If $\alpha$ and $\beta$ are disjoint, so are $\alpha$ and $\beta^k$?

Question

If $\alpha$ and $\beta$ are disjoint permutations, so are $\alpha$ and $\beta^k$.

I'm working on the group of permutations of a finite set $X$ with composition. The power is defined as usual using composition.

I don't know if it's true, I just found this while proving another thing. Disjoint meaning that if $\alpha(x)=x$ then $\beta(x)\ne x$ and vice versa.

I'm trying by induction. The base case is direct. Assuming that $\beta$ and $\alpha^{k-1}$ are disjoint, let $i\in X$. If $\beta(i)\ne i$ then $\alpha^{k-1}(i)=i$ and $\alpha(i)=i$, so aplying $\alpha$ in the first equality we get $\alpha^k(i) = i$.

But I'm a bit stuck in the case $\beta(i)=i$. We would have $\alpha^{k-1}(i)\ne i$ and $\alpha(i) \ne i$. Assuming that $\alpha^k(i) = i$ was the case then $\alpha^{k-1}(\alpha(i)) = i$ then $\beta(\alpha(i)) = \alpha(i)$ but this seems to give me nothing.

Any help will be appreciated.

Answer 1

The statement to be proved is untrue under your definition of disjoint:

Your Defintion (Disjoint): $\alpha$ and $\beta$ are disjoint if $\alpha(x)=x$ implies $\beta(x)\neq x$ and vice versa.

To see this, consider $\alpha=(1\;2)(3)(4)$ and $\beta=(1)(2)(3\;4)$, $k=2$.

You get a correct result is you instead use

Defintion (Disjoint): $\alpha$ and $\beta$ are disjoint if $\alpha(x)\neq x$ implies $\beta(x)= x$ and vice versa.

To see this, suppose $\alpha(x)\neq x$. Then disjointness implies $\beta(x)=x$, so $\beta^k(x)=\beta^{k-1}(x)=\dots=\beta(x)=x$. On the other hand, if $\beta^k(x)\neq x$, then it must be the case that $\beta(x)\neq x$ (since $\beta(x)=x\implies \beta^k(x)=x$ as shown before), so disjointness implies $\alpha(x)=x$. This proves $\alpha$ and $\beta^k$ are disjoint.

Answer 2

The set of elements that is moved by $\beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $\alpha$ (let's call it $A$) so if you apply $\beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.