Let m, n be integers. Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="58" /> </InlineMediaObject> <EquationSource Format="TEX">$C_m\square C_n$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> be the Cartesian product graph of two cycle graphs with length m and n. The 2-token graph <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="87" /> </InlineMediaObject> <EquationSource Format="TEX">$F_2(C_m\square C_n)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>F</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> of <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="58" /> </InlineMediaObject> <EquationSource Format="TEX">$C_m\square C_n$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> is a graph with vertices the 2-subsets of <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="83" /> </InlineMediaObject> <EquationSource Format="TEX">$V(C_m\square C_n)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>V</mi> <mo stretchy="false">(</mo> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> such that two 2-subsets are adjacent if and only if their symmetric difference is exactly an edge of <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="58" /> </InlineMediaObject> <EquationSource Format="TEX">$C_m\square C_n$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> </mrow> </math></EquationSource> </InlineEquation>. In this paper, the automorphism group of <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="373_2025_2940_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="87" /> </InlineMediaObject> <EquationSource Format="TEX">$F_2(C_m\square C_n)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>F</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <msub> <mi>C</mi> <mi>m</mi> </msub> <mo>□</mo> <msub> <mi>C</mi> <mi>n</mi> </msub> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> is completely determined.

Automorphism Groups of 2-Token Graph of Cartesian Product of Two Cycles

期刊论文
发表日期：

2025年6月14日

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Ju Zhang,
Young Soo Kwon,
Jin-Xin Zhou

摘要

Let m, n be integers. Let $C_m\square C_n$ $C_{m} □ C_{n}$ be the Cartesian product graph of two cycle graphs with length m and n. The 2-token graph $F_2(C_m\square C_n)$ $F_{2} (C_{m} □ C_{n})$ of $C_m\square C_n$ $C_{m} □ C_{n}$ is a graph with vertices the 2-subsets of $V(C_m\square C_n)$ $V (C_{m} □ C_{n})$ such that two 2-subsets are adjacent if and only if their symmetric difference is exactly an edge of $C_m\square C_n$ $C_{m} □ C_{n}$ . In this paper, the automorphism group of $F_2(C_m\square C_n)$ $F_{2} (C_{m} □ C_{n})$ is completely determined.