Mixing
A question about a semigroup with two elements
$begingroup$
Let $({p,q},∗)$ be a semigroup, where $p∗p=q$. I want to show that:
$$q∗q=q$$
I know semigroups are closed and associative, but I am not able to prove this equality.
abstract-algebra semigroups idempotents finite-semigroups
edited Jan 20 at 19:46
Yanior Weg
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
$endgroup$
add a comment |
$begingroup$
Let $({p,q},∗)$ be a semigroup, where $p∗p=q$. I want to show that:
$$q∗q=q$$
I know semigroups are closed and associative, but I am not able to prove this equality.
abstract-algebra semigroups idempotents finite-semigroups
edited Jan 20 at 19:46
Yanior Weg
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
$endgroup$
$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01
add a comment |
$begingroup$
Let $({p,q},∗)$ be a semigroup, where $p∗p=q$. I want to show that:
$$q∗q=q$$
I know semigroups are closed and associative, but I am not able to prove this equality.
abstract-algebra semigroups idempotents finite-semigroups
edited Jan 20 at 19:46
Yanior Weg
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
$endgroup$
Let $({p,q},∗)$ be a semigroup, where $p∗p=q$. I want to show that:
$$q∗q=q$$
I know semigroups are closed and associative, but I am not able to prove this equality.
abstract-algebra semigroups idempotents finite-semigroups
abstract-algebra semigroups idempotents finite-semigroups
edited Jan 20 at 19:46
Yanior Weg
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
edited Jan 20 at 19:46
Yanior Weg
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
edited Jan 20 at 19:46
Yanior Weg
2,31911144
edited Jan 20 at 19:46
Yanior Weg
2,31911144
edited Jan 20 at 19:46
Yanior Weg
2,31911144
2,31911144
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
asked Nov 8 '17 at 14:49
rahul sharmarahul sharma
7418
7418
$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01
add a comment |
$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01
$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Suppose to the contrary that $q*q=p$.
Case 1: $p*q=p$. Try to derive a contradiction thinking about $p*p*q$.
Case 2: $p*q=q$. Same advice as case 1.
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
$endgroup$
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Suppose to the contrary that $q*q=p$.
Case 1: $p*q=p$. Try to derive a contradiction thinking about $p*p*q$.
Case 2: $p*q=q$. Same advice as case 1.
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
$endgroup$
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
add a comment |
$begingroup$
Hint:
Suppose to the contrary that $q*q=p$.
Case 1: $p*q=p$. Try to derive a contradiction thinking about $p*p*q$.
Case 2: $p*q=q$. Same advice as case 1.
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
$endgroup$
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
add a comment |
$begingroup$
Hint:
Suppose to the contrary that $q*q=p$.
Case 1: $p*q=p$. Try to derive a contradiction thinking about $p*p*q$.
Case 2: $p*q=q$. Same advice as case 1.
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
$endgroup$
Hint:
Suppose to the contrary that $q*q=p$.
Case 1: $p*q=p$. Try to derive a contradiction thinking about $p*p*q$.
Case 2: $p*q=q$. Same advice as case 1.
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
answered Nov 8 '17 at 15:08
paw88789paw88789
29.3k12349
29.3k12349
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
add a comment |
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
$begingroup$
yes you are right. I did not think that way.Thanks
$endgroup$
– rahul sharma
Nov 8 '17 at 15:21
add a comment |
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$begingroup$
@Teddy38, your forgot to put backslashes in front of the OP's curly braces. (I can't fix this because it's only a two-character edit.)
$endgroup$
– PersonX
Nov 8 '17 at 14:58
$begingroup$
@PersonX Cheers.
$endgroup$
– Teddy38
Nov 8 '17 at 15:01