$\widehat{L}_{\varepsilon}(h^n\circ f)=-\widehat{L}_{\varepsilon}(f)$. Thus $\widehat{L}_{\varepsilon}(f)=0$, which contradicts Lemma \[k7 odd\]. Hence in fact, $K_7$ is intrinsically chiral.

\[K4n+3\] For every odd number $n$, the complete graph $K_{4n+3}$ is intrinsically chiral.

Suppose that for some embedding $f$ of $K_{4n+3}$ in $S^3$, there is an orientation reversing homeomorphism $h$ of (${S}^3$, $f(K_{4n+3})$). Even though in general the homeomorphism $h$ will not have finite order, the automorphism that $h$ induces on $K_{4n+3}$ does have finite order and its order can be expressed as $2^ab$ for some odd number $b$. Now $g=h^b$ is an orientation reversing homeomorphism of (${S}^3$, $f(K_{4n+3})$) which induces an automorphism of $K_{4n+3}$ of order $2^a$.

Observe that the number of $K_7$ subgraphs in $K_{4n+3}$ is

$$\frac{(4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2)(4n-3)}{7!}$$ $$=\frac{(4n+3)(2n+1)(4n+1)(n)(4n-1)(2n-1)(4n-3)}{315}.$$

This number is odd, since $n$ is odd. Thus $g$ leaves invariant some $K_7$ subgraph. But this is impossible since by Theorem \[k7 chiral\], $K_7$ is intrinsically chiral.

Mobius ladders {#sec:mobius .unnumbered}
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A Möbius ladder $M_n$ with $n$ rungs is the graph obtained from a circle with $2n$ vertices by adding an edge between every pair of antipodal vertices. Let $N\geq 2$, and consider the oriented labeled graph of $M_{2N+1}$ illustrated in Figure \[FigureM\_2N+1\] (note there is no vertex at the center of 