# 2020

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UPSC Algebra

$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

### 1. (a) Let $$S_3$$ and $$Z_3$$ be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of $$S_3$$ in $$Z_3$$ except the trivial homomorphism.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that there is no homomorphism of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ into ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ except the trivial homomorphism, we can use the following properties of homomorphisms:
1. A homomorphism $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \to H$\varphi :G\to H$ preserves the identity element, i.e., $\varphi \left({e}_{G}\right)={e}_{H}$$\varphi \left({e}_{G}\right)={e}_{H}$phi(e_(G))=e_(H)\phi(e_G) = e_H$\varphi \left({e}_{G}\right)={e}_{H}$.
2. A homomorphism $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \to H$\varphi :G\to H$ preserves the group operation, i.e., $\varphi \left(a\ast b\right)=\varphi \left(a\right)\ast \varphi \left(b\right)$$\varphi \left(a\ast b\right)=\varphi \left(a\right)\ast \varphi \left(b\right)$phi(a**b)=phi(a)**phi(b)\phi(a \ast b) = \phi(a) \ast \phi(b)$\varphi \left(a\ast b\right)=\varphi \left(a\right)\ast \varphi \left(b\right)$.
3. A homomorphism $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \to H$\varphi :G\to H$ preserves the order of elements, i.e., if $a$$a$aa$a$ has order $n$$n$nn$n$ in $G$$G$GG$G$, then $\varphi \left(a\right)$$\varphi \left(a\right)$phi(a)\phi(a)$\varphi \left(a\right)$ has order dividing $n$$n$nn$n$ in $H$$H$HH$H$.
Properties of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ and ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$
1. ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ is the permutation group on 3 symbols, and it has $3!=6$$3!=6$3!=63! = 6$3!=6$ elements.
2. ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ is the group of residue classes modulo 3, and it has 3 elements: $\left[0\right],\left[1\right],\left[2\right]$$\left[0\right],\left[1\right],\left[2\right]$[0],[1],[2][0], [1], [2]$\left[0\right],\left[1\right],\left[2\right]$.
Steps to Show No Non-Trivial Homomorphism Exists
1. Identity Element: Any homomorphism $\varphi :{S}_{3}\to {Z}_{3}$$\varphi :{S}_{3}\to {Z}_{3}$phi:S_(3)rarrZ_(3)\phi: S_3 \to Z_3$\varphi :{S}_{3}\to {Z}_{3}$ must map the identity element of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ (the identity permutation $e$$e$ee$e$) to the identity element of ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ ($\left[0\right]$$\left[0\right]$[0][0]$\left[0\right]$).
$\varphi \left(e\right)=\left[0\right]$$\varphi \left(e\right)=\left[0\right]$phi(e)=[0]\phi(e) = [0]$\varphi \left(e\right)=\left[0\right]$
2. Order of Elements: The order of any element in ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ divides 3. In ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$, we have elements of order 2 (e.g., transpositions) and elements of order 3 (e.g., 3-cycles). If there exists a non-trivial homomorphism $\varphi$$\varphi$phi\phi$\varphi$, then it must map elements of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ to elements of ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ in such a way that the order of the image divides the order of the original element.
However, ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ only has elements of order 1 ($\left[0\right]$$\left[0\right]$[0][0]$\left[0\right]$) and order 3 ($\left[1\right],\left[2\right]$$\left[1\right],\left[2\right]$[1],[2][1], [2]$\left[1\right],\left[2\right]$). There are no elements of order 2 in ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$.
3. Contradiction: ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ contains elements of order 2 (transpositions). Any homomorphism $\varphi$$\varphi$phi\phi$\varphi$ would have to map these elements to an element in ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ whose order divides 2. Since ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ contains no such elements (other than the identity), we reach a contradiction.
Therefore, the only homomorphism that can exist from ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ to ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$ is the trivial homomorphism that maps all elements of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ to the identity element $\left[0\right]$$\left[0\right]$[0][0]$\left[0\right]$ in ${Z}_{3}$${Z}_{3}$Z_(3)Z_3${Z}_{3}$.
5/5
$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$
$$a=b\:cos\:C+c\:cos\:B$$

### 2. (a) Let $$G$$ be a finite cyclic group of order $$n$$. Then prove that $$G$$ has $$\phi(n)$$ generators (where $$\phi$$ is Euler's $$\phi$$-function).

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
The problem asks us to prove that a finite cyclic group $G$$G$GG$G$ of order $n$$n$nn$n$ has $\varphi \left(n\right)$$\varphi \left(n\right)$phi(n)\phi(n)$\varphi \left(n\right)$ generators, where $\varphi$$\varphi$phi\phi$\varphi$ is Euler’s $\varphi$$\varphi$phi\phi$\varphi$-function. Euler’s $\varphi$$\varphi$phi\phi$\varphi$-function $\varphi \left(n\right)$$\varphi \left(n\right)$phi(n)\phi(n)$\varphi \left(n\right)$ is defined as the number of positive integers less than $n$$n$nn$n$ that are relatively prime to $n$$n$nn$n$.
Preliminaries
Let $G$$G$GG$G$ be a finite cyclic group of order $n$$n$nn$n$ generated by $a$$a$aa$a$. That is, $G=\left\{{a}^{0},{a}^{1},{a}^{2},\dots ,{a}^{n-1}\right\}$$G=\left\{{a}^{0},{a}^{1},{a}^{2},\dots ,{a}^{n-1}\right\}$G={a^(0),a^(1),a^(2),dots,a^(n-1)}G = \{ a^0, a^1, a^2, \ldots, a^{n-1} \}$G=\left\{{a}^{0},{a}^{1},{a}^{2},\dots ,{a}^{n-1}\right\}$.
Generators of $G$$G$GG$G$
A generator $g$$g$gg$g$ of $G$$G$GG$G$ is an element such that every element in $G$$G$GG$G$ can be written as a power of $g$$g$gg$g$. In other words, $G=\left\{{g}^{0},{g}^{1},{g}^{2},\dots ,{g}^{n-1}\right\}$$G=\left\{{g}^{0},{g}^{1},{g}^{2},\dots ,{g}^{n-1}\right\}$G={g^(0),g^(1),g^(2),dots,g^(n-1)}G = \{ g^0, g^1, g^2, \ldots, g^{n-1} \}$G=\left\{{g}^{0},{g}^{1},{g}^{2},\dots ,{g}^{n-1}\right\}$.
Euler’s $\varphi$$\varphi$phi\phi$\varphi$-Function
Euler’s $\varphi$$\varphi$phi\phi$\varphi$-function $\varphi \left(n\right)$$\varphi \left(n\right)$phi(n)\phi(n)$\varphi \left(n\right)$ counts the number of positive integers less than $n$$n$nn$n$ that are relatively prime to $n$$n$nn$n$.
Proof
1. Claim: An element ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ generates $G$$G$GG$G$ if and only if $gcd\left(k,n\right)=1$$gcd\left(k,n\right)=1$gcd(k,n)=1\gcd(k, n) = 1$gcd\left(k,n\right)=1$.
Proof of Claim:
• Forward Direction: Suppose ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ generates $G$$G$GG$G$. Then $⟨{a}^{k}⟩=G$$⟨{a}^{k}⟩=G$(:a^(k):)=G\langle a^k \rangle = G$⟨{a}^{k}⟩=G$, which means ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ has order $n$$n$nn$n$. By Lagrange’s theorem, the order of ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ must divide $n$$n$nn$n$. Since ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ has order $n$$n$nn$n$, $k$$k$kk$k$ and $n$$n$nn$n$ must be relatively prime.
• Backward Direction: Suppose $gcd\left(k,n\right)=1$$gcd\left(k,n\right)=1$gcd(k,n)=1\gcd(k, n) = 1$gcd\left(k,n\right)=1$. We want to show that ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ generates $G$$G$GG$G$. To do this, we need to show that every element ${a}^{m}$${a}^{m}$a^(m)a^m${a}^{m}$ in $G$$G$GG$G$ can be written as $\left({a}^{k}{\right)}^{r}$$\left({a}^{k}{\right)}^{r}$(a^(k))^(r)(a^k)^r$\left({a}^{k}{\right)}^{r}$ for some integer $r$$r$rr$r$.
Since $gcd\left(k,n\right)=1$$gcd\left(k,n\right)=1$gcd(k,n)=1\gcd(k, n) = 1$gcd\left(k,n\right)=1$, there exist integers $p$$p$pp$p$ and $q$$q$qq$q$ such that $pk+qn=1$$pk+qn=1$pk+qn=1pk + qn = 1$pk+qn=1$. For any ${a}^{m}$${a}^{m}$a^(m)a^m${a}^{m}$ in $G$$G$GG$G$, we have:
${a}^{m}={a}^{mpk}=\left({a}^{k}{\right)}^{mp}=\left({a}^{k}{\right)}^{r}$${a}^{m}={a}^{mpk}=\left({a}^{k}{\right)}^{mp}=\left({a}^{k}{\right)}^{r}$a^(m)=a^(mpk)=(a^(k))^(mp)=(a^(k))^(r)a^m = a^{mpk} = (a^k)^{mp} = (a^k)^r${a}^{m}={a}^{mpk}=\left({a}^{k}{\right)}^{mp}=\left({a}^{k}{\right)}^{r}$
where $r=mp$$r=mp$r=mpr = mp$r=mp$. This shows that ${a}^{k}$${a}^{k}$a^(k)a^k${a}^{k}$ generates $G$$G$GG$G$.
2. Counting Generators: The number of generators of $G$$G$GG$G$ is the same as the number of integers $k$$k$kk$k$ such that $0$00 < k < n0 < k < n$0 and $gcd\left(k,n\right)=1$$gcd\left(k,n\right)=1$gcd(k,n)=1\gcd(k, n) = 1$gcd\left(k,n\right)=1$. This is precisely $\varphi \left(n\right)$$\varphi \left(n\right)$phi(n)\phi(n)$\varphi \left(n\right)$.
Conclusion
We have proven that a finite cyclic group $G$$G$GG$G$ of order $n$$n$nn$n$ has $\varphi \left(n\right)$$\varphi \left(n\right)$phi(n)\phi(n)$\varphi \left(n\right)$ generators. The proof relies on the properties of Euler’s $\varphi$$\varphi$phi\phi$\varphi$-function and the definition of a cyclic group.
5/5
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$
$$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$$

### 3. (a) Let $$R$$ be a finite field of characteristic $$p(>0)$$. Show that the mapping $$f: R \rightarrow R$$ defined by $$f(a)=a^p, \forall a \in R$$ is an isomorphism.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
The problem asks us to prove that the mapping $f:R\to R$$f:R\to R$f:R rarr Rf: R \rightarrow R$f:R\to R$ defined by $f\left(a\right)={a}^{p}$$f\left(a\right)={a}^{p}$f(a)=a^(p)f(a) = a^p$f\left(a\right)={a}^{p}$ for all $a\in R$$a\in R$a in Ra \in R$a\in R$ is an isomorphism, where $R$$R$RR$R$ is a finite field of characteristic $p>0$$p>0$p > 0p > 0$p>0$.
Definitions
1. Field: A set $R$$R$RR$R$ with two operations $+$$+$++$+$ and $×$$×$xx\times$×$ that satisfy the field axioms.
2. Characteristic: A field $R$$R$RR$R$ has characteristic $p$$p$pp$p$ if $p$$p$pp$p$ is the smallest positive integer such that $p\cdot a=0$$p\cdot a=0$p*a=0p \cdot a = 0$p\cdot a=0$ for all $a\in R$$a\in R$a in Ra \in R$a\in R$. If no such $p$$p$pp$p$ exists, the characteristic is zero.
3. Isomorphism: A bijective map $f:R\to R$$f:R\to R$f:R rarr Rf: R \rightarrow R$f:R\to R$ that preserves the field operations.
Properties Needed
1. Frobenius Endomorphism: In a field of characteristic $p$$p$pp$p$, $\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}$$\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}$(a+b)^(p)=a^(p)+b^(p)(a+b)^p = a^p + b^p$\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}$ and $\left(ab{\right)}^{p}={a}^{p}{b}^{p}$$\left(ab{\right)}^{p}={a}^{p}{b}^{p}$(ab)^(p)=a^(p)b^(p)(ab)^p = a^p b^p$\left(ab{\right)}^{p}={a}^{p}{b}^{p}$.
Proof
To prove that $f$$f$ff$f$ is an isomorphism, we need to show that $f$$f$ff$f$ is a bijective map that preserves addition and multiplication.
1. $f$$f$ff$f$ is a Homomorphism
$f\left(a+b\right)=\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}=f\left(a\right)+f\left(b\right)$$f\left(a+b\right)=\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}=f\left(a\right)+f\left(b\right)$f(a+b)=(a+b)^(p)=a^(p)+b^(p)=f(a)+f(b)f(a+b) = (a+b)^p = a^p + b^p = f(a) + f(b)$f\left(a+b\right)=\left(a+b{\right)}^{p}={a}^{p}+{b}^{p}=f\left(a\right)+f\left(b\right)$
3. Preservation of Multiplication:
$f\left(ab\right)=\left(ab{\right)}^{p}={a}^{p}{b}^{p}=f\left(a\right)f\left(b\right)$$f\left(ab\right)=\left(ab{\right)}^{p}={a}^{p}{b}^{p}=f\left(a\right)f\left(b\right)$f(ab)=(ab)^(p)=a^(p)b^(p)=f(a)f(b)f(ab) = (ab)^p = a^p b^p = f(a) f(b)$f\left(ab\right)=\left(ab{\right)}^{p}={a}^{p}{b}^{p}=f\left(a\right)f\left(b\right)$
4. $f$$f$ff$f$ is Injective (One-to-One)
Suppose $f\left(a\right)=f\left(b\right)$$f\left(a\right)=f\left(b\right)$f(a)=f(b)f(a) = f(b)$f\left(a\right)=f\left(b\right)$. Then ${a}^{p}={b}^{p}$${a}^{p}={b}^{p}$a^(p)=b^(p)a^p = b^p${a}^{p}={b}^{p}$ which implies ${a}^{p}-{b}^{p}=0$${a}^{p}-{b}^{p}=0$a^(p)-b^(p)=0a^p – b^p = 0${a}^{p}-{b}^{p}=0$. Since $R$$R$RR$R$ is a field, it has no zero divisors, and we can factor ${a}^{p}-{b}^{p}$${a}^{p}-{b}^{p}$a^(p)-b^(p)a^p – b^p${a}^{p}-{b}^{p}$ as $\left(a-b{\right)}^{p}$$\left(a-b{\right)}^{p}$(a-b)^(p)(a-b)^p$\left(a-b{\right)}^{p}$. This means $a-b=0$$a-b=0$a-b=0a-b = 0$a-b=0$ or $a=b$$a=b$a=ba = b$a=b$, proving that $f$$f$ff$f$ is injective.
1. $f$$f$ff$f$ is Surjective (Onto)
Since $R$$R$RR$R$ is finite and $f$$f$ff$f$ is injective, $f$$f$ff$f$ must also be surjective. Alternatively, for any $b\in R$$b\in R$b in Rb \in R$b\in R$, $b={b}^{{p}^{2}}$$b={b}^{{p}^{2}}$b=b^(p^(2))b = b^{p^2}$b={b}^{{p}^{2}}$ (by Fermat’s Little Theorem or the fact that ${R}^{\ast }$${R}^{\ast }$R^(**)R^*${R}^{\ast }$, the multiplicative group of $R$$R$RR$R$, has order ${p}^{n}-1$${p}^{n}-1$p^(n)-1p^n – 1${p}^{n}-1$). Thus, $f\left({b}^{p-1}\right)=\left({b}^{p-1}{\right)}^{p}={b}^{p}=b$$f\left({b}^{p-1}\right)=\left({b}^{p-1}{\right)}^{p}={b}^{p}=b$f(b^(p-1))=(b^(p-1))^(p)=b^(p)=bf(b^{p-1}) = (b^{p-1})^p = b^p = b$f\left({b}^{p-1}\right)=\left({b}^{p-1}{\right)}^{p}={b}^{p}=b$, showing that $f$$f$ff$f$ is surjective.
Conclusion
We have shown that $f$$f$ff$f$ preserves both addition and multiplication, and is both injective and surjective. Therefore, $f:R\to R$$f:R\to R$f:R rarr Rf: R \rightarrow R$f:R\to R$ defined by $f\left(a\right)={a}^{p}$$f\left(a\right)={a}^{p}$f(a)=a^(p)f(a) = a^p$f\left(a\right)={a}^{p}$ is an isomorphism.
$$cot\:\theta =\frac{cos\:\theta }{sin\:\theta }$$