# Isomorphism between a quotient of the dual group and the dual of a subgroup

In Lang’s Algebra the following corollary is given:

Corollary 9.3. Let $$A$$ be a finite abelian group, $$B$$ a subgroup, $$\bar{A}$$ the dual group, and $$K$$ the set of $$\phi$$ $$\epsilon$$ $$\bar{A}$$ such that $$\phi(B)$$=0. Then we have a natural isomorphism of $$\bar{A}$$/$$K$$ with $$\bar{B}$$.

The preceding theorem says:

Theorem 9.2. Let $$A\timesA’\rightarrowC$$ be a bilinear map of two abelian groups into a cyclic group of order $$m$$. Let $$B$$, $$B’$$ be its respective kernels on the left and right. Assume that $$A’/B’$$ is finite. Then $$A/B$$ is finite, and $$A’/B’$$ is isomorphic to the dual group of $$A/B$$.

I understand why the theorem implies the corollary as long as the right kernel of the bilinear map $$\bar{A}\timesB\rightarrowC$$(where $$C$$ is a cyclic group of order $$m$$ and $$m$$ is an exponent of $$A$$) defined by ($$\phi$$,$$b$$)$$\mapsto\phi(b)$$ is trivial. However I do not understand why this is(that is, why for each non-identity $$b$$ $$\epsilon$$ $$B$$ is there some $$\phi$$ $$\epsilon$$ $$\bar{A}$$ such that $$\phi(b)$$ is not the identity in $$C$$?)

• Write $A\times B$ for $A\times B$, rather than $A$$\times$$B$. Jul 29 at 18:28

I will follow Lang and use $$A^{\wedge}$$ for the dual.
Express $$A$$ as a direct product of cyclic groups of prie power order, $$A \cong \mathbf{Z}_{p_1^{a_1}}\times\cdots\times \mathbf{Z}_{p_k^{a_k}},$$ (as done by Lang in the prior section; I would have used the invariant factor decomposition, but I didn't see it in the section). Note that because $$A$$ is of exponent $$m$$, $$p_i^{a_i}\mid m$$ for all $$i$$. Let $$\pi_i\colon A\to \mathbf{Z}_{p_i^{a_i}}$$ be the canonical projection onto the $$i$$th component.
If $$b\in B$$, $$b\neq 0$$, then there exists $$i$$ such that $$\pi_i(b)\neq 0$$. Then composing $$\pi_i$$ with an embedding $$\mathbf{Z}_{p_i^{a_i}}\hookrightarrow \mathbf{Z}_m$$ (possible since $$p_i^{a_i}\mid m$$, so $$\mathbf{Z}_m$$ has a subgroup isomorphic to $$\mathbf{Z}_{{p_i}^{a_i}}$$) yields an element of $$A^{\wedge}$$ that does not have $$b$$ in its kernel.