# 2017

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UPSC Algebra

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

### 1(b) Let $$G$$ be a group of order $$n$$. Show that $$G$$ is isomorphic to a subgroup of the permutation group $$S_n$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that a group $G$$G$GG$G$ of order $n$$n$nn$n$ is isomorphic to a subgroup of the symmetric group ${S}_{n}$${S}_{n}$S_(n)S_n${S}_{n}$, we can use Cayley’s Theorem. Cayley’s Theorem states that every group $G$$G$GG$G$ is isomorphic to a subgroup of the symmetric group acting on $G$$G$GG$G$.
Proof:
1. Define the Action: Consider the action of $G$$G$GG$G$ on itself by left multiplication. For each $g\in G$$g\in G$g in Gg \in G$g\in G$, define a function ${f}_{g}:G\to G$${f}_{g}:G\to G$f_(g):G rarr Gf_g: G \rightarrow G${f}_{g}:G\to G$ by ${f}_{g}\left(x\right)=gx$${f}_{g}\left(x\right)=gx$f_(g)(x)=gxf_g(x) = gx${f}_{g}\left(x\right)=gx$ for all $x\in G$$x\in G$x in Gx \in G$x\in G$.
2. Show that ${f}_{g}$${f}_{g}$f_(g)f_g${f}_{g}$ is a Permutation:
• Injective: Assume ${f}_{g}\left(x\right)={f}_{g}\left(y\right)$${f}_{g}\left(x\right)={f}_{g}\left(y\right)$f_(g)(x)=f_(g)(y)f_g(x) = f_g(y)${f}_{g}\left(x\right)={f}_{g}\left(y\right)$ for some $x,y\in G$$x,y\in G$x,y in Gx, y \in G$x,y\in G$. Then $gx=gy$$gx=gy$gx=gygx = gy$gx=gy$. Since $G$$G$GG$G$ is a group, every element has an inverse, so we can multiply both sides on the left by ${g}^{-1}$${g}^{-1}$g^(-1)g^{-1}${g}^{-1}$ to get $x=y$$x=y$x=yx = y$x=y$. This shows that ${f}_{g}$${f}_{g}$f_(g)f_g${f}_{g}$ is injective.
• Surjective: For any $y\in G$$y\in G$y in Gy \in G$y\in G$, let $x={g}^{-1}y$$x={g}^{-1}y$x=g^(-1)yx = g^{-1}y$x={g}^{-1}y$. Then ${f}_{g}\left(x\right)=gx=g\left({g}^{-1}y\right)=y$${f}_{g}\left(x\right)=gx=g\left({g}^{-1}y\right)=y$f_(g)(x)=gx=g(g^(-1)y)=yf_g(x) = gx = g(g^{-1}y) = y${f}_{g}\left(x\right)=gx=g\left({g}^{-1}y\right)=y$. This shows that ${f}_{g}$${f}_{g}$f_(g)f_g${f}_{g}$ is surjective.
Since ${f}_{g}$${f}_{g}$f_(g)f_g${f}_{g}$ is both injective and surjective, it is a permutation of $G$$G$GG$G$.
3. Define a Homomorphism: Define a function $\mathrm{\Phi }:G\to {S}_{n}$$\mathrm{\Phi }:G\to {S}_{n}$Phi:G rarrS_(n)\Phi: G \rightarrow S_n$\mathrm{\Phi }:G\to {S}_{n}$ by $\mathrm{\Phi }\left(g\right)={f}_{g}$$\mathrm{\Phi }\left(g\right)={f}_{g}$Phi(g)=f_(g)\Phi(g) = f_g$\mathrm{\Phi }\left(g\right)={f}_{g}$. We need to show that $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is a homomorphism, i.e., $\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$$\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$Phi(gh)=Phi(g)Phi(h)\Phi(gh) = \Phi(g)\Phi(h)$\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$ for all $g,h\in G$$g,h\in G$g,h in Gg, h \in G$g,h\in G$.
For any $x\in G$$x\in G$x in Gx \in G$x\in G$,
$\mathrm{\Phi }\left(gh\right)\left(x\right)={f}_{gh}\left(x\right)=\left(gh\right)x=g\left(hx\right)={f}_{g}\left({f}_{h}\left(x\right)\right)=\mathrm{\Phi }\left(g\right)\left(\mathrm{\Phi }\left(h\right)\left(x\right)\right)$$\mathrm{\Phi }\left(gh\right)\left(x\right)={f}_{gh}\left(x\right)=\left(gh\right)x=g\left(hx\right)={f}_{g}\left({f}_{h}\left(x\right)\right)=\mathrm{\Phi }\left(g\right)\left(\mathrm{\Phi }\left(h\right)\left(x\right)\right)$Phi(gh)(x)=f_(gh)(x)=(gh)x=g(hx)=f_(g)(f_(h)(x))=Phi(g)(Phi(h)(x))\Phi(gh)(x) = f_{gh}(x) = (gh)x = g(hx) = f_g(f_h(x)) = \Phi(g)(\Phi(h)(x))$\mathrm{\Phi }\left(gh\right)\left(x\right)={f}_{gh}\left(x\right)=\left(gh\right)x=g\left(hx\right)={f}_{g}\left({f}_{h}\left(x\right)\right)=\mathrm{\Phi }\left(g\right)\left(\mathrm{\Phi }\left(h\right)\left(x\right)\right)$
This shows that $\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$$\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$Phi(gh)=Phi(g)Phi(h)\Phi(gh) = \Phi(g)\Phi(h)$\mathrm{\Phi }\left(gh\right)=\mathrm{\Phi }\left(g\right)\mathrm{\Phi }\left(h\right)$, so $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is a homomorphism.
4. Show that $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is Injective:
Assume $\mathrm{\Phi }\left(g\right)=\mathrm{\Phi }\left(h\right)$$\mathrm{\Phi }\left(g\right)=\mathrm{\Phi }\left(h\right)$Phi(g)=Phi(h)\Phi(g) = \Phi(h)$\mathrm{\Phi }\left(g\right)=\mathrm{\Phi }\left(h\right)$ for some $g,h\in G$$g,h\in G$g,h in Gg, h \in G$g,h\in G$. Then ${f}_{g}={f}_{h}$${f}_{g}={f}_{h}$f_(g)=f_(h)f_g = f_h${f}_{g}={f}_{h}$, which means $gx=hx$$gx=hx$gx=hxgx = hx$gx=hx$ for all $x\in G$$x\in G$x in Gx \in G$x\in G$. Letting $x={g}^{-1}$$x={g}^{-1}$x=g^(-1)x = g^{-1}$x={g}^{-1}$, we get $g\left({g}^{-1}\right)=h\left({g}^{-1}\right)$$g\left({g}^{-1}\right)=h\left({g}^{-1}\right)$g(g^(-1))=h(g^(-1))g(g^{-1}) = h(g^{-1})$g\left({g}^{-1}\right)=h\left({g}^{-1}\right)$, which simplifies to $e=h{g}^{-1}$$e=h{g}^{-1}$e=hg^(-1)e = hg^{-1}$e=h{g}^{-1}$, where $e$$e$ee$e$ is the identity element in $G$$G$GG$G$. Multiplying both sides on the right by $g$$g$gg$g$, we get $g=h$$g=h$g=hg = h$g=h$. This shows that $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is injective.
5. Conclusion: Since $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is an injective homomorphism from $G$$G$GG$G$ to ${S}_{n}$${S}_{n}$S_(n)S_n${S}_{n}$, the image of $\mathrm{\Phi }$$\mathrm{\Phi }$Phi\Phi$\mathrm{\Phi }$ is a subgroup of ${S}_{n}$${S}_{n}$S_(n)S_n${S}_{n}$ that is isomorphic to $G$$G$GG$G$. This completes the proof.
5/5
$$c=a\:cos\:B+b\:cos\:A$$
$$cos\:2\theta =2\:cos^2\theta -1$$

### 2(c) Let $$F$$ be a field and $$F[X]$$ denote the ring of polynomials over $$F$$ in a single variable $$X$$. For $$f(X), g(X) \in F[X]$$ with $$g(X) \neq 0$$, show that there exist $$q(X), r(X) \in F[X]$$ such that degree $$(r(X))<\text{degree}(g(X))$$ and $$f(X)=q(X) \cdot g(X)+r(X)$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that for any two polynomials $f\left(X\right),g\left(X\right)\in F\left[X\right]$$f\left(X\right),g\left(X\right)\in F\left[X\right]$f(X),g(X)in F[X]f(X), g(X) \in F[X]$f\left(X\right),g\left(X\right)\in F\left[X\right]$ with $g\left(X\right)\ne 0$$g\left(X\right)\ne 0$g(X)!=0g(X) \neq 0$g\left(X\right)\ne 0$, there exist $q\left(X\right),r\left(X\right)\in F\left[X\right]$$q\left(X\right),r\left(X\right)\in F\left[X\right]$q(X),r(X)in F[X]q(X), r(X) \in F[X]$q\left(X\right),r\left(X\right)\in F\left[X\right]$ such that $\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$ and $f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$, we can use the division algorithm for polynomials. The proof is constructive and provides a method to find $q\left(X\right)$$q\left(X\right)$q(X)q(X)$q\left(X\right)$ and $r\left(X\right)$$r\left(X\right)$r(X)r(X)$r\left(X\right)$.
Proof:
1. Base Case: If $\text{degree}\left(f\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(f\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(f(X)) < “degree”(g(X))\text{degree}(f(X)) < \text{degree}(g(X))$\text{degree}\left(f\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$, then we can directly take $q\left(X\right)=0$$q\left(X\right)=0$q(X)=0q(X) = 0$q\left(X\right)=0$ and $r\left(X\right)=f\left(X\right)$$r\left(X\right)=f\left(X\right)$r(X)=f(X)r(X) = f(X)$r\left(X\right)=f\left(X\right)$. In this case, the condition $\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$ is satisfied.
2. Inductive Step: Assume that $\text{degree}\left(f\left(X\right)\right)\ge \text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(f\left(X\right)\right)\ge \text{degree}\left(g\left(X\right)\right)$“degree”(f(X)) >= “degree”(g(X))\text{degree}(f(X)) \geq \text{degree}(g(X))$\text{degree}\left(f\left(X\right)\right)\ge \text{degree}\left(g\left(X\right)\right)$. Let $n=\text{degree}\left(f\left(X\right)\right)$$n=\text{degree}\left(f\left(X\right)\right)$n=”degree”(f(X))n = \text{degree}(f(X))$n=\text{degree}\left(f\left(X\right)\right)$ and $m=\text{degree}\left(g\left(X\right)\right)$$m=\text{degree}\left(g\left(X\right)\right)$m=”degree”(g(X))m = \text{degree}(g(X))$m=\text{degree}\left(g\left(X\right)\right)$. Write the polynomials in the form:
$f\left(X\right)={a}_{n}{X}^{n}+{a}_{n-1}{X}^{n-1}+\dots +{a}_{1}X+{a}_{0}$$f\left(X\right)={a}_{n}{X}^{n}+{a}_{n-1}{X}^{n-1}+\dots +{a}_{1}X+{a}_{0}$f(X)=a_(n)X^(n)+a_(n-1)X^(n-1)+dots+a_(1)X+a_(0)f(X) = a_nX^n + a_{n-1}X^{n-1} + \ldots + a_1X + a_0$f\left(X\right)={a}_{n}{X}^{n}+{a}_{n-1}{X}^{n-1}+\dots +{a}_{1}X+{a}_{0}$
$g\left(X\right)={b}_{m}{X}^{m}+{b}_{m-1}{X}^{m-1}+\dots +{b}_{1}X+{b}_{0}$$g\left(X\right)={b}_{m}{X}^{m}+{b}_{m-1}{X}^{m-1}+\dots +{b}_{1}X+{b}_{0}$g(X)=b_(m)X^(m)+b_(m-1)X^(m-1)+dots+b_(1)X+b_(0)g(X) = b_mX^m + b_{m-1}X^{m-1} + \ldots + b_1X + b_0$g\left(X\right)={b}_{m}{X}^{m}+{b}_{m-1}{X}^{m-1}+\dots +{b}_{1}X+{b}_{0}$
where ${a}_{n},{b}_{m}\ne 0$${a}_{n},{b}_{m}\ne 0$a_(n),b_(m)!=0a_n, b_m \neq 0${a}_{n},{b}_{m}\ne 0$.
3. Construct the Quotient: Consider the polynomial $h\left(X\right)=\frac{{a}_{n}}{{b}_{m}}{X}^{n-m}\cdot g\left(X\right)$$h\left(X\right)=\frac{{a}_{n}}{{b}_{m}}{X}^{n-m}\cdot g\left(X\right)$h(X)=(a_(n))/(b_(m))X^(n-m)*g(X)h(X) = \frac{a_n}{b_m}X^{n-m} \cdot g(X)$h\left(X\right)=\frac{{a}_{n}}{{b}_{m}}{X}^{n-m}\cdot g\left(X\right)$. The leading term of $h\left(X\right)$$h\left(X\right)$h(X)h(X)$h\left(X\right)$ will cancel out the leading term of $f\left(X\right)$$f\left(X\right)$f(X)f(X)$f\left(X\right)$ when subtracted. Define a new polynomial ${f}_{1}\left(X\right)=f\left(X\right)-h\left(X\right)$${f}_{1}\left(X\right)=f\left(X\right)-h\left(X\right)$f_(1)(X)=f(X)-h(X)f_1(X) = f(X) – h(X)${f}_{1}\left(X\right)=f\left(X\right)-h\left(X\right)$.
4. Recursive Application: Now, $\text{degree}\left({f}_{1}\left(X\right)\right)<\text{degree}\left(f\left(X\right)\right)$$\text{degree}\left({f}_{1}\left(X\right)\right)<\text{degree}\left(f\left(X\right)\right)$“degree”(f_(1)(X)) < “degree”(f(X))\text{degree}(f_1(X)) < \text{degree}(f(X))$\text{degree}\left({f}_{1}\left(X\right)\right)<\text{degree}\left(f\left(X\right)\right)$. Apply the division algorithm to ${f}_{1}\left(X\right)$${f}_{1}\left(X\right)$f_(1)(X)f_1(X)${f}_{1}\left(X\right)$ and $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$. By induction, there exist ${q}_{1}\left(X\right),r\left(X\right)\in F\left[X\right]$${q}_{1}\left(X\right),r\left(X\right)\in F\left[X\right]$q_(1)(X),r(X)in F[X]q_1(X), r(X) \in F[X]${q}_{1}\left(X\right),r\left(X\right)\in F\left[X\right]$ such that $\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$ and ${f}_{1}\left(X\right)={q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)$${f}_{1}\left(X\right)={q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)$f_(1)(X)=q_(1)(X)*g(X)+r(X)f_1(X) = q_1(X) \cdot g(X) + r(X)${f}_{1}\left(X\right)={q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)$.
5. Combine Results: Now, we can express $f\left(X\right)$$f\left(X\right)$f(X)f(X)$f\left(X\right)$ as:
$f\left(X\right)=h\left(X\right)+{f}_{1}\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)=\left(h\left(X\right)+{q}_{1}\left(X\right)\right)\cdot g\left(X\right)+r\left(X\right)$$f\left(X\right)=h\left(X\right)+{f}_{1}\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)=\left(h\left(X\right)+{q}_{1}\left(X\right)\right)\cdot g\left(X\right)+r\left(X\right)$f(X)=h(X)+f_(1)(X)=h(X)+q_(1)(X)*g(X)+r(X)=(h(X)+q_(1)(X))*g(X)+r(X)f(X) = h(X) + f_1(X) = h(X) + q_1(X) \cdot g(X) + r(X) = (h(X) + q_1(X)) \cdot g(X) + r(X)$f\left(X\right)=h\left(X\right)+{f}_{1}\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)\cdot g\left(X\right)+r\left(X\right)=\left(h\left(X\right)+{q}_{1}\left(X\right)\right)\cdot g\left(X\right)+r\left(X\right)$
Let $q\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)$$q\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)$q(X)=h(X)+q_(1)(X)q(X) = h(X) + q_1(X)$q\left(X\right)=h\left(X\right)+{q}_{1}\left(X\right)$. Then $f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$ with $\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$.
6. Conclusion: We have shown that for any $f\left(X\right),g\left(X\right)\in F\left[X\right]$$f\left(X\right),g\left(X\right)\in F\left[X\right]$f(X),g(X)in F[X]f(X), g(X) \in F[X]$f\left(X\right),g\left(X\right)\in F\left[X\right]$ with $g\left(X\right)\ne 0$$g\left(X\right)\ne 0$g(X)!=0g(X) \neq 0$g\left(X\right)\ne 0$, there exist $q\left(X\right),r\left(X\right)\in F\left[X\right]$$q\left(X\right),r\left(X\right)\in F\left[X\right]$q(X),r(X)in F[X]q(X), r(X) \in F[X]$q\left(X\right),r\left(X\right)\in F\left[X\right]$ such that $\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$“degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))$\text{degree}\left(r\left(X\right)\right)<\text{degree}\left(g\left(X\right)\right)$ and $f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)$f\left(X\right)=q\left(X\right)\cdot g\left(X\right)+r\left(X\right)$. This completes the proof.
5/5
$$a=b\:cos\:C+c\:cos\:B$$
$$cos\:2\theta =cos^2\theta -sin^2\theta$$

### 3(a) Show that the groups $$\mathbb{Z}_5 \times \mathbb{Z}_7$$ and $$\mathbb{Z}_{35}$$ are isomorphic.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the groups ${\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$${\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$Z_(5)xxZ_(7)\mathbb{Z}_5 \times \mathbb{Z}_7${\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$ and ${\mathbb{Z}}_{35}$${\mathbb{Z}}_{35}$Z_(35)\mathbb{Z}_{35}${\mathbb{Z}}_{35}$ are isomorphic, we need to find a bijective function $f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$$f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$f:Z_(5)xxZ_(7)rarrZ_(35)f: \mathbb{Z}_5 \times \mathbb{Z}_7 \rightarrow \mathbb{Z}_{35}$f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$ such that $f\left(a+b\right)=f\left(a\right)+f\left(b\right)$$f\left(a+b\right)=f\left(a\right)+f\left(b\right)$f(a+b)=f(a)+f(b)f(a+b) = f(a) + f(b)$f\left(a+b\right)=f\left(a\right)+f\left(b\right)$ for all $a,b\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$$a,b\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$a,b inZ_(5)xxZ_(7)a, b \in \mathbb{Z}_5 \times \mathbb{Z}_7$a,b\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$.
Step 1: Define the Function $f$$f$ff$f$
Let’s define the function $f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$$f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$f:Z_(5)xxZ_(7)rarrZ_(35)f: \mathbb{Z}_5 \times \mathbb{Z}_7 \rightarrow \mathbb{Z}_{35}$f:{\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}\to {\mathbb{Z}}_{35}$ as follows:
$f\left(a,b\right)=7a+5b$$f\left(a,b\right)=7a+5b$f(a,b)=7a+5bf(a, b) = 7a + 5b$f\left(a,b\right)=7a+5b$
Step 2: Show that $f$$f$ff$f$ is Well-Defined
To show that $f$$f$ff$f$ is well-defined, we need to show that $f\left(a,b\right)$$f\left(a,b\right)$f(a,b)f(a, b)$f\left(a,b\right)$ is in ${\mathbb{Z}}_{35}$${\mathbb{Z}}_{35}$Z_(35)\mathbb{Z}_{35}${\mathbb{Z}}_{35}$ for all $\left(a,b\right)\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$$\left(a,b\right)\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$(a,b)inZ_(5)xxZ_(7)(a, b) \in \mathbb{Z}_5 \times \mathbb{Z}_7$\left(a,b\right)\in {\mathbb{Z}}_{5}×{\mathbb{Z}}_{7}$.
Let’s substitute the values:
• $a$$a$aa$a$ can be $0,1,2,3,4$$0,1,2,3,4$0,1,2,3,40, 1, 2, 3, 4$0,1,2,3,4$ (from ${\mathbb{Z}}_{5}$${\mathbb{Z}}_{5}$Z_(5)\mathbb{Z}_5${\mathbb{Z}}_{5}$)
• $b$$b$bb$b$ can be $0,1,2,3,4,5,6$$0,1,2,3,4,5,6$0,1,2,3,4,5,60, 1, 2, 3, 4, 5, 6$0,1,2,3,4,5,6$ (from ${\mathbb{Z}}_{7}$${\mathbb{Z}}_{7}$Z_(7)\mathbb{Z}_7${\mathbb{Z}}_{7}$)
After substituting these values into $f\left(a,b\right)=7a+5b$$f\left(a,b\right)=7a+5b$f(a,b)=7a+5bf(a, b) = 7a + 5b$f\left(a,b\right)=7a+5b$, we find that $f\left(a,b\right)$$f\left(a,b\right)$f(a,b)f(a, b)$f\left(a,b\right)$ can range from $0$$0$00$0$ to $34$$34$3434$34$, which is exactly the set ${\mathbb{Z}}_{35}$${\mathbb{Z}}_{35}$Z_(35)\mathbb{Z}_{35}${\mathbb{Z}}_{35}$.
Step 3: Show that $f$$f$ff$f$ is a Homomorphism
To show that $f$$f$ff$f$ is a homomorphism, we need to show that $f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=f\left({a}_{1},{b}_{1}\right)+f\left({a}_{2},{b}_{2}\right)$$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=f\left({a}_{1},{b}_{1}\right)+f\left({a}_{2},{b}_{2}\right)$f(a_(1)+a_(2),b_(1)+b_(2))=f(a_(1),b_(1))+f(a_(2),b_(2))f(a_1 + a_2, b_1 + b_2) = f(a_1, b_1) + f(a_2, b_2)$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=f\left({a}_{1},{b}_{1}\right)+f\left({a}_{2},{b}_{2}\right)$.
Let’s substitute the values:
$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=7\left({a}_{1}+{a}_{2}\right)+5\left({b}_{1}+{b}_{2}\right)$$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=7\left({a}_{1}+{a}_{2}\right)+5\left({b}_{1}+{b}_{2}\right)$f(a_(1)+a_(2),b_(1)+b_(2))=7(a_(1)+a_(2))+5(b_(1)+b_(2))f(a_1 + a_2, b_1 + b_2) = 7(a_1 + a_2) + 5(b_1 + b_2)$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=7\left({a}_{1}+{a}_{2}\right)+5\left({b}_{1}+{b}_{2}\right)$
After calculating, we get:
$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=7{a}_{1}+7{a}_{2}+5{b}_{1}+5{b}_{2}=f\left({a}_{1},{b}_{1}\right)+f\left({a}_{2},{b}_{2}\right)$$f\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2}\right)=7{a}_{1}+7{a}_{2}+5{b}_{1}+5{b}_{2}=f\left({a}_{1},{b}_{1}\right)+f\left({a}_{2},{b}_{2}\right)$f(a_(1)+a_(2),b_(1)+b_(2))=7a_(1)+7a_(2)+5b_(1)+5b_(2)=f(a_(1),b_(1))+f(a_(2),b_(2))f(a_1 + a_2, b_1 + b_2) = 7a_1 + 7a_2 + 5b_1 + 5b_2 = f(a_1, b_1) + f(a_2, b_2)