Question

question

Answer 1

yeah thats what B is

Answer 2

you know for the other question where you said add those 2 cases together

Answer 3

we look at S5 that has order 3 only right ?

Answer 4

Which problem do you want to start with?

Answer 5

lets clarify the one in the messages first ?

Answer 6

Sounds good.

We want to find the number of homomorphisms \(\varphi\) from \(\mathbb{Z}_3=\{0,1,2\}\) to \(S_5\).

Case 1: It's the identity homomorphism where everything gets sent to the identity in \(S_5\). Let's call that \(e\). There is only one of these.

Case 2: It's not the identity homomorphism. In that case, we know that \(|\varphi(1)|=|\varphi(2)|=3\) since \[\varphi(1+1+1)=\varphi(1)\varphi(1)\varphi(1)\]\[\varphi(1+1+1)=\varphi(0)=e\]So by substitution, \[\varphi(1)^3=e\]The same argument follows for \(|\varphi(2)|\).

Answer 7

Since \(\varphi\) is determined entirely by \(\varphi(0)\), we can choose an element in \(S_5\) to send \(\varphi(0)\). However, since \(|<\varphi(1)>|=3\), we must choose a 3-cycle.

How many 3-cycles are there in \(S_5\)?

Answer 8

(123) (321) (213) so only 3?

Answer 9

i mean no wth there is 5 * 2! so wouldnt there be 10 choices?

Answer 10

Those are the 3-cycles in \(S_3\). (also, (321)=(213)). We want to find the 3-cycles in \(S_5\). As it turns out, there are exactly 20 elements in \(S_5\) that are 3-cycles such as (123)(4)(5).

Answer 11

Since we already listed the identity homomorphism, and there are 20 3-cycle elements, there should be 21 homomorphisms total.

Answer 12

but do we have to list all 20 3 cycles?

Answer 13

You don't have to list all of them. You just have to know that there are 20.

Answer 14

or could we just state it has 20 and is there any mathematical way to find out how much there are

because how i got 5*2! i said to my self there are 5 choices and we can rearrange it 2! ways or whatt

Answer 15

The formula is \[\frac{5!}{[3^1(1!)][1^2(2!)]}=\frac{120}{6}=20\]See this for where I got the formula http://groupprops.subwiki.org/wiki/Conjugacy_class_size_formula_in_symmetric_group

Answer 16

OOO okayy sick makes sense

Answer 17

alright this question is done now just minor questions

Answer 18

for it to be a homomorphism that is basically saying if it is onto right like being surjective?

Answer 19

and if we show its surjective somehow we basically showed its a homomorphism right?

Answer 20

To be a homomorphism, it doesn't have to be surjective. It just has to have the property that \[\varphi(gh)=\varphi(g)*\varphi(h)\]I don't believe that showing it's surjective shows anything definitively (unless you know other things as well).

Answer 21

hmm alright makes sense

Answer 22

so i shouldnt really think of it that way then rather than try to get phi(gh) = phi(g)*phi(h)

Answer 23

Right. If you can prove \(\varphi(gh)=\varphi(g)*\varphi(h)\) it's a homomorphism.

Answer 24

alright awesome thanks a lot :):)

Answer 25

You're welcome. Let me know if this gives you any trouble.

Answer 26

yeah for sure honestly man i dont know if i wouldve been able to pass this course if it wasnt for you like its getting fun now lol

Answer 27

but right now the only question I have left is that matrix question I posted the picture I sent

Answer 28

Do you know where to start on that?

Answer 29

no these 2 questions i was completely lost like i didnt know how to approach the question im pretty sure i can tryy let me tryy

Answer 30

ok see here is what i get confused like the definition you provided and what the book have abnout homomorphism is phi(gh) = phi(g) *phi(h) you know what i am trying to get at how do i even start like phi of the matrix

Answer 31

Alright. The way \(\varphi\) is defined, it works like this. \[\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\right)=d^2\]So you need to take \[\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\begin{bmatrix} e&f\\0&h\end{bmatrix}\right)\]and show that this is equal to \(d^2\cdot h^2\)

Answer 32

uhh okayy

Answer 33

where phi[ (e f; 0 h) ] = h^2?

Answer 34

That's precisely it.

First, do the matrix multiplication \[\begin{bmatrix} a&b\\0&d\end{bmatrix}\begin{bmatrix} e&f\\0&h\end{bmatrix}=\begin{bmatrix}ae&af+bh\\0&dh\end{bmatrix}\]Can you tell me what \[\varphi\left(\begin{bmatrix}ae&af+bh\\0&dh\end{bmatrix}\right)\] is equal to?

Answer 35

thats what i dont get how on earth are you suppose to take the phi of that matrix are we just suppose to assume it has to be d^2*h^2

Answer 36

That almost it. From the definition of \(\varphi\), you get that \[\varphi\left(\begin{bmatrix}ae&af+bh\\0&dh\end{bmatrix}\right)=(dh)^2=d^2h^2\]

Answer 37

Since \[\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\right)\varphi\left(\begin{bmatrix} e&f\\0&h\end{bmatrix}\right)=d^2h^2\]as well, we know that \[\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\begin{bmatrix} e&f\\0&h\end{bmatrix}\right)=\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\right)\varphi\left(\begin{bmatrix} e&f\\0&h\end{bmatrix}\right)\]So it must be a homomorphism.

Answer 38

ooo that is clearly phi(d^2 * h^2) = d^2 * h^2

Answer 39

o wow thats all ? :\

Answer 40

yup. That's all to show that it's a homomorphism.

Answer 41

not bad just have to keep practicing then hard at it and for part B to that question i think the teacher was up to something and it may show up on our test

Answer 42

he didnt explain how to find the kernel he was just saying so by the definition blah blah

Answer 43

like all he did was state the definition but he said you would need to know how to find the kernel and what not

Answer 44

To find the kernel, you need to find all matrices such that \[\varphi\left(\begin{bmatrix} a&b\\0&d\end{bmatrix}\right)=1\]From this relation, and the definition of \(\varphi\) you know that \(d^2=1\implies d=\pm1\)

Answer 45

Thus, the kernel is merely \[\begin{bmatrix} a&b\\0&\pm1\end{bmatrix}\]

Answer 46

yeahh

Answer 47

seriously thats all we dont need to do more? :\

Answer 48

That's all I would show on one of my homework assignments. You might want to put a few more words in there to make sure it's clear, but that's all you need.

Answer 49

ooo okay so basicallly how do you fiind the kernel just make our phi =1 ?

Answer 50

To find the kernel of \(\varphi\), set \(\varphi(g)=e\) where \(e\) is the identity in group that is mapped to. Once you've don'e that, solve for \(g\).

Answer 51

alright perfect it makes sense now thanks a lot really appreciate it thank youu

Answer 52

You're welcome.