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If $\sigma , \tau $ are two permuations that disturb no common element and $\sigma \tau = e$ , prove that $\sigma = \tau =e $

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I understand that they commute since they are disjoint, and I understand it with an example, but I just don't know how to generally prove it. – user39794 Oct 16 '12 at 21:04
Suppose they are not $e$. What happens if $\tau$ moves some element? Can $\sigma$ move it back? – EuYu Oct 16 '12 at 21:08
Try proving the converse statement: If $\sigma\tau=e$ and $\sigma\neq e$, then there is some element that both $\sigma$ and $\tau$ disturb. – alex.jordan Oct 16 '12 at 21:09
up vote 1 down vote accepted

Suppose that $\sigma(k)\ne k$. By hypothesis $\tau(k)=k$, so $(\sigma\tau)(k)=\sigma\big(\tau(k)\big)=\sigma(k)\ne k$, and $\sigma\tau\ne e$.

Added: You can even get $\tau=e$ without knowing that $\sigma$ and $\tau$ commute. Suppose that $\tau(k)\ne k$; say $\tau(k)=\ell$. Then clearly $\tau(\ell)\ne\ell$, so $\sigma(\ell)=\ell$. Thus, $(\sigma\tau)(k)=\sigma\big(\tau(k)\big)=\sigma(\ell)=\ell\ne k$, and again $\sigma\tau\ne e$.

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Suppose that $\sigma \neq e$ then $\sigma$ disturbs some element $a$ sending it to $b$ and $\tau$ sends $b$ to $a$, which it cannot do without moving $a$.

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For any $x$, assume for the sake of contradiction that $\tau x \ne x$. Since $\tau$ disturbs $x$, $\sigma$ does not, so $\sigma \tau x \ne x$. This contradicts the claim that $\sigma \tau = e$. So $\tau = e$, and $\sigma \tau = e$ so $\sigma = e$.

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