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Let $G$ be a group containing $2k$ lements where $k$ is odd. Let $g \in G$ of order $2$ and define $$ \lambda_g \quad : \quad G \longmapsto G \quad : \quad x \mapsto gx $$ I had to show that $\lambda_g$ is an odd permutation in $S(G)$.

I know that $\lambda_g^2 = e$, so if we label the elements of G, we would get $$ \lambda_g \quad = \quad (a_1b_1)(a_2b_2)\cdots (a_nb_n) $$ for some $n$ smaller than $k$. Now it is enough to show that $n$ must be odd. I know that $e \in G$ and some other won't be moved, but I know how to go on from here.

It seems like those inversions appear pairwise, but I don't know why. Can you help me to solv this one?

Koenraad van Duin
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1 Answers1

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For each $x\in G$, $\lambda_g(x)=gx\neq x$, since $g$ has $2$ as its order, which means that $g\neq e$. Thus in your representation $$\lambda_g \quad = \quad (a_1b_1)(a_2b_2)\cdots (a_nb_n),$$ every element of $G$ appears once, in the form of $(x,gx)$. Therefore, $n=k$ is odd, and this completes your proof.

p.s. You said that $e$ is unchanged by $\lambda_g$, but $\lambda_g(e)=ge=g\neq e$.

Jaehyeon Seo
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