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Consider the set $\Sigma^*$ of all strings over the alphabet $\Sigma = \{0, 1\}$. $\Sigma^*$ with the concatenation operator for strings

  1. does not form a group
  2. forms a non-commutative group
  3. does not have a right identity element
  4. forms a group if the empty string is removed from $\Sigma^*$
asked in Set Theory & Algebra by Veteran (59.6k points) | 1.6k views

1 Answer

+32 votes
Best answer
Identity element for concatenation is empty string $\epsilon$. Now, we cannot concatenate any string with a given string to get empty string $\implies$ there is no inverse for string concatenation. Only other 3 group properties -- closure, associative and existence of identity -- are satisfied and hence, ans should be (a).
answered by Active (2.4k points)
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+1
U are right
0
why not inverse?
+6
Identity element for concatenation is empty string $\epsilon$. Now, we cannot concatenate any string with a given string to get empty string $\implies$ there is no inverse for string concatenation.
+2
ok. similar to division by zero. not defined. illogical. doesnt make sense.

 

thanks.
+4
think about concatinating  string with some any other string to get null string...you will not find any such string so there is no inverse thats it
0
difference between right identity and left identity element ?

There exists identity element e such that,

                1) a*e = a for all a belongs to set    (right identity)

                2) e*a = a for all a belongs to set     (left identity)

Is this correct ???
0
Can we say Commutative property also fails here?
0

@  Bhagirathi

your answer always crystal clear... :) it proves me very helpful..

0
of course Anil

$ab \neq ba$

$ab$ is string that start with a and end with b , and have length 2 while $ba$ is a string that starts with b and ends with a and have length 2 , clearly both are different strings.
0
Identity element for concatenation is empty string ϵ and that is not in group so can we say that there is no identity element in set so it is not even a monoid?
0
nice logic @ mgrwt
0
@Mk Utkarsh-No your identity element is present in the set $\sum^*$ as

$\sum^*=(0+1)^*$
Answer:

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