# 2011

Estimated reading: 65 minutes 14 views

UPSC Algebra

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

### 1(a) Show that the set $G=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\}$ of six transformations on the set of Complex numbers defined by \begin{aligned} &f_1(z)=z, f_2(z)=1-z \\ &f_3(z)=\frac{z}{(z-1)}, f_4(z)=\frac{1}{z} \\ &f_5(z)=\frac{1}{(1-z)} \text { and } f_6(z)=\frac{(z-1)}{z} \end{aligned} is a non-abelian group of order 6 w.r.t. composition of mappings.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set $G=\left\{{f}_{1},{f}_{2},{f}_{3},{f}_{4},{f}_{5},{f}_{6}\right\}$$G=\left\{{f}_{1},{f}_{2},{f}_{3},{f}_{4},{f}_{5},{f}_{6}\right\}$G={f_(1),f_(2),f_(3),f_(4),f_(5),f_(6)}G = \{ f_1, f_2, f_3, f_4, f_5, f_6 \}$G=\left\{{f}_{1},{f}_{2},{f}_{3},{f}_{4},{f}_{5},{f}_{6}\right\}$ forms a group under composition of mappings, we need to verify the following group axioms:
1. Closure
We need to show that the composition of any two functions in $G$$G$GG$G$ is also in $G$$G$GG$G$.
For example, let’s consider ${f}_{3}\circ {f}_{4}$${f}_{3}\circ {f}_{4}$f_(3)@f_(4)f_3 \circ f_4${f}_{3}\circ {f}_{4}$. The composition ${f}_{3}\left({f}_{4}\left(z\right)\right)$${f}_{3}\left({f}_{4}\left(z\right)\right)$f_(3)(f_(4)(z))f_3(f_4(z))${f}_{3}\left({f}_{4}\left(z\right)\right)$ is given by:
${f}_{3}\left({f}_{4}\left(z\right)\right)={f}_{3}\left(\frac{1}{z}\right)=\frac{\frac{1}{z}}{\frac{1}{z}-1}=\frac{1}{1-z}$${f}_{3}\left({f}_{4}\left(z\right)\right)={f}_{3}\left(\frac{1}{z}\right)=\frac{\frac{1}{z}}{\frac{1}{z}-1}=\frac{1}{1-z}$f_(3)(f_(4)(z))=f_(3)((1)/(z))=((1)/(z))/((1)/(z)-1)=(1)/(1-z)f_3(f_4(z)) = f_3\left(\frac{1}{z}\right) = \frac{\frac{1}{z}}{\frac{1}{z} – 1} = \frac{1}{1 – z}${f}_{3}\left({f}_{4}\left(z\right)\right)={f}_{3}\left(\frac{1}{z}\right)=\frac{\frac{1}{z}}{\frac{1}{z}-1}=\frac{1}{1-z}$
After calculating, we find that ${f}_{3}\circ {f}_{4}={f}_{5}$${f}_{3}\circ {f}_{4}={f}_{5}$f_(3)@f_(4)=f_(5)f_3 \circ f_4 = f_5${f}_{3}\circ {f}_{4}={f}_{5}$, which is in $G$$G$GG$G$.
Similarly, we can show that the composition of any two functions in $G$$G$GG$G$ will result in another function in $G$$G$GG$G$. Therefore, $G$$G$GG$G$ is closed under composition.
1. Associativity
Composition of functions is always associative. That is, for any $f,g,h\in G$$f,g,h\in G$f,g,h in Gf, g, h \in G$f,g,h\in G$, $\left(f\circ g\right)\circ h=f\circ \left(g\circ h\right)$$\left(f\circ g\right)\circ h=f\circ \left(g\circ h\right)$(f@g)@h=f@(g@h)(f \circ g) \circ h = f \circ (g \circ h)$\left(f\circ g\right)\circ h=f\circ \left(g\circ h\right)$.
1. Identity Element
The function ${f}_{1}\left(z\right)=z$${f}_{1}\left(z\right)=z$f_(1)(z)=zf_1(z) = z${f}_{1}\left(z\right)=z$ serves as the identity element for $G$$G$GG$G$ under composition. For any $f\in G$$f\in G$f in Gf \in G$f\in G$, ${f}_{1}\circ f=f$${f}_{1}\circ f=f$f_(1)@f=ff_1 \circ f = f${f}_{1}\circ f=f$ and $f\circ {f}_{1}=f$$f\circ {f}_{1}=f$f@f_(1)=ff \circ f_1 = f$f\circ {f}_{1}=f$.
1. Inverse Element
Each function in $G$$G$GG$G$ must have an inverse that is also in $G$$G$GG$G$.
• ${f}_{1}$${f}_{1}$f_(1)f_1${f}_{1}$ is its own inverse.
• ${f}_{2}$${f}_{2}$f_(2)f_2${f}_{2}$ is its own inverse.
• ${f}_{3}$${f}_{3}$f_(3)f_3${f}_{3}$ and ${f}_{6}$${f}_{6}$f_(6)f_6${f}_{6}$ are inverses of each other.
• ${f}_{4}$${f}_{4}$f_(4)f_4${f}_{4}$ and ${f}_{5}$${f}_{5}$f_(5)f_5${f}_{5}$ are inverses of each other.
1. Non-Abelian
To show that $G$$G$GG$G$ is non-abelian, we need to find $f,g\in G$$f,g\in G$f,g in Gf, g \in G$f,g\in G$ such that $f\circ g\ne g\circ f$$f\circ g\ne g\circ f$f@g!=g@ff \circ g \neq g \circ f$f\circ g\ne g\circ f$.
Let’s consider ${f}_{3}$${f}_{3}$f_(3)f_3${f}_{3}$ and ${f}_{4}$${f}_{4}$f_(4)f_4${f}_{4}$:
• ${f}_{3}\circ {f}_{4}={f}_{5}$${f}_{3}\circ {f}_{4}={f}_{5}$f_(3)@f_(4)=f_(5)f_3 \circ f_4 = f_5${f}_{3}\circ {f}_{4}={f}_{5}$
• ${f}_{4}\circ {f}_{3}={f}_{6}$${f}_{4}\circ {f}_{3}={f}_{6}$f_(4)@f_(3)=f_(6)f_4 \circ f_3 = f_6${f}_{4}\circ {f}_{3}={f}_{6}$
Since ${f}_{5}\ne {f}_{6}$${f}_{5}\ne {f}_{6}$f_(5)!=f_(6)f_5 \neq f_6${f}_{5}\ne {f}_{6}$, $G$$G$GG$G$ is non-abelian.
Conclusion
Since $G$$G$GG$G$ satisfies all the group axioms and is non-abelian, it is a non-abelian group of order 6 with respect to composition of mappings.
5/5
$$c=a\:cos\:B+b\:cos\:A$$
$$\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}$$

### 1(e) (i) Prove that a group of Prime order is abelian.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that a group $G$$G$GG$G$ of prime order $p$$p$pp$p$ is abelian, we need to show that for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$, $a\cdot b=b\cdot a$$a\cdot b=b\cdot a$a*b=b*aa \cdot b = b \cdot a$a\cdot b=b\cdot a$.
Step 1: Group Order is Prime
Let $p$$p$pp$p$ be a prime number, and let $G$$G$GG$G$ be a group of order $p$$p$pp$p$.
Step 2: Elements in the Group
Since $G$$G$GG$G$ has order $p$$p$pp$p$, it has $p$$p$pp$p$ elements. One of these elements is the identity element $e$$e$ee$e$.
Step 3: Properties of Subgroups
By Lagrange’s theorem, the order of any subgroup of $G$$G$GG$G$ must divide the order of $G$$G$GG$G$. Since $p$$p$pp$p$ is prime, the only divisors of $p$$p$pp$p$ are 1 and $p$$p$pp$p$ itself. Therefore, any subgroup of $G$$G$GG$G$ must have order 1 or $p$$p$pp$p$.
Step 4: Subgroups are Trivial or the Whole Group
A subgroup of order 1 is trivial and contains only the identity element $e$$e$ee$e$. A subgroup of order $p$$p$pp$p$ must be $G$$G$GG$G$ itself.
Step 5: Every Element Generates the Group
Take any element $a\in G$$a\in G$a in Ga \in G$a\in G$ where $a\ne e$$a\ne e$a!=ea \neq e$a\ne e$. The subgroup generated by $a$$a$aa$a$, denoted $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$, must have order greater than 1 (since $a\ne e$$a\ne e$a!=ea \neq e$a\ne e$). By Step 4, $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ must therefore be the whole group $G$$G$GG$G$.
Step 6: Commutativity
Take any two elements $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$. Both $a$$a$aa$a$ and $b$$b$bb$b$ generate $G$$G$GG$G$ (by Step 5). Therefore, $b$$b$bb$b$ can be written as ${a}^{n}$${a}^{n}$a^(n)a^n${a}^{n}$ for some integer $n$$n$nn$n$, and $a$$a$aa$a$ can be written as ${b}^{m}$${b}^{m}$b^(m)b^m${b}^{m}$ for some integer $m$$m$mm$m$.
Now, $a\cdot b=a\cdot {a}^{n}={a}^{n+1}$$a\cdot b=a\cdot {a}^{n}={a}^{n+1}$a*b=a*a^(n)=a^(n+1)a \cdot b = a \cdot a^n = a^{n+1}$a\cdot b=a\cdot {a}^{n}={a}^{n+1}$ and $b\cdot a={a}^{n}\cdot a={a}^{n+1}$$b\cdot a={a}^{n}\cdot a={a}^{n+1}$b*a=a^(n)*a=a^(n+1)b \cdot a = a^n \cdot a = a^{n+1}$b\cdot a={a}^{n}\cdot a={a}^{n+1}$.
Since $a\cdot b=b\cdot a$$a\cdot b=b\cdot a$a*b=b*aa \cdot b = b \cdot a$a\cdot b=b\cdot a$, $G$$G$GG$G$ is abelian.
Conclusion
We have shown that any group of prime order $p$$p$pp$p$ must be abelian.
5/5
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$
$$cot\:\theta =\frac{cos\:\theta }{sin\:\theta }$$

### 1(e) (ii) How many generators are there of the cyclic group $$(G, .)$$ of order 8?

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In a cyclic group $G$$G$GG$G$ of order $n$$n$nn$n$, the number of generators is equal to the number of integers $k$$k$kk$k$ that are relatively prime to $n$$n$nn$n$. These integers $k$$k$kk$k$ are in the range $1\le k$1\le k1 <= k < n1 \leq k < n$1\le k.
Step 1: Identify the Order of the Group
The order of the group $G$$G$GG$G$ is given as 8.
Step 2: Find Integers Relatively Prime to 8
We need to find the integers $k$$k$kk$k$ that are relatively prime to 8 and lie in the range $1\le k<8$$1\le k<8$1 <= k < 81 \leq k < 8$1\le k<8$.
The integers in this range are $\left\{1,2,3,4,5,6,7\right\}$$\left\{1,2,3,4,5,6,7\right\}${1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}$\left\{1,2,3,4,5,6,7\right\}$.
Out of these, the integers that are relatively prime to 8 are $\left\{1,3,5,7\right\}$$\left\{1,3,5,7\right\}${1,3,5,7}\{1, 3, 5, 7\}$\left\{1,3,5,7\right\}$.
Step 3: Count the Number of Generators
The number of generators of $G$$G$GG$G$ is equal to the number of integers that are relatively prime to 8, which is 4.
Conclusion
There are 4 generators of the cyclic group $G$$G$GG$G$ of order 8.
5/5
$$cos\:2\theta =1-2\:sin^2\theta$$
$$cot\:\theta =\frac{cos\:\theta }{sin\:\theta }$$

### 2(a) Give an example of a group $$G$$ in which every proper subgroup is cyclic but the group itself is not cyclic.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
A classic example of a group $G$$G$GG$G$ where every proper subgroup is cyclic but $G$$G$GG$G$ itself is not cyclic is the Klein four-group ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$.
Definition of ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$
The Klein four-group ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$ is defined as the set $\left\{e,a,b,c\right\}$$\left\{e,a,b,c\right\}${e,a,b,c}\{ e, a, b, c \}$\left\{e,a,b,c\right\}$ with the following multiplication table:
$\begin{array}{ccccc}\cdot & e& a& b& c\\ e& e& a& b& c\\ a& a& e& c& b\\ b& b& c& e& a\\ c& c& b& a& e\end{array}$$\begin{array}{ccccc}\cdot & e& a& b& c\\ e& e& a& b& c\\ a& a& e& c& b\\ b& b& c& e& a\\ c& c& b& a& e\end{array}${:[*,e,a,b,c],[e,e,a,b,c],[a,a,e,c,b],[b,b,c,e,a],[c,c,b,a,e]:}\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \end{array}$\begin{array}{ccccc}\cdot & e& a& b& c\\ e& e& a& b& c\\ a& a& e& c& b\\ b& b& c& e& a\\ c& c& b& a& e\end{array}$
Proper Subgroups are Cyclic
The proper subgroups of ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$ are:
1. $\left\{e,a\right\}$$\left\{e,a\right\}${e,a}\{ e, a \}$\left\{e,a\right\}$ generated by $a$$a$aa$a$
2. $\left\{e,b\right\}$$\left\{e,b\right\}${e,b}\{ e, b \}$\left\{e,b\right\}$ generated by $b$$b$bb$b$
3. $\left\{e,c\right\}$$\left\{e,c\right\}${e,c}\{ e, c \}$\left\{e,c\right\}$ generated by $c$$c$cc$c$
Each of these subgroups is cyclic, as they are generated by a single element.
${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$ is Not Cyclic
The group ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$ itself is not cyclic because there is no single element that generates the entire group. Each of $a,b,c$$a,b,c$a,b,ca, b, c$a,b,c$ only generates a subgroup of order 2, and $e$$e$ee$e$ generates only the trivial subgroup $\left\{e\right\}$$\left\{e\right\}${e}\{ e \}$\left\{e\right\}$.
Conclusion
The Klein four-group ${V}_{4}$${V}_{4}$V_(4)V_4${V}_{4}$ is an example of a group where every proper subgroup is cyclic, but the group itself is not cyclic.
5/5
$$2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$
$$a=b\:cos\:C+c\:cos\:B$$

### 3(a) Let $$F$$ be the set of all real-valued continuous functions defined on the closed interval $$[0,1]$$. Prove that $$(F,+,.)$$ is a Commutative Ring with unity with respect to addition and multiplication of functions defined pointwise.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that $\left(F,+,\cdot \right)$$\left(F,+,\cdot \right)$(F,+,*)(F, +, \cdot)$\left(F,+,\cdot \right)$ is a commutative ring with unity, we need to verify the following properties:
1. Closure under Addition and Multiplication
For any $f,g\in F$$f,g\in F$f,g in Ff, g \in F$f,g\in F$, $f+g$$f+g$f+gf+g$f+g$ and $f\cdot g$$f\cdot g$f*gf \cdot g$f\cdot g$ are also real-valued continuous functions defined on $\left[0,1\right]$$\left[0,1\right]$[0,1][0, 1]$\left[0,1\right]$. Therefore, $F$$F$FF$F$ is closed under addition and multiplication.
1. Associativity of Addition and Multiplication
For any $f,g,h\in F$$f,g,h\in F$f,g,h in Ff, g, h \in F$f,g,h\in F$,
$\left(f+g\right)+h=f+\left(g+h\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(f\cdot g\right)\cdot h=f\cdot \left(g\cdot h\right)$$\left(f+g\right)+h=f+\left(g+h\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(f\cdot g\right)\cdot h=f\cdot \left(g\cdot h\right)$(f+g)+h=f+(g+h)quad”and”quad(f*g)*h=f*(g*h)(f + g) + h = f + (g + h) \quad \text{and} \quad (f \cdot g) \cdot h = f \cdot (g \cdot h)$\left(f+g\right)+h=f+\left(g+h\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(f\cdot g\right)\cdot h=f\cdot \left(g\cdot h\right)$
1. Commutativity of Addition and Multiplication
For any $f,g\in F$$f,g\in F$f,g in Ff, g \in F$f,g\in F$,
$f+g=g+f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot g=g\cdot f$$f+g=g+f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot g=g\cdot f$f+g=g+f quad”and”quad f*g=g*ff + g = g + f \quad \text{and} \quad f \cdot g = g \cdot f$f+g=g+f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot g=g\cdot f$
1. Existence of Additive and Multiplicative Identity (Unity)
The zero function $f\left(x\right)=0$$f\left(x\right)=0$f(x)=0f(x) = 0$f\left(x\right)=0$ for all $x\in \left[0,1\right]$$x\in \left[0,1\right]$x in[0,1]x \in [0, 1]$x\in \left[0,1\right]$ serves as the additive identity, and the function $f\left(x\right)=1$$f\left(x\right)=1$f(x)=1f(x) = 1$f\left(x\right)=1$ for all $x\in \left[0,1\right]$$x\in \left[0,1\right]$x in[0,1]x \in [0, 1]$x\in \left[0,1\right]$ serves as the multiplicative identity (unity).
For any $f\in F$$f\in F$f in Ff \in F$f\in F$,
$f+0=f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot 1=f$$f+0=f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot 1=f$f+0=f quad”and”quad f*1=ff + 0 = f \quad \text{and} \quad f \cdot 1 = f$f+0=f\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\cdot 1=f$
For any $f\in F$$f\in F$f in Ff \in F$f\in F$, the function $-f$$-f$-f-f$-f$ defined by $-f\left(x\right)=-f\left(x\right)$$-f\left(x\right)=-f\left(x\right)$-f(x)=-f(x)-f(x) = -f(x)$-f\left(x\right)=-f\left(x\right)$ for all $x\in \left[0,1\right]$$x\in \left[0,1\right]$x in[0,1]x \in [0, 1]$x\in \left[0,1\right]$ is the additive inverse of $f$$f$ff$f$.
$f+\left(-f\right)=0$$f+\left(-f\right)=0$f+(-f)=0f + (-f) = 0$f+\left(-f\right)=0$
1. Distributive Law
For any $f,g,h\in F$$f,g,h\in F$f,g,h in Ff, g, h \in F$f,g,h\in F$,
$f\cdot \left(g+h\right)=\left(f\cdot g\right)+\left(f\cdot h\right)$$f\cdot \left(g+h\right)=\left(f\cdot g\right)+\left(f\cdot h\right)$f*(g+h)=(f*g)+(f*h)f \cdot (g + h) = (f \cdot g) + (f \cdot h)$f\cdot \left(g+h\right)=\left(f\cdot g\right)+\left(f\cdot h\right)$
Conclusion
Since all these properties are satisfied, $\left(F,+,\cdot \right)$$\left(F,+,\cdot \right)$(F,+,*)(F, +, \cdot)$\left(F,+,\cdot \right)$ is a commutative ring with unity with respect to addition and multiplication of functions defined pointwise.
5/5
$$sin\left(\theta -\phi \right)=sin\:\theta \:cos\:\phi -cos\:\theta \:sin\:\phi$$
$$sin\:3\theta =3\:sin\:\theta -4\:sin^3\:\theta$$

### 4(a) Let $$a$$ and $$b$$ be the elements of a group, with $$a^2=e$$, $$b^6=e$$ and $$a b = b^4 a$$. Find the order of $$ab$$, and express its inverse in each of the.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To find the order of $ab$$ab$abab$ab$ and its inverse, we need to consider the given relations:
1. ${a}^{2}=e$${a}^{2}=e$a^(2)=ea^2 = e${a}^{2}=e$
2. ${b}^{6}=e$${b}^{6}=e$b^(6)=eb^6 = e${b}^{6}=e$
3. $ab={b}^{4}a$$ab={b}^{4}a$ab=b^(4)aab = b^4a$ab={b}^{4}a$
Step 1: Find the Order of $ab$$ab$abab$ab$
To find the order of $ab$$ab$abab$ab$, we need to find the smallest positive integer $n$$n$nn$n$ such that $\left(ab{\right)}^{n}=e$$\left(ab{\right)}^{n}=e$(ab)^(n)=e(ab)^n = e$\left(ab{\right)}^{n}=e$.
Let’s substitute the values into the formula and simplify:
$\left(ab{\right)}^{2}=abab$$\left(ab{\right)}^{2}=abab$(ab)^(2)=abab(ab)^2 = abab$\left(ab{\right)}^{2}=abab$
Using the relation $ab={b}^{4}a$$ab={b}^{4}a$ab=b^(4)aab = b^4a$ab={b}^{4}a$, we get:
$\left(ab{\right)}^{2}={b}^{4}a\cdot {b}^{4}a={b}^{4}\cdot {b}^{4}\cdot a\cdot a={b}^{8}\cdot {a}^{2}={b}^{2}\cdot e={b}^{2}$$\left(ab{\right)}^{2}={b}^{4}a\cdot {b}^{4}a={b}^{4}\cdot {b}^{4}\cdot a\cdot a={b}^{8}\cdot {a}^{2}={b}^{2}\cdot e={b}^{2}$(ab)^(2)=b^(4)a*b^(4)a=b^(4)*b^(4)*a*a=b^(8)*a^(2)=b^(2)*e=b^(2)(ab)^2 = b^4a \cdot b^4a = b^4 \cdot b^4 \cdot a \cdot a = b^8 \cdot a^2 = b^2 \cdot e = b^2$\left(ab{\right)}^{2}={b}^{4}a\cdot {b}^{4}a={b}^{4}\cdot {b}^{4}\cdot a\cdot a={b}^{8}\cdot {a}^{2}={b}^{2}\cdot e={b}^{2}$
Similarly, we can find $\left(ab{\right)}^{3}$$\left(ab{\right)}^{3}$(ab)^(3)(ab)^3$\left(ab{\right)}^{3}$:
$\left(ab{\right)}^{3}=\left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{4}a={b}^{6}\cdot a=e\cdot a=a$$\left(ab{\right)}^{3}=\left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{4}a={b}^{6}\cdot a=e\cdot a=a$(ab)^(3)=(ab)^(2)*ab=b^(2)*b^(4)a=b^(6)*a=e*a=a(ab)^3 = (ab)^2 \cdot ab = b^2 \cdot b^4a = b^6 \cdot a = e \cdot a = a$\left(ab{\right)}^{3}=\left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{4}a={b}^{6}\cdot a=e\cdot a=a$
Finally, let’s find $\left(ab{\right)}^{6}$$\left(ab{\right)}^{6}$(ab)^(6)(ab)^6$\left(ab{\right)}^{6}$:
$\left(ab{\right)}^{6}=\left(ab{\right)}^{3}\cdot \left(ab{\right)}^{3}=a\cdot a={a}^{2}=e$$\left(ab{\right)}^{6}=\left(ab{\right)}^{3}\cdot \left(ab{\right)}^{3}=a\cdot a={a}^{2}=e$(ab)^(6)=(ab)^(3)*(ab)^(3)=a*a=a^(2)=e(ab)^6 = (ab)^3 \cdot (ab)^3 = a \cdot a = a^2 = e$\left(ab{\right)}^{6}=\left(ab{\right)}^{3}\cdot \left(ab{\right)}^{3}=a\cdot a={a}^{2}=e$
So, the smallest positive integer $n$$n$nn$n$ such that $\left(ab{\right)}^{n}=e$$\left(ab{\right)}^{n}=e$(ab)^(n)=e(ab)^n = e$\left(ab{\right)}^{n}=e$ is $n=6$$n=6$n=6n = 6$n=6$. Therefore, the order of $ab$$ab$abab$ab$ is 6.
Step 2: Find the Inverse of $ab$$ab$abab$ab$
The inverse of an element $x$$x$xx$x$ in a group is an element ${x}^{-1}$${x}^{-1}$x^(-1)x^{-1}${x}^{-1}$ such that $x\cdot {x}^{-1}=e$$x\cdot {x}^{-1}=e$x*x^(-1)=ex \cdot x^{-1} = e$x\cdot {x}^{-1}=e$.
To find $\left(ab{\right)}^{-1}$$\left(ab{\right)}^{-1}$(ab)^(-1)(ab)^{-1}$\left(ab{\right)}^{-1}$, we need to find an element that, when multiplied by $ab$$ab$abab$ab$, gives $e$$e$ee$e$.
Since $\left(ab{\right)}^{6}=e$$\left(ab{\right)}^{6}=e$(ab)^(6)=e(ab)^6 = e$\left(ab{\right)}^{6}=e$, we know that $\left(ab{\right)}^{-1}=\left(ab{\right)}^{5}$$\left(ab{\right)}^{-1}=\left(ab{\right)}^{5}$(ab)^(-1)=(ab)^(5)(ab)^{-1} = (ab)^5$\left(ab{\right)}^{-1}=\left(ab{\right)}^{5}$.
Let’s substitute the values into the formula and simplify:
$\left(ab{\right)}^{5}=\left(ab{\right)}^{2}\cdot \left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{2}\cdot {b}^{4}a={b}^{8}\cdot a={b}^{2}\cdot a$$\left(ab{\right)}^{5}=\left(ab{\right)}^{2}\cdot \left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{2}\cdot {b}^{4}a={b}^{8}\cdot a={b}^{2}\cdot a$(ab)^(5)=(ab)^(2)*(ab)^(2)*ab=b^(2)*b^(2)*b^(4)a=b^(8)*a=b^(2)*a(ab)^5 = (ab)^2 \cdot (ab)^2 \cdot ab = b^2 \cdot b^2 \cdot b^4a = b^8 \cdot a = b^2 \cdot a$\left(ab{\right)}^{5}=\left(ab{\right)}^{2}\cdot \left(ab{\right)}^{2}\cdot ab={b}^{2}\cdot {b}^{2}\cdot {b}^{4}a={b}^{8}\cdot a={b}^{2}\cdot a$
After calculating, we find that $\left(ab{\right)}^{-1}={b}^{2}a$$\left(ab{\right)}^{-1}={b}^{2}a$(ab)^(-1)=b^(2)a(ab)^{-1} = b^2a$\left(ab{\right)}^{-1}={b}^{2}a$.
Conclusion
The order of $ab$$ab$abab$ab$ is 6, and its inverse is ${b}^{2}a$${b}^{2}a$b^(2)ab^2a${b}^{2}a$.
$$cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta$$