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A characterization of a family of integral Cayley graphs

  • Sauvik Poddar
  • Sucharita Biswas
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Abstract

Let G be a finite group and H be a subgroup of G. In our earlier work, we proved that the quasi-strongly regular Cayley graph ΓH(G) is integral whenever H is normal in G, and raised the question of whether the converse also holds. In this paper, we answer this question affirmatively. We prove that if ΓH(G) is integral, then H must be a normal subgroup of G. Consequently, ΓH(G) is integral if and only if Hlhd G. Our proof uses the irreducible representations of G and the dimensions of their H-fixed subspaces. For a non-normal subgroup H, we construct a two-dimensional invariant subspace of a suitable representation operator and show that the corresponding eigenvalues are irrational.

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Cite

BibTeXRIS
BibTeX
@misc{poddar2026characterization,
  title = {{A characterization of a family of integral Cayley graphs}},
  author = {Sauvik Poddar and Sucharita Biswas},
  howpublished = {arXiv},
  year = {2026},
  month = nov,
  doi = {10.48550/arxiv.2611.06581},
  eprint = {2611.06581},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2611.06581},
}
RIS
TY  - UNPB
TI  - A characterization of a family of integral Cayley graphs
AU  - Sauvik Poddar
AU  - Sucharita Biswas
PY  - 2026
DA  - 2026/11/04
PB  - arXiv
DO  - 10.48550/arxiv.2611.06581
UR  - https://arxiv.org/abs/2611.06581
AB  - Let G be a finite group and H be a subgroup of G. In our earlier work, we proved that the quasi-strongly regular Cayley graph Γ_H(G) is integral whenever H is normal in G, and raised the question of whether the converse also holds. In this paper, we answer this question affirmatively. We prove that if Γ_H(G) is integral, then H must be a normal subgroup of G. Consequently, Γ_H(G) is integral if and only if Hlhd G. Our proof uses the irreducible representations of G and the dimensions of their H-fixed subspaces. For a non-normal subgroup H, we construct a two-dimensional invariant subspace of a suitable representation operator and show that the corresponding eigenvalues are irrational.
ER  -