Practice Questions for CSIR NET Group Theory : Basics of Group Theory I

1. If \(x, y\) and \(z\) are elements of a group such that \(x y z=1\), then

(a.) \(yzx=1\)
(b.) \(yxz=1\)
(c.) \(zxy=1\)
(d.) \(zyx=1\)

Explanation for Question 1:
\(x y z =1\)
\(\Rightarrow y z =x^{-1} \cdot 1=x^{-1}\)
\(\Rightarrow y zx =x^{-1} x=1\)(thus option a is true).
\(\Rightarrow z x=y^{-1} \cdot 1\)
\(\Rightarrow z x y=y^{-1} y=1\) which implies option c is also true. Note that b and d requires commutative property.

2. Which of the following is a subgroup of \((\mathbb{C},+)\)?

(a.) \((\mathbb{R},+)\)
(b.) \((G,+)\), where \(G=\{\pi r \mid r \in \mathbb{Q}\}\)
(c.) \((G,+)\), where \(G=\{i r \mid r \in \mathbb{R}\}\)
(d.) \((G,+)\), where \(G=\left\{\pi^n \mid n \in \mathbb{Z}\right\}\)

Explanation for Question 2:
Let \(a, b \in(\mathbb{R},+)\), then \(a^{-1}=-a \in \mathbb{R}\).
Also, \(a^{-1}*b=-a+b \in \mathbb{R}\).
\(\Rightarrow(\mathbb{R}, +)\) is a group and is a subset of \((\mathbb{C}, +)\).
\(\Rightarrow(\mathbb{R},+)\) is a subgroup of \((\mathbb{C},+)\).

Let \(G_{1}=\{\pi \cdot r / r \in Q\}\).
Let \(\pi \cdot r, \pi \cdot s \in G_{1} \Rightarrow(\pi.r)^{-1}=-\pi \cdot r \in G_{1}\),
\(-\pi r+\pi.s \in G_{1}\).
\(\Rightarrow G_{1}\) is subgroup.

Let \(G_{2}=\{i \cdot r / r \in \mathbb{R}\}\).
\(-i \cdot r+i \cdot s=i(-r+s) \in G_{2}\).
\(\Rightarrow G_{2}\) is a subgroup.

Let \(G_{3}=\left\{\pi^{n} / n \in \mathbb{Z}\right\}\).
Let \(\pi \in G_{3}\), but \(\pi+\pi=2 \pi \notin G_{3}\).
\(\Rightarrow G_{3}\) is not a subgroup. One can argue that \(G_{3}=\left\{\pi^{n} / n \in \mathbb{Z}\right\}=\{n \pi / n \in \mathbb{Z}\}\), in that case it is a gorup.

3. The value of \(\alpha\) for which \(G=\{\alpha, 1,3,9,19,27\}\) is a cyclic group under multiplication modulo 56 is

(a.) 5
(b.) 15
(c.) 25
(d.) 35

Explanation for Question 3:
\(G=\{\alpha, 1,3,9,19,27\}\) is a group under \((\bmod 56)\). However, \(5 \times 3 \equiv 15(\bmod 56)\), \(15 \times 3 \equiv 45(\bmod 56)\), and \(35 \times 3=49(\mathrm{mod} 56)\) make \(a, b,\) and \(d\) false.

4. Let \(U(n)\) be the set of all positive integers less than \(n\) and relatively prime to \(n\). Then \(U(n)\) is a group under multiplication modulo \(n\). For \(n=248\), the number of elements in \(U(n)\) is

(a.) 60
(b.) 120
(c.) 180
(d.) 240

Explanation for Question 4:
\(|U(n)|=\phi(n)\).
\(\Rightarrow|U(248)|=\phi(248)=\phi\left(2^{3}\right) \times \phi(31)=4 \times 30=120\).

5. Let \(Q^c\) be the set of irrational real numbers and let \(G=Q^c \cup\{0\}\). Then, under the usual addition of real numbers, \(G\) is

(a.) A group, since \(\mathbb{R}\) and \(\mathbb{Q}\) are groups under addition
(b.) A group, since the additive identity is in \(G\)
(c.) Not a group, since addition on \(G\) is not a binary operation
(d.) Not a group, since not all elements in \(G\) have an inverse

Explanation for Question 5:
Let \(a = \pi + 2 \quad \& b=\pi\).
\(\Rightarrow a b^{-1}=a-b \notin G\).
\(\Rightarrow(G,+)\) is not a group as addition is not binary operation.

6. Let \(G\) be a group such that \(a^2=e\) for each \(a \in G\), where \(e\) is the identity element of \(G\). Then

(a.) \(G\) is cyclic
(b.) \(G\) is finite
(c.) \(G\) is abelian
(d.) None of these

Explanation for Question 6:
Let \(a, b \in G\) be a group.\)
\(\Rightarrow a \cdot b \in G\).
\(\forall a \in G, a^{2}=e\).
\(\Rightarrow(a b)^{2}=e\).
\(\Rightarrow(a b)^{-1}=a b\).
\(\Rightarrow b^{-1} a^{-1}=a b\).
\(\Rightarrow b a=a b\left(\because b^{2}=e \Rightarrow b^{-1}=b\right)\).
\(\Rightarrow G\) is abelian group.
For \(G=K_{4}\), \(G\) is not cyclic.

7. In the group \(\{1,2, \ldots ., 16\}\) under the operation of multiplication modulo 17, the order of the element 3 is

(a.) 4
(b.) 8
(c.) 12
(d.) 16

Explanation for Question 7:
left as an exercise.

8. Let \(G\) be a finite abelian group of order \(n\) with identity \(e\). If for all \(a \in G, a^3=e\), then, by induction on \(n\), show that \(n=3^k\) for some non-negative integer \(k\).

(a.) True
(b.) False

Explanation for Question 8:
left as an exercise.

9. Let \(G=\{a \in \mathbb{R}: a>0, a \neq 1\}\), define \(a * b=a^{\log b}\) then

(a.) \((G, *)\) is a semigroup but not a group
(b.) \((G, *)\) is a monoid, but not a group
(c.) \((G, *)\) is a group
(d.) \((G, *)\) is an abelian group

Explanation for Question 9:
\(G=\left\{a \in \mathbb{R} / a>0, a \neq 1\right\}\).
\(a*b=a^{\log ^{b}} \in \mathbb{R}\).
\(b>0 \Rightarrow \log b \in \mathbb{R}\).
\(\log b \in \mathbb{R} \& \ a>0 \Rightarrow a^{\log b}>0\).
If \(a^{\log b}=1 \Rightarrow \log b=0\) which is contradiction as \(b \neq 1\).
\(\Rightarrow G\) is closed.

\( \begin{aligned} & a *(b * c)=a *\left(b^{\log c}\right)=a^{\log b^{\log c}}=a^{\log c . \log b} \\ & (a * b) * c=(a * b)^{\log c}=\left(a^{\log b})^{\log c}=a^{\log b \cdot \log c}\right. \\ & \Rightarrow * \text { is associative } \end{aligned}\)
\( \begin{aligned} & e * a=a=a*e \quad \because \log e=1\ \& \ e^{\log x}=x \\ & G \text { has identity } \end{aligned} \)
If \(b=a^{-1}\) then \(a^{\log b}=e\).
\(\Rightarrow \log a^{\log b}=1\).
\(\Rightarrow \log b \log a=1\).
\(\Rightarrow \log b=\frac{1}{\log a}\).
\(\Rightarrow b=e^{\frac{1}{\log a}} \in G\).
\(\Rightarrow \forall a \in G \exists a^{-1} \in G\) such that \(a*a^{-1}=e\).
Also \(a^{\log b}\) equal to \(b^{\log a}\), thus G is abelian group.

10. The set of real numbers is a group with respect to

(a.) Arithmetic subtraction
(b.) Arithmetic multiplication
(c.) Arithmetic division
(d.) Composition defined by \(a \circ b=a+b+1\) for all real \(a\) and \(b\)

Explanation for Question 10:
a) \((a-b)-c \neq a-(b -c)\) - not a group
b) zero does not have inverse - not a group
c) not well defined binary operaiton as \((a \div 0)\) is not defined. Also \((a \div b) \div c\) and \(a \div(b \div c)\) differ - not a group
d) \(\Rightarrow a o b=a+b+1\in \mathbb{R}\), imples G is closed, Showing associativity is trivial.
Let \(e\) be the identity.
\(\Rightarrow a+e+1=a \Rightarrow e=-1\).
If \(b=a^{-1}\), then \(a+b+1=-1 \Rightarrow b=-a-2\in \mathbb{R} \).
\(\Rightarrow (G, o)\) is a group.

11. Let \(G\) be a group and let \(a \in G\). If \(o(a)=n\) and \(k\) is any integer, then which one of the following is correct?

(a.) \(o(a^k)>n\) only
(b.) \(o(a^k) \geq n\)
(c.) \(o(a^k) < n\) only
(d.) \(o(a^k) \leq n\)

Explanation for Question 11:
Given: \(O(a^{k})=\frac{o(a)}{\operatorname{gcd}(k, o(a))} \Rightarrow o(a^{k}) \leq n\)

12. Assertion (A): Let \(G=\mathbb{N} \cup\left\{\frac{1}{n}: n \in \mathbb{N}\right\}\), where \(\mathbb{N}\) is the set of all positive integers. The \(G\) is a group with respect to the usual multiplication.
Reason (R): Multiplication in both \(\mathbb{N}\) and \(\left\{\frac{1}{n}: n \in \mathbb{N}\right\}\) is well-defined and associative.\(l \in G\) and \(a \in G \Rightarrow \frac{1}{a} \in G\)

(a.) Both A and R are individually true and R is the correct explanation of A
(b.) Both A and R are individually true but R is not the correct explanation of A
(c.) A is true but R is false
(d.) A is false but R is true

Explanation for Question 12:
Given: \(G=\mathbb{N} \cup\{\frac{1}{n}, n \in \mathbb{N}\}\) is not closed as \(2 \times \frac{1}{3}=\frac{2}{3} \notin G\). But " \(R\) " is true.

13. Consider the following statements in respect of a finite group \(G\):
1. \(O(a)=O\left(a^{-1}\right)\) for all \(a \in G\)
2. \(O(a)=O\left(b a b^{-1}\right)\) for all \(a, b \in G\)

(a.) 1 only
(b.) 2 only
(c.) Both 1 and 2
(d.) Neither 1 nor 2

Explanation for Question 13:
Given: \[ \begin{aligned} & \text { If } o(a)=n \\ & \Rightarrow a^{n}=e \Rightarrow e=a^{-n}=(a^{-1})^{n} \\ & \Rightarrow o\left(a^{-1}\right) / n \Rightarrow o\left(a^{-1}\right) / o(a) \\ & \text { Similarly } o(a) / o\left(a^{-1}\right) \Rightarrow o(a)=o\left(a^{-1}\right) \\ & \text { If } o(a)=n \\ & \Rightarrow \underbrace{b a b^{-1} b a b^{-1} \cdots b a b^{-1}}=b a^{n} b^{-1}=e \\ & \Rightarrow o\left(b a b^{-1}\right) / o(a) \\ & \text { Similarly } o(a) / o\left(b a b^{-1}\right) & 0(a) \end{aligned} \]

14. In the set \(Q\) of rational numbers defined * as follows: for \(\alpha, \beta \in \mathbb{Q}, \alpha \beta=\frac{\alpha \cdot \beta}{3}\). If \(\mathbb{Q}^+\), \(\mathbb {Q}^{-}, \mathbb{Q}^*\) respectively denote the sets of positive, negative and non-zero rational numbers, then which one of the following pairs is an abelian group?

(a.) \((\mathbb {Q}^{+}, *)\)
(b.) \((\mathbb{Q}, *)\)
(c.) \((\mathbb{Q}^{-}, *)\)
(d.) \((\mathbb{Q}^*, *)\)

Explanation for Question 14:
Given: \[ \begin{aligned} & -1,-2 \in \mathbb{Q}^{-} \text { but }-1 *-2=\frac{-1}{-2}{3}=\frac{2}{3} \notin \mathbb{Q}^{-} \text {thus c is incorrect.}\\ & \text{for option b, closure and associative are trivial to show. for identity}, \frac{\alpha \beta}{3}=\alpha \Rightarrow \beta=3 \Rightarrow 3 \text { is identity } \\ & 0 . x=0 \neq 3 \forall x, \text { Thus zero does not have inverse. } \end{aligned} \] \(\Rightarrow(\mathbb{Q}, *)\) is not a group
for a and d, Identity element is 3 and \( a^{-1}=\frac{3}{a}\).

15. Let \(M(\mathbb{R})\) be set of all matrices with real entries. The usual matrix addition '+' is

(a.) Commutative binary operation
(b.) Non-commutative binary operation
(c.) Associative binary operation
(d.) Not a binary operation

Explanation for Question 15:
If \(A\) is a \(2 \times 2\) matrix and \(B\) is a \(3 \times 3\) matrix, then \(A+B\) is not defined.