# Non-degeneracy of product of group pairings

For $G$ finite abelian group, let $\eta,\omega:G \times G \to \mathbb{C}^\times$ be group pairings. What can I say about the (non-)degeneneracy of the product pairing $\eta \cdot \omega$ in terms of the (non-)degeneneracy of $\eta$ and $\omega$? The product is defined elementwise: $(\eta \cdot \omega) (g,h):= \eta(g,h) \cdot \omega(g,h)$.

• What is a group pairing? – Nick Gill Sep 21 '16 at 11:04
• a map $\omega: G \times H \to \mathbb{C}^\times$, s.t. the induced map $G \to H^*$ is a group homomorphism, where $H^*$ is the dual group of $H$ – Bipolar Minds Sep 21 '16 at 11:16
• If product is what I think it is, it's obviously nondegenerate of and only if so are $\mu$ and $\omega$. – Alex Degtyarev Sep 21 '16 at 16:50
• the product is defined elementwise.. I dont see why this is obvious.. – Bipolar Minds Sep 21 '16 at 18:42
• @AlexDegtyarev what kind of product did you think of? – Bipolar Minds Sep 21 '16 at 19:01

If $A$ is a finite abelian group and $\chi\colon A\to A^*$ is an isomorphism, then $\chi$ determines a group isomorphism between $\langle \textrm{End}(A); +,-,0\rangle$ and the abelian group of pairings under pointwise product. Namely, if $\varphi\in\textrm{End}(A)$, then the corresponding pairing is $(a,b)\mapsto (\chi\circ \varphi(a))(b)$. Under this correpondence the property of nondegeneracy of a pairing corresponds to the property of injectivity of an endomorphism. Since $A$ is finite, nondegenerate pairings correspond to units in $\textrm{End}(A)$.
Hence the main question about pairings (What can I say about the nondegeneneracy of the product pairing $\eta\cdot\omega$ in terms of the nondegeneneracy of $\eta$ and $\omega$?) translates into the following question: When is the sum of two units in the ring $\textrm{End}(A)$ equal to a unit? Similarly, if you replace nondegeneracy with degeneracy in the original question, you should replace unit with nonunit in the reformulated question. Since $\textrm{End}(A)$ is a fairly typical finite ring, these questions are unlikely to have nontrivial answers.