Let $G$ be a group and $H$ be a subgroup with $(G:H)=m$. Let $X$ be set of all left cosets of $H$ in $G$.

Define $L_g$ to be a bijection such that $L_g(qH)=gqH$ where $qH$ and $gqH$ are left cosets.

Show the order of $L_g$ as an element of the symetric group on $X$ is a divisor of the order of $g$ in $G$.

So far I just get that the two orders are the same. Suppose the order of $L_g$ is $n$. Then taking $L_g$ on $qH$ $n$ times yields $qH=(g^n)qH$ or $q=(g^n)q$ or $i=g^n$.

Am I doing it right? Or what do you suggest?

  • 1
    $\begingroup$ So you have shown that applying $L_g$ $n$ times gives the identity. This does not mean that it has order $n$, but precisely that it has order dividing $n$ as you needed to show. $\endgroup$ – Tobias Kildetoft Nov 23 '15 at 10:58
  • $\begingroup$ To be even more pointed about this, we could take $H = G$, so that $m = 1$. Then $X = \{G\}$ (it has one element) and the only element of $\text{Sym}(X)$ is the identity mapping which sends the coset $G$ to itself. Clearly, even if $g \neq e_G$, and $|g| = n$, we have $L_g:\{G\} \to \{G\}$ has order $1$ in $\text{Sym}(X)$. $\endgroup$ – David Wheeler Nov 23 '15 at 11:47

$qH=(g^n)qH$ does not imply $q=(g^n)q$, only that $q=(g^n)qh$ for some $h\in H$.

A simple proof of the result you seek is this:

$L: g \mapsto L_g$ is a group homomorphism $G \to S(X)$ and so $g^n=1$ implies $L_g^n=1$.

This is a more general fact:

If $\phi:G\rightarrow \Gamma$ is a homomorphism and $g\in G$, then the order of $\phi(g)$ divides the order of $g$.


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