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PERMUTATION GROUP – MCQ – LEVEL-2

PERMUTATION GROUP – MCQ – LEVEL-2

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Question 1 of 10

1. Question
Let S₄ be the group of all permutations with the identity element e. Then the number of elements is S₄ that satisfy the equation x² = e is?
- 10
- 18
- 23
- 24
Correct

Incorrect
Question 2 of 10

2. Question
Which of the following statement is true?
- $S_{3}$ is an Abelian group of order 6
- The set of odd permutations in $S_{n}$ forms a subgroup of $S_{n}$
- Permutation with odd length is an odd permutation
- Alternating group $A_{4}$ has 8 elements of order 3
Correct

Incorrect
Question 3 of 10

3. Question
Let S₅ denote the symmetric group of all permutations of the five symbols {1, 2, 3, 4, 5}. The highest possible order of an element in this group in?
- 5
- 6
- 10
- 15
Correct

Incorrect
Question 4 of 10

4. Question
Number of subgroups of order 2 in S₄ are?
- 6
- 9
- 12
- 1
Correct

Incorrect
Question 5 of 10

5. Question
As the product of disjoint cycle the permutation (1 4 3 8 7) (5 4 8) in S₈ can be written as?
- (4 1 8) (5 3 2)
- (5 3 8) (4 7 1)
- (1 2 3) (4 5 6)
- (2 5 6) (5 1 4)
Correct

Incorrect
Question 6 of 10

6. Question
Given, permutation $A = \left( {\begin{array}{ccccccccccccccc}1&2&3&4\\1&3&4&2\end{array}} \right)$ is S₄ we have ?
- $A^{3} = I$
- $A^{2} = A^{3}$
- $A = A^{2}$
- $A^{2} = I$
Correct

Incorrect
Question 7 of 10

7. Question
Statement (S₁) : A₄ has a subgroup of order 2.
Statement (S₂) : A₄ has a subgroup of order 3.
- Both $S_1$ and $S_2$ are correct
- $S_1$ is correct but not $S_2$
- $S_2$ is correct but not $S_1$
- Neither $S_1$ nor $S_2$ is correct
Correct

Incorrect
Question 8 of 10

8. Question
Which of the following is correct?
- Let $G = S_{n}$ be the permutation group of n symbols, then for all $f, g \in G,~~ (f \cdot g)^2 = f^2 \cdot g^2$
- Let $G = S_{3}$ let $A=\{ f \in G:{f^3}=\mathrm{identity}$ . Then A, is a subgroup of G.
- Let $G = S_{4}$ let $B = \{ \sigma \in {S_4}:{\sigma ^4} = \mathrm{identity \ but} {\sigma ^2} \ne e\}$ Then, B contains exactly 7 elements.
- None of the above.
Correct

Incorrect
Question 9 of 10

9. Question
If $f = \left( {\begin{array}{ccccccccccccccc}1&2&3&4&5\\1&5&3&2&4\end{array}} \right)$ and $g = \left( {\begin{array}{ccccccccccccccc}1&2&3&4&5\\2&3&1&5&4\end{array}} \right)$ Then,
- ${f^{ - 1}} \circ \;{g^{ - 1}} = \left( {\begin{array}{ccccccccccccccc}1&2&3&4&5\\3&5&2&4&1\end{array}} \right)$
- ${g^{ - 1}} \circ \;{f^{ - 1}} = \left( {\begin{array}{ccccccccccccccc}1&2&3&4&5\\3&5&2&4&1\end{array}} \right)$
- ${f^{ - 1}} \circ \;{g^{ - 1}} = {g^{ - 1}} \circ \;{f^{ - 1}}$
- None of the these
Correct

Incorrect
Question 10 of 10

10. Question
Given, the permutation $\left( {\begin{array}{ccccccccccccccc}1&2&3&4\\1&3&4&2\end{array}} \right),$
- The inverse of the permutation is $\left( {\begin{array}{ccccccccccccccc}1&3&4&2\\1&2&3&4\end{array}} \right)$
- The inverse of the permutation is $\left( {\begin{array}{ccccccccccccccc}1&2&3&4\\1&2&3&4\end{array}} \right)$
- The inverse of the permutation does not exist
- None of the above
Correct

Incorrect

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