IIT JAM 2019 Abstract Algebra: Question (2)

By
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, 

 $h(ab)=g•g(ab)=g(g(ab))=g(b^{-1}a^{-1})=(a^{-1})^{-1}(b^{-1})^{-1}=ab=h(a)h(b)$ for all $a,b\in G$. 

Hence by definition of homomorphism, $h$ is a homomorphism from $G$ to itself. (Note here $h$ is identity map from $G$ to itself)

Hence answer is, option (B), (C). 
















Comments

Post a Comment

Popular posts from this blog

IIT-JAM 2018 Abstract Algebra Question (3)

By
Image
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! 
Read more

Question on Negation of Definition of limit of Sequence

By
Image
Question Press/click below hints, answer "buttons" for solution:   Do you really need hint? Give your 100% first! If still you have problem then check hint 2 Hint 1 "given that $\lim_{x \to +\infty} a_{n}= 0$. Hence by definition of limit of sequence we have, for given any $\epsilon >0$ there exists $K\in\mathbb{N}$ such that, $|a_n-0|<\epsilon$ i.e. given any $\epsilon >0$ there exists $K\in\mathbb{N}$ such that, $|a_n|<\epsilon$ "The question ask negation of above statement" Which is given by "there exists an $\epsilon>0$ such that for every $K\in\mathbb{N}$ there exists $n>K$ such that $|a_n|\geq\epsilon$ from this can you conclude the required answer ? Hint 2 By Hint2 we conclude that the answer is option (a) Can you discard the other options? yes! Find counterexample for discarding other options. Did you find any? ("I will not do spoon feeding" If You hav...
Read more