PPSC Lecturer Math Abstract Algebra Solved MCQs & Quiz

Question 1 (Part i): Let $G$ be a group and $H \le G$ such that the index $[G:H] = 2$. Prove that $H$ is a normal subgroup of $G$.

Solution:
The index $[G:H] = 2$ implies there are exactly two distinct left cosets and two distinct right cosets of $H$ in $G$.
Let $g \in G$. If $g \in H$, then $gH = H = Hg$.
If $g \notin H$, the left cosets of $H$ in $G$ are $H$ and $gH$. Since the partition of $G$ into left cosets yields $G = H \cup gH$ with $H \cap gH = \emptyset$, we have $gH = G \setminus H$.
Similarly, the right cosets are $H$ and $Hg$, giving $G = H \cup Hg$ with $H \cap Hg = \emptyset$, so $Hg = G \setminus H$.
Therefore, $gH = Hg$ for all $g \in G$, which proves $H \triangleleft G$.
Answer: $$H \triangleleft G \quad \text{(Every subgroup of index 2 is normal)}$$

Question 1 (Part ii): Find all subgroups of index 2 in the Dihedral Group $D_4 = \langle r, s \mid r^4 = e, s^2 = e, srs = r^{-1} \rangle$.

Solution:
The order of $D_4$ is 8. A subgroup of index 2 must have order $\frac{8}{2} = 4$.
The elements of $D_4$ are $\{e, r, r^2, r^3, s, sr, sr^2, sr^3\}$.
1) Subgroup generated by rotations: $H_1 = \langle r \rangle = \{e, r, r^2, r^3\}$, which is cyclic of order 4.
2) Subgroup containing $r^2$ and $s$: $H_2 = \{e, r^2, s, sr^2\} \cong V_4$ (Klein 4-group).
3) Subgroup containing $r^2$ and $sr$: $H_3 = \{e, r^2, sr, sr^3\} \cong V_4$.
All three subgroups have order 4 (index 2 in $D_4$).
Answer: $$\{ \langle r \rangle, \{e, r^2, s, sr^2\}, \{e, r^2, sr, sr^3\} \}$$

Question 2 (Part i): Calculate the number of generators of the cyclic group $\mathbb{Z}_{360}$.

Solution:
The number of generators of a finite cyclic group $\mathbb{Z}_n$ is given by Euler's totient function $\phi(n)$. Here $n = 360$.
Prime factorization of 360: $360 = 2^3 \times 3^2 \times 5^1$.
Applying Euler's formula:
$$\phi(360) = 360 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{5}\right)$$ $$\phi(360) = 360 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} = 360 \times \frac{8}{30} = 12 \times 8 = 96$$ Answer: $$96$$

Question 2 (Part ii): Find all elements of order 6 in the cyclic group $\mathbb{Z}_{30}$.

Solution:
In $\mathbb{Z}_{30}$, the order of an element $k \in \mathbb{Z}_{30}$ is given by $|k| = \frac{30}{\gcd(k, 30)}$.
We need $|k| = 6$, which means $\frac{30}{\gcd(k, 30)} = 6 \implies \gcd(k, 30) = 5$.
We find all multiples of 5 in $\{0, 1, 2, \dots, 29\}$ that satisfy $\gcd(k, 30) = 5$:
Multiples of 5: $k \in \{5, 10, 15, 20, 25\}$.
- $\gcd(5, 30) = 5$ $\checkmark$
- $\gcd(10, 30) = 10 \neq 5$ $\times$
- $\gcd(15, 30) = 15 \neq 5$ $\times$
- $\gcd(20, 30) = 10 \neq 5$ $\times$
- $\gcd(25, 30) = 5$ $\checkmark$
The elements of order 6 are 5 and 25.
Answer: $$\{5, 25\}$$

Question 3 (Part i): Determine the order of the permutation $\sigma = (1\ 3\ 5)(2\ 4\ 6\ 8)(7\ 9) \in S_9$ and express it as a product of transpositions.

Solution:
The permutation $\sigma$ is written as a product of disjoint cycles of lengths 3, 4, and 2.
The order of a product of disjoint cycles is the least common multiple (LCM) of their cycle lengths:
$$\text{order}(\sigma) = \text{lcm}(3, 4, 2) = 12$$ Now, convert each cycle into a product of transpositions using $(a_1\ a_2\ \dots\ a_k) = (a_1\ a_k)(a_1\ a_{k-1})\dots(a_1\ a_2)$:
- $(1\ 3\ 5) = (1\ 5)(1\ 3)$
- $(2\ 4\ 6\ 8) = (2\ 8)(2\ 6)(2\ 4)$
- $(7\ 9) = (7\ 9)$
Putting them together:
$$\sigma = (1\ 5)(1\ 3)(2\ 8)(2\ 6)(2\ 4)(7\ 9)$$ Answer: $$\text{Order} = 12, \quad \sigma = (1\ 5)(1\ 3)(2\ 8)(2\ 6)(2\ 4)(7\ 9)$$

Question 3 (Part ii): Determine whether $\sigma = (1\ 3\ 5)(2\ 4\ 6\ 8)(7\ 9)$ is an even or odd permutation.

Solution:
Count the total number of transpositions in the expansion of $\sigma$:
From Part (i), $\sigma = (1\ 5)(1\ 3)(2\ 8)(2\ 6)(2\ 4)(7\ 9)$, which contains 6 transpositions.
Alternatively, the number of transpositions for a cycle of length $k$ is $k - 1$.
Total transpositions = $(3-1) + (4-1) + (2-1) = 2 + 3 + 1 = 6$.
Since 6 is an even number, $\sigma$ is an even permutation.
Answer: $$\text{Even permutation}$$

Question 4 (Part i): Show that $I = \langle x^2 + 1 \rangle$ is a maximal ideal in the polynomial ring $\mathbb{R}[x]$.

Solution:
$\mathbb{R}[x]$ is a Principal Ideal Domain (PID) because $\mathbb{R}$ is a field.
In a PID, a non-zero ideal $\langle p(x) \rangle$ is maximal if and only if the generator polynomial $p(x)$ is irreducible over the base field.
Consider $p(x) = x^2 + 1 \in \mathbb{R}[x]$.
The roots of $x^2 + 1 = 0$ are $x = \pm i \notin \mathbb{R}$.
Since $p(x)$ is a quadratic polynomial with no real roots, it cannot be factored into polynomials of lower degree over $\mathbb{R}$. Thus, $x^2 + 1$ is irreducible over $\mathbb{R}$.
Consequently, $I = \langle x^2 + 1 \rangle$ is a maximal ideal of $\mathbb{R}[x]$.
Answer: $$I = \langle x^2 + 1 \rangle \text{ is a maximal ideal in } \mathbb{R}[x]$$

Question 4 (Part ii): Prove that the quotient ring $\mathbb{R}[x]/\langle x^2 + 1 \rangle$ is isomorphic to the field of complex numbers $\mathbb{C}$.

Solution:
Define an evaluation homomorphism $\phi: \mathbb{R}[x] \to \mathbb{C}$ by $\phi(f(x)) = f(i)$.
1) Homomorphism property: $\phi(f+g) = (f+g)(i) = f(i)+g(i) = \phi(f)+\phi(g)$ and $\phi(fg) = (fg)(i) = f(i)g(i) = \phi(f)\phi(g)$.
2) Surjectivity: For any $a + bi \in \mathbb{C}$ with $a, b \in \mathbb{R}$, choose $p(x) = a + bx \in \mathbb{R}[x]$. Then $\phi(p(x)) = a + bi$. Thus, $\phi$ is surjective ($\text{Im}(\phi) = \mathbb{C}$).
3) Kernel: $\ker(\phi) = \{f(x) \in \mathbb{R}[x] \mid f(i) = 0\}$.
Since $i$ is a root of $f(x)$, its complex conjugate $-i$ must also be a root because $f(x)$ has real coefficients.
Hence, $(x - i)(x + i) = x^2 + 1$ divides $f(x)$ in $\mathbb{R}[x]$. Thus, $\ker(\phi) = \langle x^2 + 1 \rangle$.
By the First Isomorphism Theorem for Rings:
$$\mathbb{R}[x]/\ker(\phi) \cong \text{Im}(\phi) \implies \mathbb{R}[x]/\langle x^2 + 1 \rangle \cong \mathbb{C}$$ Answer: $$\mathbb{R}[x]/\langle x^2 + 1 \rangle \cong \mathbb{C}$$

Question 5 (Part i): Consider the group homomorphism $\phi: \mathbb{C}^* \to \mathbb{R}^*$ defined by $\phi(z) = |z|$, where $\mathbb{C}^*$ and $\mathbb{R}^*$ are multiplicative groups. Find $\ker(\phi)$ and describe the quotient group $\mathbb{C}^*/\ker(\phi)$.

Solution:
The identity element of the target group $(\mathbb{R}^*, \cdot)$ is $1$.
The kernel of $\phi$ is:
$$\ker(\phi) = \{z \in \mathbb{C}^* \mid \phi(z) = 1\} = \{z \in \mathbb{C}^* \mid |z| = 1\} = U(1)$$ which is the unit circle group (group of complex numbers of modulus 1).
The image of $\phi$ is $\text{Im}(\phi) = \{|z| \mid z \in \mathbb{C}^*\} = \mathbb{R}^+$ (the group of positive real numbers under multiplication).
By the First Isomorphism Theorem for groups:
$$\mathbb{C}^*/U(1) \cong \mathbb{R}^+$$ Answer: $$\ker(\phi) = U(1) = \{z \in \mathbb{C} \mid |z| = 1\}, \quad \mathbb{C}^*/U(1) \cong \mathbb{R}^+$$

Question 5 (Part ii): Prove that $GL(n, \mathbb{R})/SL(n, \mathbb{R}) \cong \mathbb{R}^*$, where $GL(n, \mathbb{R})$ is the general linear group and $SL(n, \mathbb{R})$ is the special linear group.

Solution:
Define the determinant map $\det: GL(n, \mathbb{R}) \to \mathbb{R}^*$ by $A \mapsto \det(A)$.
1) Homomorphism: $\det(AB) = \det(A)\det(B)$ for all $A, B \in GL(n, \mathbb{R})$.
2) Surjectivity: For any $r \in \mathbb{R}^*$, consider the diagonal matrix $D = \text{diag}(r, 1, 1, \dots, 1) \in GL(n, \mathbb{R})$. Then $\det(D) = r$, so $\det$ is surjective.
3) Kernel: $\ker(\det) = \{A \in GL(n, \mathbb{R}) \mid \det(A) = 1\} = SL(n, \mathbb{R})$.
Applying the First Isomorphism Theorem:
$$GL(n, \mathbb{R})/\ker(\det) \cong \text{Im}(\det) \implies GL(n, \mathbb{R})/SL(n, \mathbb{R}) \cong \mathbb{R}^*$$ Answer: $$GL(n, \mathbb{R})/SL(n, \mathbb{R}) \cong \mathbb{R}^*$$

Question 6 (Part i): Using Sylow's Theorems, prove that no group of order 15 is simple.

Solution:
Let $G$ be a group with order $|G| = 15 = 3 \times 5$.
By Sylow's Third Theorem, the number of Sylow 5-subgroups, $n_5$, satisfies:
$$n_5 \equiv 1 \pmod 5 \quad \text{and} \quad n_5 \mid 3$$ The divisors of 3 are 1 and 3. Testing conditions:
- If $n_5 = 1$: $1 \equiv 1 \pmod 5$ $\checkmark$
- If $n_5 = 3$: $3 \not\equiv 1 \pmod 5$ $\times$
Thus, $n_5 = 1$. This means there is a unique Sylow 5-subgroup $P_5$ of order 5.
Since any unique Sylow $p$-subgroup is normal in $G$, $P_5 \triangleleft G$.
Because $P_5$ is a non-trivial proper normal subgroup of $G$ ($1 < |P_5| = 5 < 15$), $G$ cannot be simple.
Answer: $$n_5 = 1 \implies P_5 \triangleleft G \implies G \text{ is not simple}$$

Question 6 (Part ii): Prove that any group of order 15 is isomorphic to the cyclic group $\mathbb{Z}_{15}$.

Solution:
Let $|G| = 15 = 3 \times 5$. From Part (i), $n_5 = 1$, so $P_5 \triangleleft G$ with $|P_5| = 5$.
Now calculate the number of Sylow 3-subgroups, $n_3$:
$$n_3 \equiv 1 \pmod 3 \quad \text{and} \quad n_3 \mid 5$$ The divisors of 5 are 1 and 5.
- If $n_3 = 1$: $1 \equiv 1 \pmod 3$ $\checkmark$
- If $n_3 = 5$: $5 \equiv 2 \neq 1 \pmod 3$ $\times$
Thus, $n_3 = 1$, which implies the unique Sylow 3-subgroup $P_3 \triangleleft G$ with $|P_3| = 3$.
Since $P_3 \cap P_5 = \{e\}$ (orders 3 and 5 are coprime) and $P_3, P_5 \triangleleft G$, $G$ is the internal direct product:
$$G \cong P_3 \times P_5$$ Since $|P_3| = 3$ (prime) and $|P_5| = 5$ (prime), $P_3 \cong \mathbb{Z}_3$ and $P_5 \cong \mathbb{Z}_5$.
By the Chinese Remainder Theorem / direct product property for coprime orders:
$$G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}$$ Therefore, every group of order 15 is cyclic.
Answer: $$G \cong \mathbb{Z}_{15}$$