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

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1

$begingroup$

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 $iin X$. 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.

group-theory symmetric-groups

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asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

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1

$begingroup$

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 $iin X$. 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.

group-theory symmetric-groups

share|cite|improve this question

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

$endgroup$

add a comment | 

$begingroup$

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 $iin X$. 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.

group-theory symmetric-groups

share|cite|improve this question

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

$endgroup$

share|cite|improve this question

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

share|cite|improve this question

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

asked yesterday

AnalyticHarmonyAnalyticHarmony

657313

2 Answers
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1

$begingroup$

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 $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.

share|cite|improve this answer

answered yesterday

Mike EarnestMike Earnest

24.3k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoter, care to explain?
  $endgroup$
  – Mike Earnest
  yesterday
  

  
  
  
  

add a comment | 

3

$begingroup$

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 $A$.

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answered yesterday

gammagamma

3,39221342

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1

$begingroup$

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 $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.

share|cite|improve this answer

answered yesterday

Mike EarnestMike Earnest

24.3k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoter, care to explain?
  $endgroup$
  – Mike Earnest
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

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 $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.

share|cite|improve this answer

answered yesterday

Mike EarnestMike Earnest

24.3k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoter, care to explain?
  $endgroup$
  – Mike Earnest
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

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 $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.

share|cite|improve this answer

answered yesterday

Mike EarnestMike Earnest

24.3k22151

$endgroup$

share|cite|improve this answer

answered yesterday

Mike EarnestMike Earnest

24.3k22151

answered yesterday

Mike EarnestMike Earnest

24.3k22151

answered yesterday

Mike EarnestMike Earnest

24.3k22151

3

$begingroup$

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 $A$.

share|cite|improve this answer

answered yesterday

gammagamma

3,39221342

$endgroup$

add a comment | 

3

$begingroup$

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 $A$.

share|cite|improve this answer

answered yesterday

gammagamma

3,39221342

$endgroup$

add a comment | 

$begingroup$

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 $A$.

share|cite|improve this answer

answered yesterday

gammagamma

3,39221342

$endgroup$

share|cite|improve this answer

answered yesterday

gammagamma

3,39221342

answered yesterday

gammagamma

3,39221342

answered yesterday

gammagamma

3,39221342