i R_{t_i}^{e_i}T_i^{-1} =P\in[R,R]$$ Where $R$ is the smallest normal subgroup containing all $\{R_{t}\}$ and $\forall i$ $\exists j$ such that $e_i = - e_j$ and $T_i \equiv T_j Q_{t_i}^{c_i}mod(R)$ then $P=P'$ (another product of ${R_t}$ conjugated by $T_i \in F$) where instead of $T_j'$ we actually have $T_i'$ (meaning for every $T_i' R_{t_i}^{e_i}{T_i'}^{-1}$ in $P'$ we have $T_i' R_{t_i}^{-e_i}{T_i'}^{-1}$).

I think that because $P\in[R,R]$ it is possible for us to exchange $T_i$ with $T_j Q_{t_i}^{c_i}$ and get $$T_i R_{t_i}^{e_i}T_i^{-1}=T_j Q_{t_i}^{c_i}Q_t^{q_{t_i}}Q_{t_i}^{-c_i}T_j^{-1}=T_j Q_{t_i}^{c_i}Q_{t_i}^{q_t}Q_{t_i}^{-c_i}T_j^{-1}=T_j Q_{t_i}^{q_t}T_j^{-1}=T_j R_{t_j}^{-e_j}T_j^{-1}$$ yet I do not understand completely why this exchange is valid.

Thanks mates.

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