Groups and Subgroups Questions and Answers

Q: Which of the following statements is not true ?

The false statement is Every group of order 12 is cyclic.The alternating group \(A_{4}\), the dihedral group of order 12 and \(\mathbf{Z}_{2} \times \mathbf{Z}_{6}\) all have order 12 and none is cyclic; \(A_{4}\) is not even abelian.The other statements are true. There is a field with 4 elements, \(\mathbf{F}_{4} = \mathbf{Z}_{2}[x]/(x^{2} + x + 1)\), since a finite field exists for every prime power. A Boolean ring with 8 elements is \(\mathbf{Z}_{2} \times \mathbf{Z}_{2} \times \mathbf{Z}_{2}\), or the power set of a three-element set under symmetric difference and intersection. Every group of order 15 = 3 × 5 is cyclic, hence abelian, because 3 does not divide 5 - 1.

Q: Consider the following statements :(I) Let G be a finite group. Then every element is of finite order.(II) Let G be a group such that every element is of finite order. Then G is a finite group.Which is true ?

Only (I) is true.(I) In a finite group of order \(n\), the powers \(g, g^{2}, g^{3}, \ldots\) cannot all be distinct, so \(g^{i} = g^{j}\) for some \(i \lt j\) and \(g^{j - i} = e\). By Lagrange's theorem the order of \(g\) even divides \(n\).(II) A group in which every element has finite order (a torsion group) need not be finite. The group \(\mathbf{Q}/\mathbf{Z}\) is infinite, yet each element \(p/q + \mathbf{Z}\) has order dividing \(q\). An infinite direct sum of copies of \(\mathbf{Z}_{2}\) is another example: every non-identity element has order 2.So the options asserting (II) are wrong, and 'neither' ignores the elementary argument for (I).

Q: Consider the following statements :(I) Let H and K be subgroups of an abelian group G. Suppose \(|H| = m\) and \(|K| = n\). Let \(d = \text{lcm}(m, n)\). Then G has a subgroup of order \(d\).(II) The group of order 15 is abelian.Which of the following is true ?

Both (I) and (II) are true.(I) In an abelian group, HK is a subgroup, and it is finite of order \(mn/|H \cap K|\), a number divisible by both \(m\) and \(n\), hence by \(d = \text{lcm}(m, n)\). A finite abelian group has a subgroup of every order dividing its order (the converse of Lagrange holds for abelian groups), so HK, and therefore G, has a subgroup of order \(d\).(II) Every group of order 15 is cyclic, hence abelian. By Sylow's theorems the number of Sylow 5-subgroups divides 3 and is 1 mod 5, so it is 1; the number of Sylow 3-subgroups divides 5 and is 1 mod 3, so it is 1. Both are normal and intersect trivially, so \(G \cong \mathbf{Z}_{3} \times \mathbf{Z}_{5} \cong \mathbf{Z}_{15}\).So the options that deny either statement are wrong.

Q: Let \(S_{3}\) be the symmetric group, \(I_{n}(S_{3})\) be the group of inner automorphisms of \(S_{3}\) and \(Z(S_{3})\) be the center of \(S_{3}\). Then which is FALSE ?

The false statement is \(I_{n}(S_{3}) = \{e\}\).The center of \(S_{3}\) is trivial: no non-identity permutation commutes with all others (for example (1 2) does not commute with (1 3)). So \(Z(S_{3}) = \{e\}\), the first option, is true.The map \(g \mapsto\) (conjugation by \(g\)) is a homomorphism from \(S_{3}\) onto the inner automorphism group, with kernel the center. So \(I_{n}(S_{3}) \cong S_{3}/Z(S_{3})\), the last option, and since the center is trivial this is a copy of \(S_{3}\) itself, which the paper writes as \(I_{n}(S_{3}) = S_{3}\).Consequently the inner automorphism group has 6 elements, not 1. It would be trivial only for an abelian group, where every conjugation is the identity map, and \(S_{3}\) is not abelian.

Q: Let G be a finite abelian group. If \(n \in \mathbf{N}\) and |G| are relatively prime, then the function \(Q : G \to G\) defined by \(Q(a) = a^{n}\) is :

The map is an isomorphism.In an abelian group \((ab)^{n} = a^{n}b^{n}\), so \(Q(a) = a^{n}\) is a homomorphism.Its kernel consists of elements with \(a^{n} = e\). The order of such an a divides both n and |G| (Lagrange), and these are coprime, so the order is 1 and \(a = e\). The kernel is trivial, so Q is one-one, and a one-one map of a finite set into itself is onto. Hence Q is an isomorphism, which rules out 'not 1 − 1', 'not onto' and 'not a homomorphism'.

Q: Let G be a group and a map \(Q : G \to G\) defined by \(Q(a) = a^{-1}\). Then which of the following is false ?

The false statement is Q is a homomorphism if G is any permutation group.Inversion satisfies \(Q(ab) = (ab)^{-1} = b^{-1}a^{-1}\), while \(Q(a)Q(b) = a^{-1}b^{-1}\). These agree for all a, b exactly when \(ab = ba\) for all a, b. So Q is a homomorphism if and only if G is abelian, which makes the second and third statements true. Q is always bijective, being its own inverse.A permutation group such as \(S_{3}\) is not abelian: with \(a = (1\,2)\) and \(b = (1\,3)\), \((ab)^{-1} \neq a^{-1}b^{-1}\). So Q is not a homomorphism for every permutation group.

Q: Consider the following statements :(I) If G is a finite group that has only 2 conjugate classes, then |G| = 2.(II) If G is a group and \(a \in Z(G)\), the center of G, then the conjugacy class of a is {e, a}, where e is the identity of GWhich of the following is true ?

Only (I) is true.(I) The class equation of a group with 2 classes reads \(|G| = 1 + k\), where k is the size of the other class, and k divides |G|. So \(k \mid k + 1\), forcing \(k = 1\) and \(|G| = 2\).(II) If a is in the centre, then \(gag^{-1} = a\) for every g, so its conjugacy class is {a} alone, not {e, a}. The identity is in its own class, separate from a unless \(a = e\).

Q: The special linear group SL2(Z/2Z) of 2 × 2 matrices over Z/2Z :

\(SL_{2}(\mathbf{Z}/2\mathbf{Z})\) is isomorphic to \(S_{3}\).Over \(\mathbf{Z}_{2}\) the determinant of an invertible matrix is always 1, so \(SL_{2} = GL_{2}\). Count the invertible matrices: the first column is any non-zero vector (3 choices) and the second any vector off the line it spans (2 choices), giving \(3 \times 2 = 6\) elements.The group permutes the three non-zero vectors of \(\mathbf{Z}_{2}^{2}\) faithfully, which gives an injective map into \(S_{3}\); with 6 elements on both sides it is an isomorphism. The group is non-abelian, so it is not \(\mathbf{Z}_{6}\).Cardinalities 8 and 16 are wrong (16 counts all 2 × 2 matrices), and \(A_{4}\) has order 12.

Q: Consider the following statements :I : If G is a group such that every cyclic group of G is a normal subgroup of G, then every subgroup of G is a normal subgroup of G.II : If H is a subgroup of a group G such that the index of H in G is 2, then H is a normal subgroup of G.Which of the following statement(s) is/are true ?

Both I and II are true.I: any subgroup H is the union of the cyclic subgroups \(\langle h \rangle\), \(h \in H\). If each \(\langle h \rangle\) is normal, then \(g h g^{-1} \in \langle h \rangle \subseteq H\) for every g, so H is normal. (Groups like this are called Dedekind groups; the quaternion group is a non-abelian example.)II: if H has index 2, its two left cosets are H and \(G \setminus H\), and so are its two right cosets. So \(gH = Hg\) for every g, which is normality.

Q: Which of the folowing is correct ?

The correct statement is that \(A_{4}\) has only one proper non-trivial normal subgroup.That subgroup is the Klein four-group \(V = \{e, (12)(34), (13)(24), (14)(23)\}\). The conjugacy classes of \(A_{4}\) have sizes 1, 3, 4 and 4, and a normal subgroup is a union of classes containing the identity whose size divides 12; only \(1 + 3 = 4\) works besides 1 and 12.The others fail. \(A_{4}\) has no subgroup of order 6, the standard counterexample to the converse of Lagrange's theorem. Having V as a normal subgroup, \(A_{4}\) is not simple. And conjugacy in \(S_{n}\) need not survive in \(A_{n}\): the 3-cycles (123) and (132) are conjugate in \(S_{4}\) but not in \(A_{4}\), where the 3-cycles split into two classes of 4.

21.

Which of the following statements is not true ?

A.There is a field with 4 elements
B.There is a Boolean ring with 8 elements
C.Every group of order 12 is cyclic
D.Every group of order 15 is abelian

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The false statement is Every group of order 12 is cyclic.

The alternating group \(A_{4}\), the dihedral group of order 12 and \(\mathbf{Z}_{2} \times \mathbf{Z}_{6}\) all have order 12 and none is cyclic; \(A_{4}\) is not even abelian.

The other statements are true. There is a field with 4 elements, \(\mathbf{F}_{4} = \mathbf{Z}_{2}[x]/(x^{2} + x + 1)\), since a finite field exists for every prime power. A Boolean ring with 8 elements is \(\mathbf{Z}_{2} \times \mathbf{Z}_{2} \times \mathbf{Z}_{2}\), or the power set of a three-element set under symmetric difference and intersection. Every group of order 15 = 3 × 5 is cyclic, hence abelian, because 3 does not divide 5 - 1.

Discuss in Forum

22.

Consider the following statements :

(I) Let G be a finite group. Then every element is of finite order.
(II) Let G be a group such that every element is of finite order. Then G is a finite group.

Which is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

Only (I) is true.

(I) In a finite group of order \(n\), the powers \(g, g^{2}, g^{3}, \ldots\) cannot all be distinct, so \(g^{i} = g^{j}\) for some \(i \lt j\) and \(g^{j - i} = e\). By Lagrange's theorem the order of \(g\) even divides \(n\).

(II) A group in which every element has finite order (a torsion group) need not be finite. The group \(\mathbf{Q}/\mathbf{Z}\) is infinite, yet each element \(p/q + \mathbf{Z}\) has order dividing \(q\). An infinite direct sum of copies of \(\mathbf{Z}_{2}\) is another example: every non-identity element has order 2.

So the options asserting (II) are wrong, and 'neither' ignores the elementary argument for (I).

Discuss in Forum

23.

Consider the following statements :

(I) Let H and K be subgroups of an abelian group G. Suppose \(|H| = m\) and \(|K| = n\). Let \(d = \text{lcm}(m, n)\). Then G has a subgroup of order \(d\).
(II) The group of order 15 is abelian.

Which of the following is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Both (I) and (II) are true.

(I) In an abelian group, HK is a subgroup, and it is finite of order \(mn/|H \cap K|\), a number divisible by both \(m\) and \(n\), hence by \(d = \text{lcm}(m, n)\). A finite abelian group has a subgroup of every order dividing its order (the converse of Lagrange holds for abelian groups), so HK, and therefore G, has a subgroup of order \(d\).

(II) Every group of order 15 is cyclic, hence abelian. By Sylow's theorems the number of Sylow 5-subgroups divides 3 and is 1 mod 5, so it is 1; the number of Sylow 3-subgroups divides 5 and is 1 mod 3, so it is 1. Both are normal and intersect trivially, so \(G \cong \mathbf{Z}_{3} \times \mathbf{Z}_{5} \cong \mathbf{Z}_{15}\).

So the options that deny either statement are wrong.

Discuss in Forum

24.

Let \(S_{3}\) be the symmetric group, \(I_{n}(S_{3})\) be the group of inner automorphisms of \(S_{3}\) and \(Z(S_{3})\) be the center of \(S_{3}\). Then which is FALSE ?

A.\(Z(S_{3}) = \{e\}\), where \(e\) is the identity of \(S_{3}\)
B.\(I_{n}(S_{3}) = S_{3}\)
C.\(I_{n}(S_{3}) = \{e\}\), where \(e\) is the identity of \(S_{3}\)
D.\(I_{n}(S_{3}) \cong S_{3} / Z(S_{3})\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The false statement is \(I_{n}(S_{3}) = \{e\}\).

The center of \(S_{3}\) is trivial: no non-identity permutation commutes with all others (for example (1 2) does not commute with (1 3)). So \(Z(S_{3}) = \{e\}\), the first option, is true.

The map \(g \mapsto\) (conjugation by \(g\)) is a homomorphism from \(S_{3}\) onto the inner automorphism group, with kernel the center. So \(I_{n}(S_{3}) \cong S_{3}/Z(S_{3})\), the last option, and since the center is trivial this is a copy of \(S_{3}\) itself, which the paper writes as \(I_{n}(S_{3}) = S_{3}\).

Consequently the inner automorphism group has 6 elements, not 1. It would be trivial only for an abelian group, where every conjugation is the identity map, and \(S_{3}\) is not abelian.

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25.

Let G be a finite abelian group. If \(n \in \mathbf{N}\) and |G| are relatively prime, then the function \(Q : G \to G\) defined by \(Q(a) = a^{n}\) is :

A.homomorphism but not 1 − 1
B.homomorphism but not onto
C.isomorphism
D.not a homomorphism

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The map is an isomorphism.

In an abelian group \((ab)^{n} = a^{n}b^{n}\), so \(Q(a) = a^{n}\) is a homomorphism.

Its kernel consists of elements with \(a^{n} = e\). The order of such an a divides both n and |G| (Lagrange), and these are coprime, so the order is 1 and \(a = e\). The kernel is trivial, so Q is one-one, and a one-one map of a finite set into itself is onto. Hence Q is an isomorphism, which rules out 'not 1 − 1', 'not onto' and 'not a homomorphism'.

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26.

Let G be a group and a map \(Q : G \to G\) defined by \(Q(a) = a^{-1}\). Then which of the following is false ?

A.Q is bijective
B.Q is a homomorphism if G is abelian
C.Q is a homomorphism only if G is abelian
D.Q is a homomorphism, if G is any permutation group

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The false statement is Q is a homomorphism if G is any permutation group.

Inversion satisfies \(Q(ab) = (ab)^{-1} = b^{-1}a^{-1}\), while \(Q(a)Q(b) = a^{-1}b^{-1}\). These agree for all a, b exactly when \(ab = ba\) for all a, b. So Q is a homomorphism if and only if G is abelian, which makes the second and third statements true. Q is always bijective, being its own inverse.

A permutation group such as \(S_{3}\) is not abelian: with \(a = (1\,2)\) and \(b = (1\,3)\), \((ab)^{-1} \neq a^{-1}b^{-1}\). So Q is not a homomorphism for every permutation group.

Discuss in Forum

27.

Consider the following statements :

(I) If G is a finite group that has only 2 conjugate classes, then |G| = 2.
(II) If G is a group and \(a \in Z(G)\), the center of G, then the conjugacy class of a is {e, a}, where e is the identity of G

Which of the following is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only (I) is true.

(I) The class equation of a group with 2 classes reads \(|G| = 1 + k\), where k is the size of the other class, and k divides |G|. So \(k \mid k + 1\), forcing \(k = 1\) and \(|G| = 2\).

(II) If a is in the centre, then \(gag^{-1} = a\) for every g, so its conjugacy class is {a} alone, not {e, a}. The identity is in its own class, separate from a unless \(a = e\).

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28.

The special linear group SL₂(Z/2Z) of 2 × 2 matrices over Z/2Z :

A.has cardinality 8
B.has cardinality 16
C.is isomorphic to \(S_{3}\)
D.is isomorphic to \(A_{4}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

\(SL_{2}(\mathbf{Z}/2\mathbf{Z})\) is isomorphic to \(S_{3}\).

Over \(\mathbf{Z}_{2}\) the determinant of an invertible matrix is always 1, so \(SL_{2} = GL_{2}\). Count the invertible matrices: the first column is any non-zero vector (3 choices) and the second any vector off the line it spans (2 choices), giving \(3 \times 2 = 6\) elements.

The group permutes the three non-zero vectors of \(\mathbf{Z}_{2}^{2}\) faithfully, which gives an injective map into \(S_{3}\); with 6 elements on both sides it is an isomorphism. The group is non-abelian, so it is not \(\mathbf{Z}_{6}\).

Cardinalities 8 and 16 are wrong (16 counts all 2 × 2 matrices), and \(A_{4}\) has order 12.

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29.

Consider the following statements :

I : If G is a group such that every cyclic group of G is a normal subgroup of G, then every subgroup of G is a normal subgroup of G.
II : If H is a subgroup of a group G such that the index of H in G is 2, then H is a normal subgroup of G.

Which of the following statement(s) is/are true ?

A.Only I is true
B.Only II is true
C.Both I and II are true
D.Neither I nor II is true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

Both I and II are true.

I: any subgroup H is the union of the cyclic subgroups \(\langle h \rangle\), \(h \in H\). If each \(\langle h \rangle\) is normal, then \(g h g^{-1} \in \langle h \rangle \subseteq H\) for every g, so H is normal. (Groups like this are called Dedekind groups; the quaternion group is a non-abelian example.)

II: if H has index 2, its two left cosets are H and \(G \setminus H\), and so are its two right cosets. So \(gH = Hg\) for every g, which is normality.

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30.

Which of the folowing is correct ?

A.\(A_{4}\) has a subgroup of order 6
B.\(A_{4}\) has only one proper non-trivial normal subgroup
C.\(A_{4}\) is simple
D.Two elements of \(A_{n}\) that are conjugates in \(S_{n}\) are always conjugates in \(A_{n}\) also

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The correct statement is that \(A_{4}\) has only one proper non-trivial normal subgroup.

That subgroup is the Klein four-group \(V = \{e, (12)(34), (13)(24), (14)(23)\}\). The conjugacy classes of \(A_{4}\) have sizes 1, 3, 4 and 4, and a normal subgroup is a union of classes containing the identity whose size divides 12; only \(1 + 3 = 4\) works besides 1 and 12.

The others fail. \(A_{4}\) has no subgroup of order 6, the standard counterexample to the converse of Lagrange's theorem. Having V as a normal subgroup, \(A_{4}\) is not simple. And conjugacy in \(S_{n}\) need not survive in \(A_{n}\): the 3-cycles (123) and (132) are conjugate in \(S_{4}\) but not in \(A_{4}\), where the 3-cycles split into two classes of 4.

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