TT's Maths Thread

[TrueTears]

:
Let , then for all by closure. Then, since was arbitrary, .

:
For any , does there exist such that ? By construction, yes: . Hence, for every element in , that element exists in , so .

Cheers man

[TrueTears]

How do i show that any two left cosets of a subgroup H of a group G is either disjoint or equal?

I mean it's so trivial but how do you actually prove it?

I kinda get it but i don't think my proof is that complete:

consider g1H and g2H, clearly g1H n g2H = null set or g1H n g2H = {some non empty set}

Now we just have to show that if g1H n g2H = {some non empty set} then g1H = g2H

so then what...?


Also how is the symmetric group on 3 letters S_3 isomorphic to Z_6?

Well i know that it is isomorphic if S_3 is cyclic, however how can i show it is (or is not?)?

Well actually nvm it's not isomorphic since S_3 is not abelian while Z_6 is, however i'd still like to know: is S_3 cyclic? is there a systematic way of doing this rather than testing all 6 elements of S_3 and then raising them to integral powers to see if they generate S_3 or not?

[kamil9876]

If you already know it's not isomorphic to Z_6 then it cannot be cyclic.

==============================================================

if they have a non-empty intersection, that means we have s.t



and so

.

But any element in can be written as but by closure the thing inside the bracket is in . hence . You can get the other inclusion with the same argument.

[TrueTears]

cheers kamil, actually i found an even easier way, consider an element x in the non empty set, then so xH = g1H. But so xH = g2H. Thus g1H=xH=g2H

this is a valid argument right?

[kamil9876]

You should justify why does imply ? I think you may be using circular reasoning (ie. using the result we are actually trying to prove) but I cannot say because I don't know your reason.

[TrueTears]

You should justify why does imply ? I think you may be using circular reasoning (ie. using the result we are actually trying to prove) but I cannot say because I don't know your reason.

it's trivial to prove isn't it? but i guess the reasoning would be like, x would be of the g1h for some h in H, so xH = g1hH = g1(hH) = g1H

[kamil9876]

yep that's fine.

[TrueTears]

[IMG]http://img838.imageshack.us/img838/6016/finalquestions.jpg[/img]

Not sure how to do the first question...


Second question, is this right? D_8 has an order of 2*8 = 16 so it can't be isomorphic to Z_8 which only has an order of 8.

I read in my book that D_n has an order of 2n, but how do i prove this?


Third question, is my reasoning for this question right?

Since both are cyclic groups of finite order, then G and H are both isomorphic to Z_2003, thus they are isomorphic to each other.


Fourth question, again can someone please check my reasoning.

None of the 3 sentences are true, in other words all of the D_i's for i = 10, 8, 6 are not isomorphic to their claimed counterparts, since their orders aren't the same.

So again it comes back to the question, how do i show D_n has order 2n?


Fifth question, dw i figured it out


lol dw about this question


7th question: i kinda know how to do this question but dono how to complete it.

basically we just have to show that any right coset of G has the same number of elements of H, the subgroup.

so we can define a one-to-one, onto function mapping from H to Hx where x is an element of G.

So define phi:H -> Hx, phi(h) = hx for all h in H.

to show 1-to-1, assume phi(h1) = phi(h2) -> h1x = h2x -> h1=h2

to show onto, assume for some y in Hx we must show there exists some h in H such that hx = y. But by the definition of right coset, if y is in Hx then there has to exist some h such that hx = y.

So all right cosets of H in G has the same number of elements.


Not sure how to do this question, help please


i) ive already done

ii) just to confirm that's just the same Q as as q 7 right, except its for left cosets.

[kamil9876]

First question: C'mon I thought you liked combinatorics! What is a binary operation? if is a function of two variables f(a,b). there are different pairs (a,b) and we must assign to each some element (m choices). So how many possible different assignments?

Second: Actually your using a different definition of , (their is what your textbook would call ), ie there are symmetries of the square so they must be referring to what you would call . So a simple cardinality check won't suffice. (Hint, Z_8 has an element of order 8, does any symmetry of the square have such an order ?)

Third: Yes in general any two cyclic groups of the same finite cardinality are isomorphic, i guess they want you to prove this? Find an isomorphism which maps the generator from one to the generator of the other.

cbf... this is enough for you to chew for now anyway.

[TrueTears]

Q1: I get what you mean, but I dono how do the q, a binary operation on a set S is just a function mapping SxS into S. How to work out how many binary operations?

Q2: O i see... well then my Q4 is wrong too, so in general how do i show that these aren't isomorphic? First how do you show that D_4 doesn't have an element of order 8? And how did you know to pick order 8? I would have never thought of that...

Same with question 4, how do i do them?

[kamil9876]

Q1: i pretty much spelt it out for you, it's combinatorics.

Q2: Well a group with 8 elements is isomorphic to Z_8 iff it has an element of order 8 so it's always the perfect strategy actually. To show D_4 doesn't have an element of order 8 just work out the order of each symmetry in D_4: every rotation is a multiply of 2pi/4 ie but if you multiply that by 4 you get and so every rotation has order at most 4. Every reflection has order 2. Hence no element is of order 8.

For the first part of q4, the same strategy works.

for the second part just do a simply cardinality check.

as for the third one, try to find an isomorphism. (hint: label the vertices 1,2,3 and try to create a map from there)

[TrueTears]

Q1: i pretty much spelt it out for you, it's combinatorics.

Q2: Well a group with 8 elements is isomorphic to Z_8 iff it has an element of order 8 so it's always the perfect strategy actually. To show D_4 doesn't have an element of order 8 just work out the order of each symmetry in D_4: every rotation is a multiply of 2pi/4 ie but if you multiply that by 4 you get and so every rotation has order at most 4. Every reflection has order 2. Hence no element is of order 8.

For the first part of q4, the same strategy works.

for the second part just do a simply cardinality check.

as for the third one, try to find an isomorphism. (hint: label the vertices 1,2,3 and try to create a map from there)

oh i see, so its just m^(m^2) right?

for Q 2 does that mean for ANY dihedral group, the maximum order of any rotation permutation is equal to n where n corresponds to the regular n-gon that D_n is acting upon. And also for ANY dihedral group, any reflection permutation has order 2, since if you reflected it "twice" then you just end up with the original again.

[kamil9876]

yes.

[TrueTears]

cheers i get it now, so how about this question, how many elements of order 7 does S_4 have?

[TrueTears]

[IMG]http://img94.imageshack.us/img94/3196/idempotent.jpg[/img]

I can do i) but just a bit stuck on ii).

my working for i) is: E^2-E=0 so that means is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then and so \lambda = 0 or 1

cheers


[IMG]http://img64.imageshack.us/img64/5682/adjacencymatrix.jpg[/img]

Denote the vertices of G G1, G2, G3, does the 2 in the matrix just means there's 2 paths of length 1 from G2 to G1? ie, look like this:

[IMG]http://img4.imageshack.us/img4/9931/graphgq.jpg[/img]

thx


[IMG]http://img854.imageshack.us/img854/1174/eigenvaluesqfrom2008.jpg[/img]

I sorta have an idea but dono how to finish this Q

If we let x be a eigenvector of corresponding to , then we have

Now just playing around:

but dono how to get to the result


[IMG]http://img718.imageshack.us/img718/1669/gramschmitprocess.png[/img]

Just stuck on part c)

my answer to part b) is