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1

1

$begingroup$

Let $(M,.)$ be an associative operation such that there are two natural numbers $m,ninmathbbN$, different from zero, such that $m$ is bigger or equal to $n$ and $x^my^n=yx$, for any $x,yinmathbbM$. Show that $x^mn=x^m$ for any $x,yin M$ and an invertible element from the set commutes with every element from the set. I observed that $x^m+n=x^2$, but I think I should show that $M$ is finite to accomplish the task.

abstract-algebra semigroups

share|cite|improve this question

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

asked Mar 15 at 17:54

user651692

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that for $ainM$, $-ainM$ but as $M$ does not have every element of it invertible, it is not a group.
  $endgroup$
  – user651692
  Mar 15 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $M$ does not have to be finite. For example, the assumption is true about addition of polynomials modulo $73$ with $m=n=74$.
  $endgroup$
  – Henning Makholm
  Mar 15 at 18:21
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let $(M,.)$ be an associative operation such that there are two natural numbers $m,ninmathbbN$, different from zero, such that $m$ is bigger or equal to $n$ and $x^my^n=yx$, for any $x,yinmathbbM$. Show that $x^mn=x^m$ for any $x,yin M$ and an invertible element from the set commutes with every element from the set. I observed that $x^m+n=x^2$, but I think I should show that $M$ is finite to accomplish the task.

abstract-algebra semigroups

share|cite|improve this question

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

asked Mar 15 at 17:54

user651692

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that for $ainM$, $-ainM$ but as $M$ does not have every element of it invertible, it is not a group.
  $endgroup$
  – user651692
  Mar 15 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $M$ does not have to be finite. For example, the assumption is true about addition of polynomials modulo $73$ with $m=n=74$.
  $endgroup$
  – Henning Makholm
  Mar 15 at 18:21
  

  
  
  
  

add a comment | 

$begingroup$

Let $(M,.)$ be an associative operation such that there are two natural numbers $m,ninmathbbN$, different from zero, such that $m$ is bigger or equal to $n$ and $x^my^n=yx$, for any $x,yinmathbbM$. Show that $x^mn=x^m$ for any $x,yin M$ and an invertible element from the set commutes with every element from the set. I observed that $x^m+n=x^2$, but I think I should show that $M$ is finite to accomplish the task.

abstract-algebra semigroups

share|cite|improve this question

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

asked Mar 15 at 17:54

user651692

$endgroup$

share|cite|improve this question

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

asked Mar 15 at 17:54

user651692

share|cite|improve this question

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

asked Mar 15 at 17:54

user651692

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

edited Mar 16 at 9:41

J.-E. Pin

18.6k21754

2 Answers
2

active

oldest

votes

1

$begingroup$

For $n=1$ it's obvious.

But for $n>1$ we have the following reasoning for all $yinmathbb M$.

Let $x=y^n-1$.

Thus, $$y^mn-my^n=y^n$$ or
$$y^mn-m+n=y^n.$$
Now, if $m=n$, we are done, but for $m>n$ we obtain:
$$y^mn-m+ny^m-n=y^ny^m-n$$ or
$$y^mn=y^m.$$
Now, let $a$ be invertible.

Thus, $$a^mx^n=xa$$ gives
$$(xa)^mx^n=x^m+1a$$ and since $$(xa)^mx^n=xxa=x^2a,$$ we obtain
$$x^m+1a=x^2a$$ or $$x^m+1=x^2$$ for any $xinmathbb M,$ which gives $$a^m+1=a^2$$ or
$$a^m-1=e$$ and since $$a^mx^n=xa,$$ we obtain
$$ax^n=xa$$ and we can assume that $n>1$, otherwise our statement is proven.

Thus, $$a^n+1=a^2,$$ which gives $$a^n-1=e$$ and from here $$a^gcd(m-1,n-1)=e.$$
Now, $$x^ma^n=ax$$ gives $$x^ma=ax$$ or
$$x^m=axa^-1,$$ which gives
$$x^mn=ax^na^-1$$ or $$x^m=ax^na^-1$$ or
$$x^ma=ax^n.$$
Id est, $$ax^n=ax,$$ which gives $x^n=x$ for any $xinmathbb M$ and since
$$a^mx^n=xa,$$ we obtain
$$ax=xa$$ and we are done!

share|cite|improve this answer

edited Mar 16 at 8:12

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! What about the second part: show that an,, invertible element form the set comutes with every element from the set"?
  $endgroup$
  – user651692
  Mar 15 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jacob Denicula I added something. See now.
  $endgroup$
  – Michael Rozenberg
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Taking $y = x^m-1$ in $x^my^n = yx$ gives $x^m(x^m-1)^n = x^m-1x$, whence $x^mn = x^m$.

If $n geqslant 2$, you could use the answer to this question to show that your semigroup is commutative. However, it is not necessarily a monoid, and in this case, there are no invertible element. Take for instance $S = a, 0$ with $aa = 0$ and $a0 = 0 = 0a$.

For $n = 1$, a semigroup satisfying the identity $x^my = yx$ is not necessarily commutative as shown in this answer: the quaternion group $Q_8$
satisfies the identity $x^3y = yx$ but is not commutative.

share|cite|improve this answer

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

$endgroup$

add a comment | 

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1

$begingroup$

For $n=1$ it's obvious.

But for $n>1$ we have the following reasoning for all $yinmathbb M$.

Let $x=y^n-1$.

Thus, $$y^mn-my^n=y^n$$ or
$$y^mn-m+n=y^n.$$
Now, if $m=n$, we are done, but for $m>n$ we obtain:
$$y^mn-m+ny^m-n=y^ny^m-n$$ or
$$y^mn=y^m.$$
Now, let $a$ be invertible.

Thus, $$a^mx^n=xa$$ gives
$$(xa)^mx^n=x^m+1a$$ and since $$(xa)^mx^n=xxa=x^2a,$$ we obtain
$$x^m+1a=x^2a$$ or $$x^m+1=x^2$$ for any $xinmathbb M,$ which gives $$a^m+1=a^2$$ or
$$a^m-1=e$$ and since $$a^mx^n=xa,$$ we obtain
$$ax^n=xa$$ and we can assume that $n>1$, otherwise our statement is proven.

Thus, $$a^n+1=a^2,$$ which gives $$a^n-1=e$$ and from here $$a^gcd(m-1,n-1)=e.$$
Now, $$x^ma^n=ax$$ gives $$x^ma=ax$$ or
$$x^m=axa^-1,$$ which gives
$$x^mn=ax^na^-1$$ or $$x^m=ax^na^-1$$ or
$$x^ma=ax^n.$$
Id est, $$ax^n=ax,$$ which gives $x^n=x$ for any $xinmathbb M$ and since
$$a^mx^n=xa,$$ we obtain
$$ax=xa$$ and we are done!

share|cite|improve this answer

edited Mar 16 at 8:12

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! What about the second part: show that an,, invertible element form the set comutes with every element from the set"?
  $endgroup$
  – user651692
  Mar 15 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jacob Denicula I added something. See now.
  $endgroup$
  – Michael Rozenberg
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

1

$begingroup$

For $n=1$ it's obvious.

But for $n>1$ we have the following reasoning for all $yinmathbb M$.

Let $x=y^n-1$.

Thus, $$y^mn-my^n=y^n$$ or
$$y^mn-m+n=y^n.$$
Now, if $m=n$, we are done, but for $m>n$ we obtain:
$$y^mn-m+ny^m-n=y^ny^m-n$$ or
$$y^mn=y^m.$$
Now, let $a$ be invertible.

Thus, $$a^mx^n=xa$$ gives
$$(xa)^mx^n=x^m+1a$$ and since $$(xa)^mx^n=xxa=x^2a,$$ we obtain
$$x^m+1a=x^2a$$ or $$x^m+1=x^2$$ for any $xinmathbb M,$ which gives $$a^m+1=a^2$$ or
$$a^m-1=e$$ and since $$a^mx^n=xa,$$ we obtain
$$ax^n=xa$$ and we can assume that $n>1$, otherwise our statement is proven.

Thus, $$a^n+1=a^2,$$ which gives $$a^n-1=e$$ and from here $$a^gcd(m-1,n-1)=e.$$
Now, $$x^ma^n=ax$$ gives $$x^ma=ax$$ or
$$x^m=axa^-1,$$ which gives
$$x^mn=ax^na^-1$$ or $$x^m=ax^na^-1$$ or
$$x^ma=ax^n.$$
Id est, $$ax^n=ax,$$ which gives $x^n=x$ for any $xinmathbb M$ and since
$$a^mx^n=xa,$$ we obtain
$$ax=xa$$ and we are done!

share|cite|improve this answer

edited Mar 16 at 8:12

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! What about the second part: show that an,, invertible element form the set comutes with every element from the set"?
  $endgroup$
  – user651692
  Mar 15 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jacob Denicula I added something. See now.
  $endgroup$
  – Michael Rozenberg
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

$begingroup$

For $n=1$ it's obvious.

But for $n>1$ we have the following reasoning for all $yinmathbb M$.

Let $x=y^n-1$.

Thus, $$y^mn-my^n=y^n$$ or
$$y^mn-m+n=y^n.$$
Now, if $m=n$, we are done, but for $m>n$ we obtain:
$$y^mn-m+ny^m-n=y^ny^m-n$$ or
$$y^mn=y^m.$$
Now, let $a$ be invertible.

Thus, $$a^mx^n=xa$$ gives
$$(xa)^mx^n=x^m+1a$$ and since $$(xa)^mx^n=xxa=x^2a,$$ we obtain
$$x^m+1a=x^2a$$ or $$x^m+1=x^2$$ for any $xinmathbb M,$ which gives $$a^m+1=a^2$$ or
$$a^m-1=e$$ and since $$a^mx^n=xa,$$ we obtain
$$ax^n=xa$$ and we can assume that $n>1$, otherwise our statement is proven.

Thus, $$a^n+1=a^2,$$ which gives $$a^n-1=e$$ and from here $$a^gcd(m-1,n-1)=e.$$
Now, $$x^ma^n=ax$$ gives $$x^ma=ax$$ or
$$x^m=axa^-1,$$ which gives
$$x^mn=ax^na^-1$$ or $$x^m=ax^na^-1$$ or
$$x^ma=ax^n.$$
Id est, $$ax^n=ax,$$ which gives $x^n=x$ for any $xinmathbb M$ and since
$$a^mx^n=xa,$$ we obtain
$$ax=xa$$ and we are done!

share|cite|improve this answer

edited Mar 16 at 8:12

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

share|cite|improve this answer

edited Mar 16 at 8:12

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

answered Mar 15 at 18:16

Michael RozenbergMichael Rozenberg

109k1895200

0

$begingroup$

Taking $y = x^m-1$ in $x^my^n = yx$ gives $x^m(x^m-1)^n = x^m-1x$, whence $x^mn = x^m$.

If $n geqslant 2$, you could use the answer to this question to show that your semigroup is commutative. However, it is not necessarily a monoid, and in this case, there are no invertible element. Take for instance $S = a, 0$ with $aa = 0$ and $a0 = 0 = 0a$.

For $n = 1$, a semigroup satisfying the identity $x^my = yx$ is not necessarily commutative as shown in this answer: the quaternion group $Q_8$
satisfies the identity $x^3y = yx$ but is not commutative.

share|cite|improve this answer

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

$endgroup$

add a comment | 

0

$begingroup$

Taking $y = x^m-1$ in $x^my^n = yx$ gives $x^m(x^m-1)^n = x^m-1x$, whence $x^mn = x^m$.

If $n geqslant 2$, you could use the answer to this question to show that your semigroup is commutative. However, it is not necessarily a monoid, and in this case, there are no invertible element. Take for instance $S = a, 0$ with $aa = 0$ and $a0 = 0 = 0a$.

For $n = 1$, a semigroup satisfying the identity $x^my = yx$ is not necessarily commutative as shown in this answer: the quaternion group $Q_8$
satisfies the identity $x^3y = yx$ but is not commutative.

share|cite|improve this answer

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

$endgroup$

add a comment | 

$begingroup$

Taking $y = x^m-1$ in $x^my^n = yx$ gives $x^m(x^m-1)^n = x^m-1x$, whence $x^mn = x^m$.

If $n geqslant 2$, you could use the answer to this question to show that your semigroup is commutative. However, it is not necessarily a monoid, and in this case, there are no invertible element. Take for instance $S = a, 0$ with $aa = 0$ and $a0 = 0 = 0a$.

For $n = 1$, a semigroup satisfying the identity $x^my = yx$ is not necessarily commutative as shown in this answer: the quaternion group $Q_8$
satisfies the identity $x^3y = yx$ but is not commutative.

share|cite|improve this answer

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

$endgroup$

share|cite|improve this answer

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754

answered Mar 16 at 10:06

J.-E. PinJ.-E. Pin

18.6k21754