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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 $A$, $a*b*c=a*c$
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3 Answers

Best answer
33 votes
33 votes
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 $(a * b)*a = c \\ \implies ((a * b) * a) * a = c * a \\ \implies (a * b) * (a * a) = c * a \\ \implies (a * b) * a = c * a \\ \implies c = c * a \qquad \to (1)$

Similarly, $a * (a * b * a ) = a * c \\ \implies a * (a * (b * a)) = a * c  \\ \implies (a * a) * (b * a) = a * c\\ \implies a * (b * a) = a * c  \\ \implies (a * b) * a = a * c  \\ \implies c = a * c \qquad \to (2).$

From $(1)$ and $(2)$ we get $c * a = a*c \implies c = a$.

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, $a*b*c = a * c.$
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16 votes
16 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
7 votes
7 votes
(a) Let $a*a \neq a$.

So $a*a=b \text{ (say)}$, where $a\neq b$

$$a*(a*a)=a*b$$

$$\implies (a*a)*a=a*b$$

$$\implies b*a=a*b$$

So we reach at a contraction of $a\neq b \implies b*a\neq a*b$ given in question.  Hence $a*a=a$ is true.

(b) Let $a*b*a\neq a$

Then by the problem statement, $(a*b*a)*a\neq a*(a*b*a) \tag 1$

$$\text{LHS of }(1) = (a*b*a)*a$$ $$ = (a*b)*(a*a) \text{ [Associative property]}$$

$$=a*b*a \text{ [since $a^2=a$]}\tag 2$$

$$\text{RHS of }(1) = a*(a*b*a)$$ $$= (a*a)*(b*a) \text{ [Associative property]}$$

$$=a*b*a \text{ [since $a^2=a$]} \tag 3$$

From $(2)$ and $(3)$ we see that the LHS and RHS of $(1)$ are equal and hence we reach at a contradiction and out assumption $a*b*a\neq a$ was false.

(c) Let $a*b*c\neq a*c$

Then by the problem statement, $(a*b*c)*(a*c)\neq (a*c)*(a*b*c) \tag 4$

$$\text{LHS of }(4) = (a*b*c)*(a*c)$$ $$ = a*b*(c*a*c) \text{ [Associative property]}$$

$$=a*b*c \text{ [since $c*a*c=c$]}\tag 5$$

$$\text{RHS of }(4) = (a*c)*(a*b*c)$$ $$= (a*c*a)*b*c\text{ [Associative property]}$$

$$=a*b*c \text{ [since $a*c*a=a$]} \tag 6$$

From $(5)$ and $(6)$ we see that the LHS and RHS of $(4)$ are equal and hence we reach at a contradiction and out assumption $a*b*c\neq a*c$ was false.

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