# 2010

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UPSC Algebra

$$2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

### 1(a) Let $$G=\mathbb{I}-\{-1\}$$ be the set of all real numbers omitting $$-1$$. Define the binary relation $$*$$ on $$G$$ by $$a * b=a+b+a b$$. Show $$(G, *)$$ is a group and it is abelian.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that $\left(G,\ast \right)$$\left(G,\ast \right)$(G,**)(G, *)$\left(G,\ast \right)$ is a group, we need to verify the following group axioms:
1. Closure under $\ast$$\ast$***$\ast$
For any $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$, $a\ast b=a+b+ab$$a\ast b=a+b+ab$a**b=a+b+aba * b = a + b + ab$a\ast b=a+b+ab$ is also a real number. However, we need to ensure that $a\ast b\ne -1$$a\ast b\ne -1$a**b!=-1a * b \neq -1$a\ast b\ne -1$.
If $a\ast b=-1$$a\ast b=-1$a**b=-1a * b = -1$a\ast b=-1$, then $a+b+ab=-1$$a+b+ab=-1$a+b+ab=-1a + b + ab = -1$a+b+ab=-1$, which simplifies to $ab+a+b+1=0$$ab+a+b+1=0$ab+a+b+1=0ab + a + b + 1 = 0$ab+a+b+1=0$ or $\left(a+1\right)\left(b+1\right)=0$$\left(a+1\right)\left(b+1\right)=0$(a+1)(b+1)=0(a + 1)(b + 1) = 0$\left(a+1\right)\left(b+1\right)=0$.
Since $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$, $a\ne -1$$a\ne -1$a!=-1a \neq -1$a\ne -1$ and $b\ne -1$$b\ne -1$b!=-1b \neq -1$b\ne -1$, so $a+1\ne 0$$a+1\ne 0$a+1!=0a + 1 \neq 0$a+1\ne 0$ and $b+1\ne 0$$b+1\ne 0$b+1!=0b + 1 \neq 0$b+1\ne 0$. Therefore, $a\ast b\ne -1$$a\ast b\ne -1$a**b!=-1a * b \neq -1$a\ast b\ne -1$, and $G$$G$GG$G$ is closed under $\ast$$\ast$***$\ast$.
1. Associativity
For any $a,b,c\in G$$a,b,c\in G$a,b,c in Ga, b, c \in G$a,b,c\in G$,
$\left(a\ast b\right)\ast c=\left(a+b+ab\right)\ast c=a+b+ab+c+\left(a+b+ab\right)c=a+b+c+ab+ac+bc+abc$$\left(a\ast b\right)\ast c=\left(a+b+ab\right)\ast c=a+b+ab+c+\left(a+b+ab\right)c=a+b+c+ab+ac+bc+abc$(a**b)**c=(a+b+ab)**c=a+b+ab+c+(a+b+ab)c=a+b+c+ab+ac+bc+abc(a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + c + ab + ac + bc + abc$\left(a\ast b\right)\ast c=\left(a+b+ab\right)\ast c=a+b+ab+c+\left(a+b+ab\right)c=a+b+c+ab+ac+bc+abc$
$a\ast \left(b\ast c\right)=a\ast \left(b+c+bc\right)=a+b+c+bc+a\left(b+c+bc\right)=a+b+c+ab+ac+bc+abc$$a\ast \left(b\ast c\right)=a\ast \left(b+c+bc\right)=a+b+c+bc+a\left(b+c+bc\right)=a+b+c+ab+ac+bc+abc$a**(b**c)=a**(b+c+bc)=a+b+c+bc+a(b+c+bc)=a+b+c+ab+ac+bc+abca * (b * c) = a * (b + c + bc) = a + b + c + bc + a(b + c + bc) = a + b + c + ab + ac + bc + abc$a\ast \left(b\ast c\right)=a\ast \left(b+c+bc\right)=a+b+c+bc+a\left(b+c+bc\right)=a+b+c+ab+ac+bc+abc$
Since $\left(a\ast b\right)\ast c=a\ast \left(b\ast c\right)$$\left(a\ast b\right)\ast c=a\ast \left(b\ast c\right)$(a**b)**c=a**(b**c)(a * b) * c = a * (b * c)$\left(a\ast b\right)\ast c=a\ast \left(b\ast c\right)$, the operation $\ast$$\ast$***$\ast$ is associative.
1. Existence of Identity Element
We need to find an element $e\in G$$e\in G$e in Ge \in G$e\in G$ such that for any $a\in G$$a\in G$a in Ga \in G$a\in G$, $a\ast e=a$$a\ast e=a$a**e=aa * e = a$a\ast e=a$.
Let $a\ast e=a$$a\ast e=a$a**e=aa * e = a$a\ast e=a$, then $a+e+ae=a$$a+e+ae=a$a+e+ae=aa + e + ae = a$a+e+ae=a$, which simplifies to $e+ae=0$$e+ae=0$e+ae=0e + ae = 0$e+ae=0$ or $e\left(1+a\right)=0$$e\left(1+a\right)=0$e(1+a)=0e(1 + a) = 0$e\left(1+a\right)=0$.
Since $a\ne -1$$a\ne -1$a!=-1a \neq -1$a\ne -1$, $1+a\ne 0$$1+a\ne 0$1+a!=01 + a \neq 0$1+a\ne 0$, and therefore $e=0$$e=0$e=0e = 0$e=0$.
1. Existence of Inverse Element
For each $a\in G$$a\in G$a in Ga \in G$a\in G$, we need to find an element ${a}^{-1}\in G$${a}^{-1}\in G$a^(-1)in Ga^{-1} \in G${a}^{-1}\in G$ such that $a\ast {a}^{-1}=e$$a\ast {a}^{-1}=e$a**a^(-1)=ea * a^{-1} = e$a\ast {a}^{-1}=e$.
Let $a\ast {a}^{-1}=e$$a\ast {a}^{-1}=e$a**a^(-1)=ea * a^{-1} = e$a\ast {a}^{-1}=e$, then $a+{a}^{-1}+a{a}^{-1}=0$$a+{a}^{-1}+a{a}^{-1}=0$a+a^(-1)+aa^(-1)=0a + a^{-1} + aa^{-1} = 0$a+{a}^{-1}+a{a}^{-1}=0$, which simplifies to ${a}^{-1}\left(1+a\right)=-a$${a}^{-1}\left(1+a\right)=-a$a^(-1)(1+a)=-aa^{-1}(1 + a) = -a${a}^{-1}\left(1+a\right)=-a$.
Since $a\ne -1$$a\ne -1$a!=-1a \neq -1$a\ne -1$, $1+a\ne 0$$1+a\ne 0$1+a!=01 + a \neq 0$1+a\ne 0$, and therefore ${a}^{-1}=-\frac{a}{a+1}$${a}^{-1}=-\frac{a}{a+1}$a^(-1)=-(a)/(a+1)a^{-1} = -\frac{a}{a + 1}${a}^{-1}=-\frac{a}{a+1}$.
1. Commutativity
For any $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$,
$a\ast b=a+b+ab=b+a+ba=b\ast a$$a\ast b=a+b+ab=b+a+ba=b\ast a$a**b=a+b+ab=b+a+ba=b**aa * b = a + b + ab = b + a + ba = b * a$a\ast b=a+b+ab=b+a+ba=b\ast a$
Since $a\ast b=b\ast a$$a\ast b=b\ast a$a**b=b**aa * b = b * a$a\ast b=b\ast a$, $\left(G,\ast \right)$$\left(G,\ast \right)$(G,**)(G, *)$\left(G,\ast \right)$ is abelian.
Conclusion
Since all these properties are satisfied, $\left(G,\ast \right)$$\left(G,\ast \right)$(G,**)(G, *)$\left(G,\ast \right)$ is an abelian group.
5/5
$$cos\:2\theta =1-2\:sin^2\theta$$
$$cot\:\theta =\frac{cos\:\theta }{sin\:\theta }$$

### 1(b) Show that a cyclic group of order 6 is isomorphic to the product of a cyclic group of order 2 and a cyclic group of order 3. Can you generalize this? Justify.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Define the Groups
Let $G$$G$GG$G$ be a cyclic group of order 6, generated by an element $a$$a$aa$a$, i.e., $G=⟨a⟩=\left\{e,a,{a}^{2},{a}^{3},{a}^{4},{a}^{5}\right\}$$G=⟨a⟩=\left\{e,a,{a}^{2},{a}^{3},{a}^{4},{a}^{5}\right\}$G=(:a:)={e,a,a^(2),a^(3),a^(4),a^(5)}G = \langle a \rangle = \{ e, a, a^2, a^3, a^4, a^5 \}$G=⟨a⟩=\left\{e,a,{a}^{2},{a}^{3},{a}^{4},{a}^{5}\right\}$.
Let $H$$H$HH$H$ be a cyclic group of order 2, generated by an element $b$$b$bb$b$, i.e., $H=⟨b⟩=\left\{e,b\right\}$$H=⟨b⟩=\left\{e,b\right\}$H=(:b:)={e,b}H = \langle b \rangle = \{ e, b \}$H=⟨b⟩=\left\{e,b\right\}$.
Let $K$$K$KK$K$ be a cyclic group of order 3, generated by an element $c$$c$cc$c$, i.e., $K=⟨c⟩=\left\{e,c,{c}^{2}\right\}$$K=⟨c⟩=\left\{e,c,{c}^{2}\right\}$K=(:c:)={e,c,c^(2)}K = \langle c \rangle = \{ e, c, c^2 \}$K=⟨c⟩=\left\{e,c,{c}^{2}\right\}$.
Step 2: Define the Product Group
Consider the product group $H×K$$H×K$H xx KH \times K$H×K$, consisting of elements $\left({b}^{i},{c}^{j}\right)$$\left({b}^{i},{c}^{j}\right)$(b^(i),c^(j))(b^i, c^j)$\left({b}^{i},{c}^{j}\right)$ where $0\le i<2$$0\le i<2$0 <= i < 20 \leq i < 2$0\le i<2$ and $0\le j<3$$0\le j<3$0 <= j < 30 \leq j < 3$0\le j<3$.
Step 3: Define the Isomorphism
Define a function $\varphi :G\to H×K$$\varphi :G\to H×K$phi:G rarr H xx K\phi: G \to H \times K$\varphi :G\to H×K$ by $\varphi \left({a}^{n}\right)=\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$$\varphi \left({a}^{n}\right)=\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$phi(a^(n))=(b^(nmod2),c^(nmod3))\phi(a^n) = (b^{n \mod 2}, c^{n \mod 3})$\varphi \left({a}^{n}\right)=\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$.
Step 4: Show $\varphi$$\varphi$phi\phi$\varphi$ is an Isomorphism
1. Well-defined: $\varphi$$\varphi$phi\phi$\varphi$ is well-defined because $n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2$$n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2$nmod2n \mod 2$n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2$ and $n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3$$n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3$nmod3n \mod 3$n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3$ are always in the correct ranges for $H$$H$HH$H$ and $K$$K$KK$K$, respectively.
2. Bijective: $\varphi$$\varphi$phi\phi$\varphi$ is bijective because each element ${a}^{n}$${a}^{n}$a^(n)a^n${a}^{n}$ in $G$$G$GG$G$ maps to a unique element $\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$$\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$(b^(nmod2),c^(nmod3))(b^{n \mod 2}, c^{n \mod 3})$\left({b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)$ in $H×K$$H×K$H xx KH \times K$H×K$, and vice versa.
3. Homomorphism: $\varphi$$\varphi$phi\phi$\varphi$ is a homomorphism because for any ${a}^{m},{a}^{n}\in G$${a}^{m},{a}^{n}\in G$a^(m),a^(n)in Ga^m, a^n \in G${a}^{m},{a}^{n}\in G$,
$\varphi \left({a}^{m}\ast {a}^{n}\right)=\varphi \left({a}^{m+n}\right)=\left({b}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\left({b}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2}\ast {b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\ast {c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\varphi \left({a}^{m}\right)\ast \varphi \left({a}^{n}\right)$$\varphi \left({a}^{m}\ast {a}^{n}\right)=\varphi \left({a}^{m+n}\right)=\left({b}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\left({b}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2}\ast {b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\ast {c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\varphi \left({a}^{m}\right)\ast \varphi \left({a}^{n}\right)$phi(a^(m)**a^(n))=phi(a^(m+n))=(b^((m+n)mod2),c^((m+n)mod3))=(b^(mmod2)**b^(nmod2),c^(mmod3)**c^(nmod3))=phi(a^(m))**phi(a^(n))\phi(a^m * a^n) = \phi(a^{m+n}) = (b^{(m+n) \mod 2}, c^{(m+n) \mod 3}) = (b^{m \mod 2} * b^{n \mod 2}, c^{m \mod 3} * c^{n \mod 3}) = \phi(a^m) * \phi(a^n)$\varphi \left({a}^{m}\ast {a}^{n}\right)=\varphi \left({a}^{m+n}\right)=\left({b}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{\left(m+n\right)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\left({b}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2}\ast {b}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2},{c}^{m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\ast {c}^{n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3}\right)=\varphi \left({a}^{m}\right)\ast \varphi \left({a}^{n}\right)$
Generalization
This result can be generalized to say that a cyclic group of order $mn$$mn$mnmn$mn$ where $m$$m$mm$m$ and $n$$n$nn$n$ are coprime is isomorphic to the product of a cyclic group of order $m$$m$mm$m$ and a cyclic group of order $n$$n$nn$n$.
Justification for Generalization
1. Chinese Remainder Theorem: If $m$$m$mm$m$ and $n$$n$nn$n$ are coprime, then the map $\varphi :{\mathbb{Z}}_{mn}\to {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$$\varphi :{\mathbb{Z}}_{mn}\to {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$phi:Z_(mn)rarrZ_(m)xxZ_(n)\phi: \mathbb{Z}_{mn} \to \mathbb{Z}_m \times \mathbb{Z}_n$\varphi :{\mathbb{Z}}_{mn}\to {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$ defined by $\varphi \left(a\right)=\left(a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m,a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\right)$$\varphi \left(a\right)=\left(a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m,a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\right)$phi(a)=(amodm,amodn)\phi(a) = (a \mod m, a \mod n)$\varphi \left(a\right)=\left(a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m,a\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\right)$ is an isomorphism.
2. Cyclic Groups: A cyclic group of order $mn$$mn$mnmn$mn$ is isomorphic to ${\mathbb{Z}}_{mn}$${\mathbb{Z}}_{mn}$Z_(mn)\mathbb{Z}_{mn}${\mathbb{Z}}_{mn}$, and cyclic groups of orders $m$$m$mm$m$ and $n$$n$nn$n$ are isomorphic to ${\mathbb{Z}}_{m}$${\mathbb{Z}}_{m}$Z_(m)\mathbb{Z}_m${\mathbb{Z}}_{m}$ and ${\mathbb{Z}}_{n}$${\mathbb{Z}}_{n}$Z_(n)\mathbb{Z}_n${\mathbb{Z}}_{n}$, respectively.
3. Isomorphism: Combining these, we get that a cyclic group of order $mn$$mn$mnmn$mn$ is isomorphic to the product of a cyclic group of order $m$$m$mm$m$ and a cyclic group of order $n$$n$nn$n$.
Therefore, we can generalize that a cyclic group of order $mn$$mn$mnmn$mn$ is isomorphic to the product of a cyclic group of order $m$$m$mm$m$ and a cyclic group of order $n$$n$nn$n$ when $m$$m$mm$m$ and $n$$n$nn$n$ are coprime.
5/5
$$sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta$$
$$a=b\:cos\:C+c\:cos\:B$$

### 2(a) Let $$\left(\mathbb{R}^*, \cdot\right)$$ be the multiplicative group of nonzero reals and $$(G L(n, I R), X)$$ be the multiplicative group of $$n \times n$$ non-singular matrices. Show that the quotient group $$G L(n, \mathbb{R}) / S L(n, \mathbb{R})$$ and $$\left(\mathbb{R}^* \cdot \cdot\right)$$ are isomorphic where $$S L(n, \mathrm{IR})=\{A \in G L(n, \mathrm{IR}) / \operatorname{det} A=1\}$$. What is the centre of $$G L(n$$, IR ) ?

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Define the Groups and Quotient Group
• ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$ is the set of all nonzero real numbers with multiplication as the operation.
• $GL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)$GL(n,R)GL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)$ is the set of all $n×n$$n×n$n xx nn \times n$n×n$ invertible matrices with matrix multiplication as the operation.
• $SL\left(n,\mathbb{R}\right)$$SL\left(n,\mathbb{R}\right)$SL(n,R)SL(n, \mathbb{R})$SL\left(n,\mathbb{R}\right)$ is the set of all $n×n$$n×n$n xx nn \times n$n×n$ matrices with determinant 1, i.e., $SL\left(n,\mathbb{R}\right)=\left\{A\in GL\left(n,\mathbb{R}\right)\mid \text{det}\left(A\right)=1\right\}$$SL\left(n,\mathbb{R}\right)=\left\{A\in GL\left(n,\mathbb{R}\right)\mid \text{det}\left(A\right)=1\right\}$SL(n,R)={A in GL(n,R)∣”det”(A)=1}SL(n, \mathbb{R}) = \{ A \in GL(n, \mathbb{R}) \mid \text{det}(A) = 1 \}$SL\left(n,\mathbb{R}\right)=\left\{A\in GL\left(n,\mathbb{R}\right)\mid \text{det}\left(A\right)=1\right\}$.
The quotient group $GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$ consists of cosets $A\cdot SL\left(n,\mathbb{R}\right)$$A\cdot SL\left(n,\mathbb{R}\right)$A*SL(n,R)A \cdot SL(n, \mathbb{R})$A\cdot SL\left(n,\mathbb{R}\right)$ where $A\in GL\left(n,\mathbb{R}\right)$$A\in GL\left(n,\mathbb{R}\right)$A in GL(n,R)A \in GL(n, \mathbb{R})$A\in GL\left(n,\mathbb{R}\right)$.
Step 2: Define the Isomorphism
Define a function $\varphi :GL\left(n,\mathbb{R}\right)\to {\mathbb{R}}^{\ast }$$\varphi :GL\left(n,\mathbb{R}\right)\to {\mathbb{R}}^{\ast }$phi:GL(n,R)rarrR^(**)\phi: GL(n, \mathbb{R}) \to \mathbb{R}^*$\varphi :GL\left(n,\mathbb{R}\right)\to {\mathbb{R}}^{\ast }$ by $\varphi \left(A\right)=\text{det}\left(A\right)$$\varphi \left(A\right)=\text{det}\left(A\right)$phi(A)=”det”(A)\phi(A) = \text{det}(A)$\varphi \left(A\right)=\text{det}\left(A\right)$.
Step 3: Show $\varphi$$\varphi$phi\phi$\varphi$ Induces an Isomorphism between $GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$ and ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$
1. Well-defined: $\varphi$$\varphi$phi\phi$\varphi$ is well-defined because the determinant of any $n×n$$n×n$n xx nn \times n$n×n$ invertible matrix is a nonzero real number.
2. Homomorphism: $\varphi$$\varphi$phi\phi$\varphi$ is a homomorphism because $\text{det}\left(AB\right)=\text{det}\left(A\right)\text{det}\left(B\right)$$\text{det}\left(AB\right)=\text{det}\left(A\right)\text{det}\left(B\right)$“det”(AB)=”det”(A)”det”(B)\text{det}(AB) = \text{det}(A) \text{det}(B)$\text{det}\left(AB\right)=\text{det}\left(A\right)\text{det}\left(B\right)$.
3. Kernel: The kernel of $\varphi$$\varphi$phi\phi$\varphi$ is $SL\left(n,\mathbb{R}\right)$$SL\left(n,\mathbb{R}\right)$SL(n,R)SL(n, \mathbb{R})$SL\left(n,\mathbb{R}\right)$ because $\varphi \left(A\right)=1$$\varphi \left(A\right)=1$phi(A)=1\phi(A) = 1$\varphi \left(A\right)=1$ if and only if $\text{det}\left(A\right)=1$$\text{det}\left(A\right)=1$“det”(A)=1\text{det}(A) = 1$\text{det}\left(A\right)=1$, i.e., $A\in SL\left(n,\mathbb{R}\right)$$A\in SL\left(n,\mathbb{R}\right)$A in SL(n,R)A \in SL(n, \mathbb{R})$A\in SL\left(n,\mathbb{R}\right)$.
4. Induced Isomorphism: By the First Isomorphism Theorem, $GL\left(n,\mathbb{R}\right)/\text{Ker}\left(\varphi \right)$$GL\left(n,\mathbb{R}\right)/\text{Ker}\left(\varphi \right)$GL(n,R)//”Ker”(phi)GL(n, \mathbb{R}) / \text{Ker}(\phi)$GL\left(n,\mathbb{R}\right)/\text{Ker}\left(\varphi \right)$ is isomorphic to the image of $\varphi$$\varphi$phi\phi$\varphi$, which is ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$. Therefore, $GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$ is isomorphic to ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$.
Step 4: Find the Center of $GL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)$GL(n,R)GL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)$
The center $Z\left(GL\left(n,\mathbb{R}\right)\right)$$Z\left(GL\left(n,\mathbb{R}\right)\right)$Z(GL(n,R))Z(GL(n, \mathbb{R}))$Z\left(GL\left(n,\mathbb{R}\right)\right)$ of $GL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)$GL(n,R)GL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)$ consists of all matrices $A$$A$AA$A$ such that $AB=BA$$AB=BA$AB=BAAB = BA$AB=BA$ for all $B\in GL\left(n,\mathbb{R}\right)$$B\in GL\left(n,\mathbb{R}\right)$B in GL(n,R)B \in GL(n, \mathbb{R})$B\in GL\left(n,\mathbb{R}\right)$.
For $n>1$$n>1$n > 1n > 1$n>1$, the center consists only of scalar multiples of the identity matrix, i.e., $Z\left(GL\left(n,\mathbb{R}\right)\right)=\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}$$Z\left(GL\left(n,\mathbb{R}\right)\right)=\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}$Z(GL(n,R))={cI∣c inR^(**)}Z(GL(n, \mathbb{R})) = \{ cI \mid c \in \mathbb{R}^* \}$Z\left(GL\left(n,\mathbb{R}\right)\right)=\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}$.
For $n=1$$n=1$n=1n = 1$n=1$, $GL\left(1,\mathbb{R}\right)$$GL\left(1,\mathbb{R}\right)$GL(1,R)GL(1, \mathbb{R})$GL\left(1,\mathbb{R}\right)$ is essentially ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$, and every element commutes with every other element, so the center is $GL\left(1,\mathbb{R}\right)$$GL\left(1,\mathbb{R}\right)$GL(1,R)GL(1, \mathbb{R})$GL\left(1,\mathbb{R}\right)$ itself.
Conclusion
The quotient group $GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)/SL\left(n,\mathbb{R}\right)$ is isomorphic to ${\mathbb{R}}^{\ast }$${\mathbb{R}}^{\ast }$R^(**)\mathbb{R}^*${\mathbb{R}}^{\ast }$, and the center of $GL\left(n,\mathbb{R}\right)$$GL\left(n,\mathbb{R}\right)$GL(n,R)GL(n, \mathbb{R})$GL\left(n,\mathbb{R}\right)$ is $\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}$$\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}${cI∣c inR^(**)}\{ cI \mid c \in \mathbb{R}^* \}$\left\{cI\mid c\in {\mathbb{R}}^{\ast }\right\}$ for $n>1$$n>1$n > 1n > 1$n>1$ and $GL\left(1,\mathbb{R}\right)$$GL\left(1,\mathbb{R}\right)$GL(1,R)GL(1, \mathbb{R})$GL\left(1,\mathbb{R}\right)$ for $n=1$$n=1$n=1n = 1$n=1$.
5/5
$$2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$
$$Sin^2\left(\theta \:\right)+Cos^2\left(\theta \right)=1$$