# 2019

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UPSC Algebra

$$cos\:2\theta =1-2\:sin^2\theta$$

### 1. (a) Let $$G$$ be a finite group, $$H$$ and $$K$$ subgroups of $$G$$ such that $$K \subset H$$. Show that $$(G: K)=(G: H)(H: K)$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction:
We are given a finite group $G$$G$GG$G$ and two subgroups $H$$H$HH$H$ and $K$$K$KK$K$ such that $K\subset H$$K\subset H$K sub HK \subset H$K\subset H$. We are asked to prove that $\left(G:K\right)=\left(G:H\right)\left(H:K\right)$$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$(G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$, where $\left(G:K\right)$$\left(G:K\right)$(G:K)(G: K)$\left(G:K\right)$, $\left(G:H\right)$$\left(G:H\right)$(G:H)(G: H)$\left(G:H\right)$, and $\left(H:K\right)$$\left(H:K\right)$(H:K)(H: K)$\left(H:K\right)$ are the indices of the subgroups $K$$K$KK$K$, $H$$H$HH$H$, and $K$$K$KK$K$ in $G$$G$GG$G$, $G$$G$GG$G$, and $H$$H$HH$H$ respectively.
Work/Calculations:
Definition of Index:
The index $\left(A:B\right)$$\left(A:B\right)$(A:B)(A: B)$\left(A:B\right)$ of a subgroup $B$$B$BB$B$ in a group $A$$A$AA$A$ is defined as the number of distinct left cosets of $B$$B$BB$B$ in $A$$A$AA$A$. Mathematically, $\left(A:B\right)=|A/B|$$\left(A:B\right)=|A/B|$(A:B)=|A//B|(A: B) = |A/B|$\left(A:B\right)=|A/B|$, where $|A/B|$$|A/B|$|A//B||A/B|$|A/B|$ is the number of distinct left cosets.
Step 1: Express $\left(G:K\right)$$\left(G:K\right)$(G:K)(G: K)$\left(G:K\right)$ in terms of cosets
The index $\left(G:K\right)$$\left(G:K\right)$(G:K)(G: K)$\left(G:K\right)$ is the number of distinct left cosets of $K$$K$KK$K$ in $G$$G$GG$G$. Let’s denote this set of cosets as $G/K$$G/K$G//KG/K$G/K$.
Step 2: Express $\left(G:H\right)$$\left(G:H\right)$(G:H)(G: H)$\left(G:H\right)$ and $\left(H:K\right)$$\left(H:K\right)$(H:K)(H: K)$\left(H:K\right)$ in terms of cosets
Similarly, $\left(G:H\right)$$\left(G:H\right)$(G:H)(G: H)$\left(G:H\right)$ is the number of distinct left cosets of $H$$H$HH$H$ in $G$$G$GG$G$, denoted as $G/H$$G/H$G//HG/H$G/H$, and $\left(H:K\right)$$\left(H:K\right)$(H:K)(H: K)$\left(H:K\right)$ is the number of distinct left cosets of $K$$K$KK$K$ in $H$$H$HH$H$, denoted as $H/K$$H/K$H//KH/K$H/K$.
Step 3: Relate $G/K$$G/K$G//KG/K$G/K$ with $G/H$$G/H$G//HG/H$G/H$ and $H/K$$H/K$H//KH/K$H/K$
Each coset $gK$$gK$gKgK$gK$ in $G/K$$G/K$G//KG/K$G/K$ can be uniquely expressed as $hK$$hK$hKhK$hK$ where $h$$h$hh$h$ is in some coset $gH$$gH$gHgH$gH$ in $G/H$$G/H$G//HG/H$G/H$. Furthermore, each coset $gH$$gH$gHgH$gH$ in $G/H$$G/H$G//HG/H$G/H$ contains exactly $\left(H:K\right)$$\left(H:K\right)$(H:K)(H: K)$\left(H:K\right)$ distinct cosets of the form $hK$$hK$hKhK$hK$ in $H/K$$H/K$H//KH/K$H/K$.
Therefore, the total number of distinct cosets $gK$$gK$gKgK$gK$ in $G/K$$G/K$G//KG/K$G/K$ can be obtained by multiplying the number of distinct cosets $gH$$gH$gHgH$gH$ in $G/H$$G/H$G//HG/H$G/H$ by the number of distinct cosets $hK$$hK$hKhK$hK$ in $H/K$$H/K$H//KH/K$H/K$.
Step 4: Mathematical Expression
This relationship can be mathematically expressed as:
$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$(G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$
Conclusion:
We have successfully proven that $\left(G:K\right)=\left(G:H\right)\left(H:K\right)$$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$(G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)$\left(G:K\right)=\left(G:H\right)\left(H:K\right)$ by relating the number of distinct left cosets of $K$$K$KK$K$ in $G$$G$GG$G$ with the number of distinct left cosets of $H$$H$HH$H$ in $G$$G$GG$G$ and $K$$K$KK$K$ in $H$$H$HH$H$. This proves the statement for finite groups $G$$G$GG$G$ and subgroups $H$$H$HH$H$ and $K$$K$KK$K$ such that $K\subset H$$K\subset H$K sub HK \subset H$K\subset H$.
5/5
$$cos^2\left(\frac{\theta }{2}\right)=\frac{1+cos\:\theta }{2}$$
$$2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

### 2. (a) If $$G$$ and $$H$$ are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from $$G$$ to $$H$$, the trivial one.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction:
We are given two finite groups $G$$G$GG$G$ and $H$$H$HH$H$ whose orders are relatively prime. We are asked to prove that there is only one homomorphism from $G$$G$GG$G$ to $H$$H$HH$H$, which is the trivial homomorphism.
Work/Calculations:
Step 1: Definitions and Assumptions
Let $|G|=n$$|G|=n$|G|=n|G| = n$|G|=n$ and $|H|=m$$|H|=m$|H|=m|H| = m$|H|=m$. Since $G$$G$GG$G$ and $H$$H$HH$H$ are finite groups with relatively prime orders, $gcd\left(n,m\right)=1$$gcd\left(n,m\right)=1$gcd(n,m)=1\gcd(n, m) = 1$gcd\left(n,m\right)=1$.
Let $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \to H$\varphi :G\to H$ be a homomorphism.
Step 2: Properties of Homomorphism
We know that the order of $\varphi \left(g\right)$$\varphi \left(g\right)$phi(g)\phi(g)$\varphi \left(g\right)$ must divide the order of $g$$g$gg$g$ for any $g\in G$$g\in G$g in Gg \in G$g\in G$. This is because if ${g}^{k}={e}_{G}$${g}^{k}={e}_{G}$g^(k)=e_(G)g^k = e_G${g}^{k}={e}_{G}$ in $G$$G$GG$G$, then $\varphi \left(g{\right)}^{k}=\varphi \left({e}_{G}\right)={e}_{H}$$\varphi \left(g{\right)}^{k}=\varphi \left({e}_{G}\right)={e}_{H}$phi(g)^(k)=phi(e_(G))=e_(H)\phi(g)^k = \phi(e_G) = e_H$\varphi \left(g{\right)}^{k}=\varphi \left({e}_{G}\right)={e}_{H}$ in $H$$H$HH$H$.
Step 3: Relatively Prime Orders
Since $gcd\left(n,m\right)=1$$gcd\left(n,m\right)=1$gcd(n,m)=1\gcd(n, m) = 1$gcd\left(n,m\right)=1$, the only possible order for $\varphi \left(g\right)$$\varphi \left(g\right)$phi(g)\phi(g)$\varphi \left(g\right)$ that divides both $n$$n$nn$n$ and $m$$m$mm$m$ is 1. This means that $\varphi \left(g\right)$$\varphi \left(g\right)$phi(g)\phi(g)$\varphi \left(g\right)$ must be the identity element in $H$$H$HH$H$ for all $g\in G$$g\in G$g in Gg \in G$g\in G$.
Step 4: The Trivial Homomorphism
The only homomorphism $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \to H$\varphi :G\to H$ that satisfies these conditions is the trivial homomorphism, which maps every element of $G$$G$GG$G$ to the identity element of $H$$H$HH$H$:
$\varphi \left(g\right)={e}_{H}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}g\in G$$\varphi \left(g\right)={e}_{H}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}g\in G$phi(g)=e_(H)quad”for all”quad g in G\phi(g) = e_H \quad \text{for all} \quad g \in G$\varphi \left(g\right)={e}_{H}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}g\in G$
Conclusion:
We have shown that if $G$$G$GG$G$ and $H$$H$HH$H$ are finite groups with relatively prime orders, then the only homomorphism from $G$$G$GG$G$ to $H$$H$HH$H$ is the trivial one, which maps every element in $G$$G$GG$G$ to the identity element in $H$$H$HH$H$.
5/5
$$\sec ^2 \theta=1+\tan ^2 \theta$$
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$

### 2. (b) Write down all quotient groups of the group $$Z_{12}$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction:
We are asked to find all the quotient groups of the group ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$, which is the group of integers modulo 12. The group ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$ consists of the elements $\left\{0,1,2,\dots ,11\right\}$$\left\{0,1,2,\dots ,11\right\}${0,1,2,dots,11}\{0, 1, 2, \ldots, 11\}$\left\{0,1,2,\dots ,11\right\}$ under addition modulo 12.
Work/Calculations:
Step 1: Identify Subgroups of ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$
First, let’s identify the subgroups of ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$. The subgroups are generated by the divisors of 12. The divisors of 12 are $\left\{1,2,3,4,6,12\right\}$$\left\{1,2,3,4,6,12\right\}${1,2,3,4,6,12}\{1, 2, 3, 4, 6, 12\}$\left\{1,2,3,4,6,12\right\}$.
• $⟨1⟩={\mathbb{Z}}_{12}$$⟨1⟩={\mathbb{Z}}_{12}$(:1:)=Z_(12)\langle 1 \rangle = \mathbb{Z}_{12}$⟨1⟩={\mathbb{Z}}_{12}$
• $⟨2⟩=\left\{0,2,4,6,8,10\right\}$$⟨2⟩=\left\{0,2,4,6,8,10\right\}$(:2:)={0,2,4,6,8,10}\langle 2 \rangle = \{0, 2, 4, 6, 8, 10\}$⟨2⟩=\left\{0,2,4,6,8,10\right\}$
• $⟨3⟩=\left\{0,3,6,9\right\}$$⟨3⟩=\left\{0,3,6,9\right\}$(:3:)={0,3,6,9}\langle 3 \rangle = \{0, 3, 6, 9\}$⟨3⟩=\left\{0,3,6,9\right\}$
• $⟨4⟩=\left\{0,4,8\right\}$$⟨4⟩=\left\{0,4,8\right\}$(:4:)={0,4,8}\langle 4 \rangle = \{0, 4, 8\}$⟨4⟩=\left\{0,4,8\right\}$
• $⟨6⟩=\left\{0,6\right\}$$⟨6⟩=\left\{0,6\right\}$(:6:)={0,6}\langle 6 \rangle = \{0, 6\}$⟨6⟩=\left\{0,6\right\}$
Step 2: Find Quotient Groups
To find the quotient groups ${\mathbb{Z}}_{12}/H$${\mathbb{Z}}_{12}/H$Z_(12)//H\mathbb{Z}_{12} / H${\mathbb{Z}}_{12}/H$, where $H$$H$HH$H$ is a subgroup of ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$, we need to find the cosets of each subgroup $H$$H$HH$H$.
1. ${\mathbb{Z}}_{12}/⟨1⟩$${\mathbb{Z}}_{12}/⟨1⟩$Z_(12)//(:1:)\mathbb{Z}_{12} / \langle 1 \rangle${\mathbb{Z}}_{12}/⟨1⟩$ has one coset, $\left\{0\right\}$$\left\{0\right\}${0}\{0\}$\left\{0\right\}$, so it is isomorphic to ${\mathbb{Z}}_{1}$${\mathbb{Z}}_{1}$Z_(1)\mathbb{Z}_1${\mathbb{Z}}_{1}$.
2. ${\mathbb{Z}}_{12}/⟨2⟩$${\mathbb{Z}}_{12}/⟨2⟩$Z_(12)//(:2:)\mathbb{Z}_{12} / \langle 2 \rangle${\mathbb{Z}}_{12}/⟨2⟩$ has two cosets, $\left\{0,2,4,6,8,10\right\}$$\left\{0,2,4,6,8,10\right\}${0,2,4,6,8,10}\{0, 2, 4, 6, 8, 10\}$\left\{0,2,4,6,8,10\right\}$ and $\left\{1,3,5,7,9,11\right\}$$\left\{1,3,5,7,9,11\right\}${1,3,5,7,9,11}\{1, 3, 5, 7, 9, 11\}$\left\{1,3,5,7,9,11\right\}$, so it is isomorphic to ${\mathbb{Z}}_{2}$${\mathbb{Z}}_{2}$Z_(2)\mathbb{Z}_2${\mathbb{Z}}_{2}$.
3. ${\mathbb{Z}}_{12}/⟨3⟩$${\mathbb{Z}}_{12}/⟨3⟩$Z_(12)//(:3:)\mathbb{Z}_{12} / \langle 3 \rangle${\mathbb{Z}}_{12}/⟨3⟩$ has three cosets, so it is isomorphic to ${\mathbb{Z}}_{3}$${\mathbb{Z}}_{3}$Z_(3)\mathbb{Z}_3${\mathbb{Z}}_{3}$.
4. ${\mathbb{Z}}_{12}/⟨4⟩$${\mathbb{Z}}_{12}/⟨4⟩$Z_(12)//(:4:)\mathbb{Z}_{12} / \langle 4 \rangle${\mathbb{Z}}_{12}/⟨4⟩$ has four cosets, so it is isomorphic to ${\mathbb{Z}}_{4}$${\mathbb{Z}}_{4}$Z_(4)\mathbb{Z}_4${\mathbb{Z}}_{4}$.
5. ${\mathbb{Z}}_{12}/⟨6⟩$${\mathbb{Z}}_{12}/⟨6⟩$Z_(12)//(:6:)\mathbb{Z}_{12} / \langle 6 \rangle${\mathbb{Z}}_{12}/⟨6⟩$ has six cosets, so it is isomorphic to ${\mathbb{Z}}_{6}$${\mathbb{Z}}_{6}$Z_(6)\mathbb{Z}_6${\mathbb{Z}}_{6}$.
Step 3: List All Quotient Groups
The quotient groups are ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$Z_(1),Z_(2),Z_(3),Z_(4),\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4,${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ and ${\mathbb{Z}}_{6}$${\mathbb{Z}}_{6}$Z_(6)\mathbb{Z}_6${\mathbb{Z}}_{6}$.
Conclusion:
The quotient groups of ${\mathbb{Z}}_{12}$${\mathbb{Z}}_{12}$Z_(12)\mathbb{Z}_{12}${\mathbb{Z}}_{12}$ are ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$Z_(1),Z_(2),Z_(3),Z_(4),\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4,${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ and ${\mathbb{Z}}_{6}$${\mathbb{Z}}_{6}$Z_(6)\mathbb{Z}_6${\mathbb{Z}}_{6}$.
5/5
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$
$$2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)$$

### 3. (d) Let $$a$$ be an irreducible element of the Euclidean ring $$R$$, then prove that $$R /(a)$$ is a field.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
In this problem, we are given that $a$$a$aa$a$ is an irreducible element in a Euclidean ring $R$$R$RR$R$. We are tasked with proving that the quotient ring $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ is a field.
Work/Calculations
Step 1: Definitions and Preliminaries
• A Euclidean ring $R$$R$RR$R$ is an integral domain with a Euclidean function $d:R\to \mathbb{N}$$d:R\to \mathbb{N}$d:R rarrNd: R \to \mathbb{N}$d:R\to \mathbb{N}$ satisfying certain properties.
• An element $a\in R$$a\in R$a in Ra \in R$a\in R$ is said to be irreducible if it is not a unit and cannot be expressed as a product of two non-unit elements.
• A field is a commutative ring with unity where every non-zero element has a multiplicative inverse.
Step 2: $a$$a$aa$a$ is Irreducible $⇒a$$⇒a$=>a\Rightarrow a$⇒a$ is Prime
In a Euclidean domain, every irreducible element is prime. Therefore, $a$$a$aa$a$ is a prime element in $R$$R$RR$R$.
Step 3: $a$$a$aa$a$ is Prime $⇒\left(a\right)$$⇒\left(a\right)$=>(a)\Rightarrow (a)$⇒\left(a\right)$ is a Prime Ideal
For a prime element $a$$a$aa$a$, the ideal generated by $a$$a$aa$a$, denoted $\left(a\right)$$\left(a\right)$(a)(a)$\left(a\right)$, is a prime ideal. This means that if $ab\in \left(a\right)$$ab\in \left(a\right)$ab in(a)ab \in (a)$ab\in \left(a\right)$, then either $a\in \left(a\right)$$a\in \left(a\right)$a in(a)a \in (a)$a\in \left(a\right)$ or $b\in \left(a\right)$$b\in \left(a\right)$b in(a)b \in (a)$b\in \left(a\right)$.
Step 4: $\left(a\right)$$\left(a\right)$(a)(a)$\left(a\right)$ is a Prime Ideal $⇒R/\left(a\right)$$⇒R/\left(a\right)$=>R//(a)\Rightarrow R/(a)$⇒R/\left(a\right)$ is an Integral Domain
The quotient ring $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ is an integral domain if $\left(a\right)$$\left(a\right)$(a)(a)$\left(a\right)$ is a prime ideal.
Step 5: $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ is an Integral Domain $⇒R/\left(a\right)$$⇒R/\left(a\right)$=>R//(a)\Rightarrow R/(a)$⇒R/\left(a\right)$ is a Field
In a Euclidean domain, every non-zero, non-unit element can be uniquely factored into irreducible elements. Since $a$$a$aa$a$ is irreducible, the only elements in $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ are the cosets $0+\left(a\right),1+\left(a\right),\dots ,\left(a-1\right)+\left(a\right)$$0+\left(a\right),1+\left(a\right),\dots ,\left(a-1\right)+\left(a\right)$0+(a),1+(a),dots,(a-1)+(a)0 + (a), 1 + (a), \ldots, (a-1) + (a)$0+\left(a\right),1+\left(a\right),\dots ,\left(a-1\right)+\left(a\right)$.
Every non-zero element in $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ has an inverse, making $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ a field.
Conclusion
We have shown that if $a$$a$aa$a$ is an irreducible element in a Euclidean ring $R$$R$RR$R$, then the quotient ring $R/\left(a\right)$$R/\left(a\right)$R//(a)R/(a)$R/\left(a\right)$ is a field. This completes the proof.
$$2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)$$