If $\alpha, \beta$ are disjoint permutations and $\alpha\beta=1$, then $\alpha=\beta=1$.

Question

I have just begun the study of group theory.

This is a question from Rotman's "An introduction to the theory of groups", Ch1, Ex 1.9. We already know by now that if $\alpha, \beta$ are two disjoint permutations, then they commute, i.e. $\alpha\beta=\beta\alpha$. Now we move on to

If $\alpha, \beta\in S_X$ are disjoint permutations and $\alpha\beta=1$, then $\alpha=\beta=1$.

Let $x\in X$ be an element. Then by the given information, $\alpha(\beta(x))=x$. Now suppose $\beta\neq1$. Then there must be a nonempty set $Q\in X$ such that $\beta$ moves all the elements $q\in Q$.

Let $\beta(q)\neq q; \ \ \beta(q_1)=\ell$ where $\ell\in X$. Then, $\beta$ MUST also move $\ell$, because otherwise $$\beta(q_1)=\beta(\ell)=\ell$$ would mean that $\beta$ is many-one, which is a contradiction to the fact that $\beta$ is a permutation. Thus, by the criteria of disjointness, $\alpha$ MUST fix $\ell$. Thus, $$\alpha\beta(q_1)=\alpha(\ell)=\ell\neq q_1$$ which is a contradiction. Thus $Q$ must be empty. Thus, $\beta=1$.

Now, from the given, $\alpha\beta=1$ and $\beta=1$ means that $\alpha 1=1\implies \alpha=1$. Hence proved.

Being a beginner, I am not getting the confidence for the proof I came up with. Could someone please verify the solution?

EDIT: I thought about this problem and found another, shorter solution. Please comment:

If $\alpha\beta(x)=x$ then if $\beta$ fixes $k$, then $\alpha(\beta(k))=k\implies \alpha(k)=k.$ So $\alpha$ MUST fix everything $\beta$ fixes. And vice-versa.

Now suppose that $\beta$ moves $u$. Then, $\alpha$ fixes $u$. Then, $\beta$ must FIX $u$. This is a contradiction. Thus $\beta$ cannot move anything and hence it is the identity. Similar arguments can be made for $\alpha$.

Answer 1

Let $X$ be any non empty set.

Let $\sigma \in S_X$ and define $$\textrm{supp}{(\sigma)}=\{x\in X: \sigma(x)\neq x\}$$

$\sigma, \tau\in S_X$ are said to be disjoint if $$\textrm{supp}{(\sigma)}\cap \textrm{supp}{(\tau)}=\emptyset$$

Claim:

1. $\sigma\tau=\tau\sigma $ if $\sigma, \tau$ are disjoint.

2. $\textrm{supp}{(\sigma)}=\textrm{supp}{(\sigma^{-1})}$

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Given $\sigma\tau=\tau\sigma=1$ implies $\tau=\sigma^{-1}$

Also $\sigma, \tau(=\sigma^{-1})$ disjoint implies $\textrm{supp}{(\sigma)}\cap\textrm{supp}{(\sigma^{-1})}=\emptyset$

implies $\textrm{supp}{(\sigma)}=\emptyset$

implies $\textrm{Fix}(\sigma) =\{x\in X:\sigma(x)=x\}=X$

Implies $\sigma=1$