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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$$\times$$A’$$\rightarrow$$C$ 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}$$\times$$B$$\rightarrow$$C$(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 $C$?)

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

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