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Let $(A, *)$ be a semigroup, Furthermore, for every $a$ and $b$ in $A$, if $a \neq b$, then $a*b \neq b*a$.

  1. Show that for every $a$ in $A$, $a*a=a$
  2. Show that for every $a$, $b$ in $A$, $a*b*a=a$
  3. Show that for every $a,b,c$ in $A$, $a*b*c=a*c$
asked in Set Theory & Algebra by Veteran (59.7k points)
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2 Answers

+13 votes
Best answer

a. Let $a * a = b$. $(a * a) * a = b*a$. Since $(A, *)$ is a semigroup, $*$ is closed and associative. So, $(a *a) * a = a * (a*a) \implies a * b = b * a$, which is possible only if $a = b$. Thus we proved $a * a = a.$

b. Let $(a * b)*a = c \\ \implies (a * b) * a * a = c * a \\ \implies a * b * a = c * a \\ \implies c * a = a.$

Similarly, $a * (a * b * a ) = a * c \\ \implies a * a * (b * a) = a * c \\ \implies a * (b * a) = a * c \\ \implies a * c = a = c * a.$

So, $c = a$.

c.  Let $(a * b)*c = d. \\ \implies (a * b) * c * c = d * c \\ \implies a * b * c = d * c \\ \implies d * c = d.$

Similarly, $a * (a * b * c ) = a * d \\ \implies a * a * (b * c) = a * d \\ \implies a * (b * c) = a * d \\ \implies a * d = d.$

Thus $d * c = a * d = d $

Now $c * d *c = c * a * d = c * d \\\implies c = c * a * d = c * d$

and 

$d * c * a = a * d * a = d * a \\\implies d * c * a = a = d * a$

So,

$a * c =   (d*a)*(c*d) \\= d*(a*c)* d = d.$

Thus, $a*b*c = a * c.$

answered by Veteran (368k points)
+2
Suppose this type of question comes in exam

how to solve means how to start initially ???..

I am finding it difficult...:(
+3

Watch nptel video lectures on mathematics by prof Kamala krithivasan.If you are weak only in group theory just watch lectures 35,36 and 37.

+3

a∗b∗a=c∗a⟹c∗a=a????????????????

+3
I think for b) the proof should be as follows:

$Let, a*b*a=c\\ \bullet (a*b*a)*a = c*a\\\implies (a*b)*a*a =c*a\\\implies (a*b)*a =c*a\\\implies c=c*a ----\ (1)\\\bullet a*(a*b*a) = a*c\\\implies (a*b*a)=a*c\\\implies c=a*c----\ (2)\\ \text{from (1) and (2): }a*c=c*a\\\therefore c=a \implies a*b*c = a$
+1

@krish

yes right :)

in solution c   given in best answer                    

Thus d∗c = a∗d  =d

Now c∗d∗c = c∗a∗d = c∗d

⟹c= c∗a∗d= c∗d ( here how the value c*d*c in previous step become c ? )

+1
My proof would be:
Consider $a*b*c = d$
$d*(a*c) = a*b*(c*a*c) = a*b*c$
$(a*c)*d=a*c*(a*b*c)=(a*c*a)*b*a=a*b*c$
So, $d*(a*c)=(a*c)*d\implies d=a*c=a*b*c$
0
@Arjun sir

how in solution c

$c * d *c = c * a * d = c * d \\\implies c = c * a * d = c * d$
0 votes
a. Lets assume a*a!=a

So, if our assumption is true then, (a*a)*a !=a*(a*a)  [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

but , (a*a)*a=a*(a*a) because (A,∗) is a semigroup hence associative, which proves that our initial assumption is wrong,

hence a*a=a

b.Lets assume a∗b∗a!=a

So, if our assumption is true then , (a*b*a)*a!=a*(a*b*a)   [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

Now,(a*b*a)*a= a*b*(a*a) [since (A,*) is associative]

         or, (a*b*a)*a=a*b*a  [since a*a=a (already proved) ]------------------ (i)

Now, a*(a*b*a)=(a*a)*b*a  [since (A,*) is associative]

         or, a*(a*b*a)=a*b*a   [since a*a=a (already proved) ]-------------------(ii)

using (i) and (ii) (a*b*a)*a= a*(a*b*a) , hence proved that our initial assumption was wrong hence,a∗b∗a=a

c.  Lets assume a∗b∗c !=a∗c

So, if our assumption is true then , (a*b*c)*(a*c)!=(a*c)*(a*b*c)   [Since it is given that for every a,b in A, if a≠b, then a∗b≠b∗a ]

Now,(a*b*c)*(a*c)= a*b*(c*a*c) [since (A,*) is associative]

         or, (a*b*c)*(a*c)= a*b*c  [since (c*a*c)=c ( for any a,b , a*b*a=a  already proved as part of question b) ]------------------ (i)

Now, (a*c)*(a*b*c)=(a*c*a)*b*c  [since (A,*) is associative]

         or, (a*c)*(a*b*c)=a*b*c  [since (a*c*a)=c ( for any a,b , a*b*a=a  already proved as part of question b) ]-------------------(ii)

using (i) and (ii)  (a*b*c)*(a*c)=(a*c)*(a*b*c) , hence proved that our initial assumption was wrong hence,a∗b∗c =a∗c
answered by Active (2.2k points)

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