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I see a remarkable answer about this question in Why is the Petersen graph no Cayley graph?.

In this proof. If $G=\mathbb{Z}_{10}$ and generator set is $C=\lbrace a,b,c\rbrace$, then one can find a 4-cycle in Cayley graph $X(G,C)$. The original word in above proof is “$a^{-1}b^{-1}ab$ gives a cycle of length 4”. But I have a little confusion about this. What’s $a^{-1}b^{-1}ab$ mean? A 4-cycle or others?

Petersen graph is undirected and has no loop, impilying that $C$ is inverse-closed and $0\notin C$. So, we may assume $b=-b$ and $a+c=0$. Consequently $0-a-(a+b)-b-0$ is an obvious 4-cycle.

Dehai
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    The cycle you have is basically the same as the cycle in the question, but that one is written multiplicatively instead of additively. – Misha Lavrov Jan 14 '24 at 16:34

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The idea is that when we have a Cayley graph of a group $G$, any product of the generators, like $ac^2b^{-2}c$ or $a^{-1}b^{-1}ab$, gives a walk in the graph. The general rule is that a product $g_1g_2 \cdots g_n$ corresponds to the walk $$1, g_1, g_1 g_2, g_1 g_2g_3, \dots, g_1 g_2 \cdots g_n.$$ So, the claim "$a^{-1}b^{-1}ab$ gives a $4$-cycle" is really saying that the walk $$1, a^{-1}, a^{-1}b^{-1}, a^{-1}b^{-1}a, a^{-1}b^{-1}ab$$ is a $4$-cycle. And this is true - when the group is abelian, which we're assuming for this part of the proof. Canceling out generators and their inverses, the walk turns into $$1, a^{-1}, a^{-1}b^{-1}, b^{-1}, 1.$$ This is indeed a $4$-cycle.

Since the group is abelian, we could also rewrite all of this in additive notation, which would give us $0, -a, -(a+b), -b, 0.$ This is just the negative of the $4$-cycle $0, a, (a+b), b, 0$ mentioned in the question.

Misha Lavrov
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