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Mathematics 14 Online
OpenStudy (anonymous):

here is the question

OpenStudy (agl202):

Hum...

OpenStudy (anonymous):

Wheres the question?

OpenStudy (anonymous):

like tf

OpenStudy (anonymous):

XD

OpenStudy (anonymous):

@everyone please help

OpenStudy (anonymous):

@perl

OpenStudy (perl):

i might need an abstract algebra book

OpenStudy (perl):

we know that $$ \Large \log x + \log y = \log ( x * y )$$

OpenStudy (perl):

A function \( \bf f: G \to H \) from a group G to a group H is a homomorphism (or a group map) if \( \bf f(ab) = f(a)f(b)\) for all \( \bf a, b \in G \).

OpenStudy (perl):

actually the only way to make this work would be to define g(x) = e^x , thats why im confused

OpenStudy (perl):

exp(x + y) = exp(x) * exp(y)

OpenStudy (perl):

exp(x) means \( \bf e^x \)

OpenStudy (anonymous):

@perl

OpenStudy (anonymous):

@perl please number 2,6,8

OpenStudy (anonymous):

@perl

OpenStudy (anonymous):

@perl

OpenStudy (anonymous):

@perl

OpenStudy (perl):

i dont understand how these questions are entered

OpenStudy (anonymous):

you have the file

OpenStudy (perl):

you want the kernel of e^x ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

is it 0 or 1 or e

OpenStudy (perl):

The kernel of a group homomorphism is the set of all elements of which are mapped to the identity element of

OpenStudy (anonymous):

so what is the element then?

OpenStudy (perl):

kernel is {0}

OpenStudy (perl):

because e^0 = 1

OpenStudy (anonymous):

what about number 5

OpenStudy (anonymous):

i guess A

OpenStudy (anonymous):

and al so , A bijective homomorphism is called endomorphism Exomorphism Isomorphism Automorphism

OpenStudy (anonymous):

is it D

OpenStudy (perl):

for five, did you test the possibiloities

OpenStudy (perl):

let g(a) = 2a now test g(a+b) = g(a)g(b)

OpenStudy (anonymous):

waw, i really wish you are my real teacher

OpenStudy (anonymous):

thanks.

OpenStudy (perl):

$$ \Large g: \mathbb{(Z,+) \to (2\mathbb{Z, +}) \\g(x) = 2x \\ \\ \therefore \ \\ g(a+b) = 2(a+b) \\g(a)g(b) = 2a + 2b } $$

OpenStudy (perl):

you have to think about operations abstractly

OpenStudy (anonymous):

thank you so much @perl .. you are a genius lol

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