(d) is True.

For option (a), even I am not sure. If somebody knows, please comment regarding option (a)

26 votes

Which one of the following is false?

- The set of all bijective functions on a finite set forms a group under function composition.
- The set $\{1, 2, \dots p-1\}$ forms a group under multiplication mod $p$, where $p$ is a prime number.
- The set of all strings over a finite alphabet forms a group under concatenation.
- A subset $S \neq \emptyset$ of $G$ is a subgroup of the group $\langle G, * \rangle$ if and only if for any pair of elements $a, b \in S, a * b^{-1} \in S$.

0

option (b) and (c) are false. For (b) you can check manually by taking p=5 or whatever you like. And for option (c), there does not exist inverse for any given string because you can't concatenate two strings to form 'null' i.e identity element of the group.

(d) is True.

For option (a), even I am not sure. If somebody knows, please comment regarding option (a)

(d) is True.

For option (a), even I am not sure. If somebody knows, please comment regarding option (a)

30 votes

Best answer

$(a)$ Let set = ${1, 2, 3, 4}$

We can have identity function as $\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ) \right \}$

Since function is bijective and mapping to same set, we can have an inverse for any function by inverting the relation (changing the mapping $a \to b\text{ to } b \to a$)

Since the function maps to the same set, it must be closed and associative also. So, all four properties of group satisfied. So, $(a)$ is true.

$(b)$ Let $p = 5$. So, set $=$ $\left \{ 1,2,3,4 \right \}$

Identity element is $1$.

$$ \begin{array}{|c|c|c|c|c|}\hline *& 1 &2 &3& 4\\\hline 1& 1& 2& 3& 4\\\hline 2& 2& 4& 1& 3\\\hline 3& 3& 1& 4& 2\\\hline 4& 4& 3& 2& 1\\\hline \end{array}$$This forms a group. Similarly for any p, we get a group. So, $(b)$ is also true.

$(c)$ is false as string concatenation operation is a monoid (doesn't have inverse to become a group).

http://en.wikipedia.org/wiki/Concatenation

$(d)$ is True.

http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/32.html

8

For option '**a**', just giving a more explanatory proof, consider

Let be any other function such that .

Now and

Hence , hence stands as identity element.

For inverse, it is much obvious, since we are taking all bijective functions on a finite so for we must have a function and so

Associativity and closure is also obvious.

0

i agree, option (c) is false,

but explain option (d) here, is it **Proposition 3.2.2 **

**in the source link?**

12

yes. Actually its basic group definition. For any element $b$, $b^{-1}$ must be there (inverse property). Now $a$ and $b^{-1}$ means $a*b^{-1}$ must be there (closure property).

0

**hence **** stands**** as identity element**

what is ??

if i have 2 sets A{1,2,3}->B{a,b,c} with bijection function {<1,a><2,b><3,c>} then what will be the identity element?

0

For d option, it is not saying that s is subset of G,if my group is such that the given condition is followed but it is not subset of G,so how can it be subgroup?

0

Arjun Sir, how you can say about option (a) by taking a mapping of a function but not a composition of a function.sir can u plz give the generalize answer.

1 vote

$A.$ **TRUE**

Define a set $S_{3}=$ Set of all bijections from $\{1,2,3\}$ to $\{1,2,3\}$

$f_{1}: 1\rightarrow 1, 2\rightarrow 2,3\rightarrow 3$

$f_{2}: 1\rightarrow 2, 2\rightarrow 1,3\rightarrow 3$

$f_{3}: 1\rightarrow 3, 2\rightarrow 2,3\rightarrow 1$

$f_{4}: 1\rightarrow 1, 2\rightarrow 3,3\rightarrow 2$

$f_{5}: 1\rightarrow 2, 2\rightarrow 3,3\rightarrow 1$

$f_{6}: 1\rightarrow 3, 2\rightarrow 1,3\rightarrow 2$

$S_{3}=\{f_{1},f_{2},f_{3},f_{4},f_{5},f_{6}\}$

Consider the **composition** of functions on $S_{3}$

**Few Ex:** $f_{2}(f_{5})=f_{4}$ $f_{5}(f_{2})=f_{3}$

**Now,**

1. $S_{3}$ is closed under composition

2. There is an identity element of $S_{3}:f_{1}$

3. Is there an inverse for every element of $S_{3}?$ **Yes!!**

$f_{2}(f_{2})=f_{1},f_{3}(f_{3})=f_{1},f_{4}(f_{4})=f_{1},f_{5}(f_{6})=f_{1}$

4. Composition of functions is associative.

$B.$ **TRUE**

Go through the Best answer.

$C.$ **FALSE**

Inverse is not possible $\times$

$D.$ **TRUE**

First assume that H is a subgroup of G. We wish to show that $gh^ {−1} ∈ H$ whenever g and h are in H. Since h is in H, its inverse $h^{−1}$ must also be in H. Because of the closure of the group operation, $gh^{−1} ∈ H.$