Proposition 3

If t \in \mathscr{T}^{(\infty)}(G) and g \in G, then $\gamma(g)_* t = \varepsilon_g * t, \delta(g)_* t = t * \varepsilon_{g^{-1}}$, (Int g)_* t = $\varepsilon_g * t * \varepsilon_{g^{-1}}$. In particular, $\varepsilon_e$ is the unit element of \mathscr{T}^{(\infty)}(G).

Consider the diagram

where $\phi$ is the mapping $h \mapsto (g, h)$ and $\psi$ is the mapping $(h', h) \mapsto h'h$. Then $\gamma(g) = \psi \circ \phi$ and hence $\gamma(g)_* t = \psi_*(\phi_*(t))$. But $\phi_*(t) = \varepsilon_g \otimes t$ and hence $\psi_*(\phi_*(t)) = \varepsilon_g * t$. The argument is similar for $\delta(g)_* t$. Finally,

and hence $(\operatorname{Int} g)_* = \gamma(g)_* \circ \delta(g)_*$.

It is therefore seen that, for $t \in T(G)$, $\varepsilon_g * t$ and $t * \varepsilon_g$ are equal to $gt$ and $tg$ calculated in the group T(G) (§ 2, no. 2). But it should be noted that, for $t, t'$ in T(G), the product $tt'$ in the sense of § 2 is in general different from $t * t'$.