Corollary 2

— If G is a finite commutative group annihilated by n, then the groups G and $\mathrm{Hom}(G, T)$ have the same order.

If G is cyclic of order r, with generator g, then the mapping $f \mapsto f(g)$ is a bijection of $\mathrm{Hom}(G, T)$ onto the set of elements t of T such that $t^r = 1$, hence the assertion follows in this case. On the other hand, if H is a cyclic subgroup of G, we have $\mathrm{Card}(G) = \mathrm{Card}(H) \cdot \mathrm{Card}(G/H)$; further, we have an exact sequence

(II, p. 227, Th. 1 and Prop. 7 above), hence

Since $\mathrm{Card}(\mathrm{Hom}(H, T)) = \mathrm{Card}(H)$, it comes to the same to prove the corollary for G or for $G/H$. Now the result follows by induction on $\mathrm{Card}(G)$.

Now let G and H be two groups and $f : G \times H \to T$ a bimultiplicative mapping, that is, such that for any $g, g' \in G, h, h' \in H$ we have

We define group homomorphisms

by $s_f(g)(h) = d_f(h)(g) = f(g, h)$ (cf. II, p. 268, Cor. to Prop. 1, when G and H are commutative).