100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART 3

100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART 3

Question 10: Let $H=\{e,(12)(34)\}$ and $K=\{e,(12)(34),(13)(24),(14)(23)\}$ be subgroup of $S_4$, where $e$ denotes the identify element of $S_4$. Then

1. $H$ and $K$ are normal subgroup of $S_4$
2. $H$ is normal in $K$ and $K$ are normal in $A_4$
3. $H$ is normal in $A_4$ but not in $S_4$
4. $H$ is normal in $S_4$ but $H$ is not

Answer: Recall

Conjugation preserve disjoint cyclic structure in permutation group, $i.e.$, for all $\sigma, \tau \in S_n$, the disjoint cyclic decomposition of $\tau$ and $\sigma \tau \sigma^{-1}$ will be the same.

Since $K$ contain all element with disjoint cyclic decompositon of type $(ab)(cd)$, hence it will be normal in $S_4$ and it will be normal in $A_4( \leq S_4)$.
$K$ is Abelian so all it’s subgroup will be normal. $H \trianglelefteq K$.
Hence 2 is correct option.

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Question 11: Let $S_n$ denote the symmetric group on $n$ symbols. The group $S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ is isomorphic to which of the following groups?

1. $\mathbb{Z}/12\mathbb{Z}$
2. $\mathbb{Z}/6\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z}$
3. $A_4$, the alternating group of order $12$
4. $D_6$, the dihedral group of order $12$

Answer: Note,

• The direct sum of a group will be Abelian iff all it’s constituent groups will be Abelian.
• The direct sum contains a normal subgroup isomorphic to each of it’s constituent group.

$S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ is a non-abelian group of order $12$, hence it will be either $A_4$ or $D_6$.
Also $S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ has a normal subgroup of order $2(\cong \mathbb{Z}_2)$, since $A_4$ does not contain a normal subgroup of order $2$, hence it has to be $D_6$.
Hence, 4 is correct choice.

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Question12: Which of the following numbers can be orders of permutations $\sigma$ of $11$ symbols such that $\sigma$ does not fix any symbol?
$1.~18 \quad 2.~30 \quad 3.~15 \quad 4.~28$
Answer: The order of permutation is determined as follows

• The order of a $n-$cylce is $n$.
• The order any permutation is lcm of each lenght of individual cycle in the disjoint cyclic decomposition.

If a permutation does not fix any element of set, then it’s disjoint cyclic decomposition will contain all the elements of the set. So here we have to partition $11$, with cycle length($> 1$, as $1$ length cycle fixed the element inside it.) such that the lcm of it’s constituent will be given in the option.
Option 1: For example $2 + 9 = 11$ and $[2 , 9] = 11$, we can write a permutation consist of two disjoint cycle of lengths $2$ and $9$.
Option 2: Similarly, $5 + 6 = 11$ and $[5, 6]=30$.
Option 3: Here, $3+3+5 \neq 11$ and $[3,3,5]=15$.
Option 4: $4 + 7 = 11$ and $[4 , 7] = 28$.
Hence all are correct.

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Question 13: Let $\sigma = (1~2)(3~4~5)$ and $\tau = (1~2~3~4~5~6)$ be permutation in $S_6$, the group of permutations on six symbols. Which of the following statements are true?

1. The subgroup $\langle \sigma \rangle$ and $\langle \tau \rangle$ are isomorphic to each other.
2. $\langle \sigma \rangle$ and $\langle \tau \rangle$ are conjugate in $S_6$.
3. $\langle \sigma \rangle \cap \langle \tau \rangle$ is trivial group.
4. $\sigma$ and $\tau$ commute.

Answer: Since $|\sigma| = |\tau| = 6$, hence both will generate a group isomorphic to $\mathbb{Z}_6$.
Both $\sigma$ and $\tau$ has different disjoint cyclic structure, hence can’t be conjugate.
Also, $\langle \sigma \rangle \cap \langle \tau \rangle$ will be trivial.

Any two permutation will commute if and only if their disjoint cyclic decomposition has no elements in common.

Hence, $\sigma$ and $\tau$ will not-commute.
Hence 1 and 3 are correct choice.

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Question 14: Given the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 5 & 4 \end{pmatrix}$ the matrix $A$ is defined to be the one whose $i-$th column is the $\sigma(i)-$th column of the identity matrix $I$. Which of the following is correct?
$1.~A = A^{-2} \quad 2.~A = A^{-4} \quad 3.~A = A^{-5} \quad 4.~A = A^{-1}$
Answer: Note,

There is a natural isomorphism between $S_n$ and set of all permutation matrices of dimension $n \times n$, given by $f(\sigma) =$ the matrix whose $i-$th column is the $\sigma(i)-$th column of the identity matrix.
Then, $|\sigma| = |f(\sigma)|$

Also, $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 5 & 4 \end{pmatrix} = (132)(45)$,
Hence $|\sigma| =[2,3] = 6$, $i.e.$, A^6 = I \Rightarrow A = A^{-5}

.
Hence 3 is correct choice.

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