Group Theory Help Please

e^x

15

Can some explain the method to solve b)ii in question 1 and a)ii in question 2 please?

Thank you.

Question 1.pdf124.4KB

Reply 1

ghostwalker

17

Original post by e^x

Can some explain the method to solve b)ii in question 1 and a)ii in question 2 please?

Thank you.

I'm not aware of any particular method, as somewhat rusty on these, but I'd work out the cyclic subgroup generated by each of the elements and then consider interactions between the two subgroups.

1(b)ii.

The above covers it.

2((a)ii

Both generators are actually the same one. Why/how?

(edited 6 years ago)

Reply 2

RichE

15

Original post by e^x

Can some explain the method to solve b)ii in question 1 and a)ii in question 2 please?

Thank you.

1bii - can you show that r is in H?

2aii - what is r^6?

Reply 3

EnglishMuon

16

For dihedral groups it is sometimes nice to draw a picture...

Reply 4

e^x

OP

15

Original post by ghostwalker

I'm not aware of any particular method, as somewhat rusty on these, but I'd work out the cyclic subgroup generated by each of the elements and then consider interactions between the two subgroups.

1(b)ii.

The above covers it.
Do you mean work out the group generated by s and the group generated by r^3*s and then the group generated by r^3*s*s = r^3?

2((a)ii

Both generators are actually the same one. Why/how?
Is it because r^6 = e?

..

(edited 6 years ago)

Reply 5

e^x

OP

15

Original post by RichE

1bii - can you show that r is in H?
Yes, because I know that r^3*s*s = r^3 belongs to H so (r^3)^2 = r^6 belongs to H, that is r^6 = r^-1 = r belongs to H
2aii - what is r^6?

r^6= e

For the first one are you trying to say that if you can show that s belongs to H and r belongs to H then H=G?

Still not sure for about the second one

Reply 6

ghostwalker

17

Original post by e^x

1(b)ii.
Do you mean work out the group generated by s and the group generated by r^3*s and then the group generated by r^3*s*s = r^3?

First two bits, yes. For the last bit you need to look at all interactions between the two subgroups, and see what arises. As I said, I'm rusty, so there may be a better method. Yep - see RichE's posts.

2((a)ii

Both generators are actually the same one. Why/how?
Is it because r^6 = e?

Yes.

(edited 6 years ago)

Reply 7

RichE

15

Original post by e^x

For the first one are you trying to say that if you can show that s belongs to H and r belongs to H then H=G?

Yes, if you can show r is indeed in H.

Reply 8

Louisb19

14

( r^3 s )^{5} = rs

Do

r

and

rs

together generate

D_{14}

?

You will need to use a theorem about 'words' of generators.

(edited 6 years ago)

Reply 9

Louisb19

14

As for the second question

recall

r^6 = e

, we then have that

\langle s, r^6 s \rangle = \langle s \rangle

. What elements are in this group?

Reply 10

ghostwalker

17

Original post by Louisb19

( r^3 s )^{5} = rs

Do

r

and

rs

together generate

D_{14}

?

You will need to use a theorem about 'words' of generators.

( r^3 s )^{2} = (r^3s)r^3s[br]=(sr^4)r^3s[br]=sr^7s[br]=e[br]

Hence

( r^3 s )^{5} = r^3s \not= rs

Or am I misunderstanding something?

Reply 11

RichE

15

Original post by Louisb19

( r^3 s )^{5} = rs

.

Is that true? Isn't r^3s a reflection and so order 2?

Reply 12

Louisb19

14

Original post by ghostwalker

( r^3 s )^{2} = (r^3s)r^3s[br]=(sr^4)r^3s[br]=sr^7s[br]=e[br]

Hence

( r^3 s )^{5} = r^3s \not= rs

Or am I misunderstanding something?

Yeah sorry, you are right. Too late at night be be doing maths LMAO.

For some reason I was treating it as if it was abelian.

Reply 13

e^x

OP

15

Original post by RichE

Yes, if you can show r is indeed in H.

I get how to do the first question now, I never learned that if you can show s and r are in H then H=G.For some reason my lecturer seemed to have changed the second Q which i have attached again. I get how to obtain the answer for this Q using ghostwalker's method but do u have an alternative method for this?

Question 2.pdf105.8KB

Reply 14

e^x

OP

15

Original post by ghostwalker

First two bits, yes. For the last bit you need to look at all interactions between the two subgroups, and see what arises. As I said, I'm rusty, so there may be a better method. Yep - see RichE's posts.

Yes.

Original post by Louisb19

( r^3 s )^{5} = rs

Do

r

and

rs

together generate

D_{14}

?

You will need to use a theorem about 'words' of generators.

Original post by RichE

Yes, if you can show r is indeed in H.

Thank you.

Reply 15

RichE

15

Original post by e^x

... but do u have an alternative method for this?

If you start working out products and inverses of r^3 and s (always simplifying your answers) you'll see you can't generate that many elements. With dihedral groups it's also good to appreciate visually what these elements are.

Reply 16

e^x

OP

15

Original post by ghostwalker

I'm not aware of any particular method, as somewhat rusty on these, but I'd work out the cyclic subgroup generated by each of the elements and then consider interactions between the two subgroups.

1(b)ii.

The above covers it.

2((a)ii

Both generators are actually the same one. Why/how?

Original post by RichE

1bii - can you show that r is in H?

2aii - what is r^6?

I've got another question, how do you get to the second line starting from U(x,tau)?

In line one, the first exponential has power 4y and the second exponential has power (-(x-y)^2)/4*tau.

It seems like working backwards after completing the square.

The questions is to do with Gauss's solution for the heat equation and using the error function.

No idea what you're asking. What's the question? And important lines of the image seem to be be omitted or partially cropped.

Or looking at it now, does "line one" actually refer to line three of that image?

PS now that I've a sense what you're asking, it's just algebra. Expand the exponent in the second integral.

(edited 6 years ago)

Group Theory Help Please

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