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IIT JAM 2019 Abstract Algebra: Question (2)

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General Mathematics Question : Let $G$ be a non abelian group, $y\in G$. let the maps $f, g, h$ from $G$ to itself is defined by  $$f(x)=yxy^{-1}, g(x)=x^{-1}\text{ and }h=g•g$$ Then, (A) $g$ and $h$ are homomorphism and $f$  is not homomorphism. (B) $h$ is homomorphism, $g$ is not homomorphism. (C) $f$ is homomorphism, $g$ is not homomorphism. (D) $f,g,h$ all are homomorphism. Solution :  Since $y^{-1}y=e\in G$ where $e$ is identity element in $G$. Hence we have, $f(ab)=yaby^{-1}=yay^{-1}yby^{-1}=(yay^{-1})(yby^{-1})=f(a)f(b)$ for all $a,b\in G$. Hence by definition of homomorphism $f$ is homomorphism.  Note that, $g$ is not a homomorphism since  $g(ab)=(ab)^{-1}=b^{-1}a^{-1}$. Whereas,  $g(a)g(b)=a^{-1}b^{-1}$ so that  $g(ab)$ not always equal to $g(a)g(b)$ ( since there are some $a,b\in G$ such that $b^{-1}a^{-1}≠a^{-1}b^{-1}$ as $G$ is non-abelian, otherwise $G$ will be abelian) Finally as,   $h(ab)=g•g(ab)=g(g(ab))=g(b^{-1...
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IIT-JAM 2018 Abstract Algebra Question (3)

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General Mathematics Question Solution : we know that,  Number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is $gcd(m,n)$ Now, as $221=13\times 17$. Hence, $gcd(51,221)≠1$, $gcd(91,221)≠1$, $gcd(119,221)≠1$, But $gcd(21,221)=1$ Hence number of homomorphisms from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$ is $1$ that is, there is only one homomorphism from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$. So that this must be trivial homomorphism( zero map).  Hence if $f:\mathbb{Z}_{21}\rightarrow\mathbb{Z}_{221}$ is homomorphism then it is trivial homomorphism ( zero map) i.e. $f(a)=0$ for all $a\in\mathbb{Z}_{21}$. Hence option (A) is correct.  (Note other options are not correct because in those case there exists non-trivial homomorphisms, because gcd of numbers $51,91,119$ with $221$ is not $1$.) Hence answer is "option (A)" only. Book recommend : see below, Must buy it. Nice offer must vist! 
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Series Question IIT JAM 2018

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General Mathematics Question :  Solution :   $\sum_{n=1}^\infty (-1)^{n+1}\frac{a_{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n+1+\frac{1}{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!(n+1)}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n-1)!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n+1)!}$ $=[1-\frac{1}{1!}+\frac{1}{2!}-...]+ [1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[\mathbf{1-1}+1-\frac{1}{2!}+\frac{1}{3!}-...]+[\mathbf{1-\frac{1}{1!}}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-(1-\frac{1}{1!}+\frac{1}{2!}-...)]+e^{-1}$ $=e^{-1}+[1-e^{-1}]+ e^{-1}$ $=e^{-1}+1$ Hence answer is option (D) Book recommended : Must purchase it nice offer!   For our visitors : Please Support us By...
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