Question

question on groups and symmetries please helpp

Answer 1

i need help on question 1 in that picture pleaseee

Answer 2

@amistre64

Answer 3

@KingGeorge

Answer 4

@satellite73

Answer 5

just running thru the rolodex eh ....

Answer 6

noo, these tutors are pro in this course and helped me before

Answer 7

eww! i cant even begin to read that stuff .... im out :)

Answer 8

lol its okay :P

Answer 9

@TuringTest

Answer 10

soo far i found the order of all elements in A4

Answer 11

YESS kinggeorge heyy

Answer 12

i got the the order of (1) = 1 the order of (13)(24) = (14)(23) = 2

Answer 13

So you're having trouble with the first problem?

Answer 14

yeah and the order of (123) = (243) = (142) = 3 and the order of (134) = 4 and i still am doing the 9, 10 , 11 , 12 orders and doing it but i dont know how to map it

Answer 15

like you said map it to Z6 but Z6 = {1, 2, 3, 4, 5} i didnt get how you got the order of 2 to be 3 as well

Answer 16

Wait, I was using \(\mathbb{Z}_6=\{\{0,1,2,3,4,5\}, +_6\}\) Where \(+_6\) stands for addition modulo 6.

Answer 17

yeah thats what we have to use

Answer 18

yeah sorry i forgot the 0

Answer 19

Well then \(|2|=3\) since \(2+2+2\equiv0\pmod{6}\).

Answer 20

oo okay

Answer 21

the order of 1is just 6, order of 3 is just 2

Answer 22

whats the order of 0, 4, and 5

Answer 23

The order of 0 is 1 since it's the identity, \(|4|=3\), and \(|5|=6\).

Answer 24

hmm okay alright just give me a second let me map it to them

Answer 25

and you also said it shoulnt be the same as the mapping before

Answer 26

I'm not entirely sure what I meant by that statement. Show ,e what mapping you end up with, and we'll see where to go from there.

Answer 27

alright should i list it for all 12 elements?

Answer 28

Yes please.

Answer 29

ooo and by the way what is (2)(2) is it just e

Answer 30

I believe that would just be e.

Answer 31

ok so i got

Answer 32

(1) --> 0

(12)(34) = (13)(24) = (14)(23) --> 2

(123) = (243) = (142) --> 3

and i cant list the other elements because the order of (134) = (132) = (143) = (234) = (124) = 4 and there is no element in Z_6 that has order 4

Answer 33

ok wait i just messed it up

Answer 34

(1) --> 0

(12)(34) = (13)(24) = (14)(23) --> 3

(123) = (243) = (142) --> 2

and i cant list the other elements because the order of (134) = (132) = (143) = (234) = (124) = 4 and there is no element in Z_6 that has order 4

Answer 35

So first, let's make a complete list of all the elements in \(A_4\).
\[A_4=\left\{\begin{matrix}e,(12)(34),(13)(24),(14)(23),\\(123),(234),(134),(124),\\(132),(243),(143),(142)\end{matrix}\right\}\]Every single element in the top row is mapped to 0 since they're in the kernel. The other elements must be mapped to some number that has order 2.

Answer 36

Thus, start like this\[\begin{align} e,(12)(34),(13)(24),(14)(23)&\longmapsto 0 \\(123)&\longmapsto2\\(132)&\longmapsto4\end{align}\]After this, we need to be careful that we preserve the homomorphism property I.e.\(\varphi(gh)=\varphi(g)\varphi(h)\).

Answer 37

yes i noticed you know the other elements that i didnt include i was looking at your message if you square them i get them to go to 4

Answer 38

Now look at \((123)(234)\). This is \((12)(34)\). So \(2+\varphi((234))=0\). This means that \[(234)\longmapsto4\]and\[(243)\longmapsto2\]

Answer 39

In case it wasn't clear, \(2+\varphi((234))=0\) since \((12)(34)\) is in the kernel.

Answer 40

wohh

Answer 41

We continue to check similar things.\[(123)(134)=(234)\implies(134)\longmapsto2\]which then implies \[(143)\longmapsto4\]

Answer 42

ok just one minute you know your fourth last post from right now you said (12)(34) --> 0 but the order of (12)(34) = 2 and the order of 0 is 1?

Answer 43

\[(123)(124)=(13)(24)\implies(124)\longmapsto4\]and finally\[(142)\longmapsto2\]
To answer your question, it's correct that the order of \((12)(34)\) in \(A_4\) is 2, but since it's defined to be part of the kernel, it must be that \[(12)(34)\longmapsto0\]By the definition of the kernel.

Answer 44

alright and you know how you said (123) --> 2 should i list all of them like (123), (243), (142) --> 2 or it doesnt matter

Answer 45

ok i am actually sorry this is so much to take in. Is there a reason why we are only using those permuations and not all of them?

Answer 46

You should list it all out. So in total, you should get \[\{e,(12)(23),(13)(24),(14)(23)\}\longmapsto0\]\[\{(123),(134),(142),(243)\}\longmapsto2\]\[\{(132),(143),(124),(234)\}\longmapsto4\]

We're using these permutations because these are all of the elements of \(A_4\).

Answer 47

ok alright makes sense and that is our homomorphism then

Answer 48

that should be the homomorphism.

Answer 49

ok awesome makes sense to me

Answer 50

should i ask the other question same on here or make a new post?

Answer 51

A new post would be better.