Mathematics

Kasmir Khaan

12:04 AM

@LeakyNun hi =p

Leaky Nun

@KasmirKhaan ?

Kasmir Khaan

Let $x ,y \in G$

$\tau_{g}(xy)=g^{-1}xyg =(g^{-1}xg) (g^{-1}yg)=\tau_{g}(x) \tau_{g}(y)$ thus homomorphism.

$\tau_{g}(x) = \tau_{g}(y) \Leftrightarrow (g^{-1}xg) = (g^{-1}yg) \Leftrightarrow x=y$ , multiplication by $g$ on the left and $g^{-1}$ on the right. this proves that $\tau_{g}$ is injective.

$x=gg^{-1}xgg^{-1} =\tau_{g}(g^{-1}xg)$ this proves that $\tau_{g}$ is surjective.

and since $\tau_{g} : G \longrightarrow G$ this is an automorphism.

$\tau_{(gh)^{-1}}(x)= ghxh^{-1}g^{-1} =\tau_{g}^{-1} \tau_{h}^{-1}$ this defines an anti homomorphism.\vspace{5mm}

Leaky Nun

did you change anything?

Kasmir Khaan

di dyou mean that (gh)^-1

he told us not to shwo abvious steps =p

Leaky Nun

obvious. o-b-v-i-o-u-s.

Kasmir Khaan

12:07 AM

oh

><

I correct one thing and messed up the other

obvious *

@LeakyNun that is the full Q_2 , can you please out a ) b ) c ? in a good way ?

I can put them now and send ,

i mean where they should be

Leaky Nun

where should they be?

Kasmir Khaan

ill send now

\begin{document}
	a)
	Let \(x ,y \in G\)


	\(\tau_{g}(xy)=g^{-1}xyg =(g^{-1}xg) (g^{-1}yg)=\tau_{g}(x) \tau_{g}(y)\) thus homomorphism.

	\(\tau_{g}(x) = \tau_{g}(y) \Leftrightarrow (g^{-1}xg) = (g^{-1}yg) \Leftrightarrow x=y\) , multiplication by \(g\) on the left and \(g^{-1}\) on the right. this proves that \(\tau_{g}\) is injective.

	\(x=gg^{-1}xgg^{-1} =\tau_{g}(g^{-1}xg)\) this proves that \(\tau_{g}\) is surjective.

	and since \(\tau_{g} : G \longrightarrow G\) this is an automorphism.

@LeakyNun Done =P

Leaky Nun

170

A: How do I change the `enumerate` list format to use letters instead of the default Arabic numerals?

Without any package you could do it by redefining the command \theenumi for formatting the enumi counter. (Also enumii, etc., for nested lists.) \renewcommand{\theenumi}{\Alph{enumi}} inside the environment.... Or better, you could use a package like enumitem which allows, e.g., \usepackage{e...

Kasmir Khaan

12:30 AM

@LeakyNun \(G_{tor}\) is not empty because \(e\in G_{tor}\)\vspace{5mm}


	Let \(a,b\in G_{tor} \) with order of \(a =n\) and order of \(b=m\)\vspace{5mm}

	\(a\) and \(a^{-1}\) have the same order. so order of \(a^{-1}\) is \(n\)\vspace{5mm}

	Let \(k\) =lcm \((m,n)\)\vspace{5mm}

	\((ab)^{k} =a^{k}b^{k}\) because \(G\) is abelian we can split the powers.\vspace{5mm}

	this proves that the order of \(ab\) is finite so \(ab \in G_{tor}\)\vspace{5mm}

	Close under taking inverse and operation implies \(G_{tor}\) is a subgroup of \(G\)\vspace{5mm}

@LeakyNun this is the third Q

Leaky Nun

ok

Simply Beautiful Art

1:18 AM

Say, why is @Deedlit here?

And do relative extrema include limit points?

Deedlit

1:29 AM

@SimplyBeautifulArt No reason, just looking around

Faust

2:29 AM

Find three positive integers which, upon being multiplied by 3, 5 and 7 respectively and the products divided by 20 , have remainders in arithmetic progression with common difference 1 and quotients equal to remainders.

Anyone know what that says?

or can give me an example of one such integer?

nitsua60

@Faust Are you asking for help parsing the line above?

Faust

Its in my book and i can't read what it says

a translation from that to english would be great

or an example so i can try and reverse engineer its meaning

i understand up to products divided by 20

after that turns gibberish

nitsua60

Consider a triple of natural integers {x,y,z}. Further consider the quotients and remainders of 3x/20, 5y/20, and 7z/20. The remainders and quotients are the same consecutive integers. Find {x,y,z}.

Faust

so 3x/20 =x ?

nitsua60

"in arithmetic progression with common difference 1" is an unnecessarily-complicated way of saying "consecutive integers," to my thinking.

@Faust Nope. The (integer) quotient of 3x/20 = the remainder of 3x/20.

Faust

2:37 AM

whats integer quotient mean?

nitsua60

For instance, consider x=12. Then 3x/20 = 1 with a remainder of 16. Those aren't equal, though....

Faust

so the number of times 20 divides 3x needs to = tthe remainder upon its division?