A Summer of Maths (ASoM) 2016

Original post by EnglishMuon

oops yea, I forgot which notations I was using again (

D_{2n}

instead of

D_{n}

) so yea if it were

D_{n}

then it'd be for the square :tongue:

D_{n}

is the dihedral group, which has 2n elements. You are correct in that

D_{4}

can represent the symmetries of a square, but this will have 8 elements in total. Four of these are rotations and the other four are reflections in various axes. As an exercise, you may wish to sketch a composition table for the symmetry group of a square and verify that it is, in fact, non-abelian.

Proposition: An abelian group and a non-abelian group cannot be isomorphic.

Proof: By definition, a group isomorphism is a bijective group homomorphism. A group homomorphism is a map

\phi : G \mapsto H

between groups, such that for any

g_{1}, g_{2} \in G

,

\phi(g_{1}g_{2})=\phi(g_{1})\phi(g_{2})

.
Now suppose that

G

is the abelian group. Then for any

g_{1}, g_{2} \in G

,

g_{1}g_{2} = g_{2}g_{1}

. Thus

\phi(g_{1}g_{2})= \phi(g_{2}g_{1})=\phi(g_{1})\phi(g_{2})=\phi(g_{2})\phi(g_{1})

Since

\phi

is bijective, we have that if

G

is abelian and

G \simeq H

then

H

is also abelian. Thus, an abelian group is always isomorphic to an abelian group, which gives another reason why the symmetry group of a square cannot be isomorphic to the original group.

(edited 7 years ago)

Reply 542

Original post by A Slice of Pi

.

yeah yeah well thats sort of my point, there are different ways of representing the dihedral group depending on what your source is. Some represent the dihedral group of the regular n-gon by Dn, some by D{2n}. But D4 is abelian though (using the 2nd definition).

Reply 543

Original post by EnglishMuon

yeah yeah well thats sort of my point, there are different ways of representing the dihedral group depending on what your source is. Some represent the dihedral group of the regular n-gon by Dn, some by D{2n}. But D4 is abelian though (using the 2nd definition).

That is true but I have defined Dn in the way I have because that is the one you are more likely to come across on a maths course at university.

Reply 544

Original post by A Slice of Pi

That is true but I have defined Dn in the way I have because that is the one you are more likely to come across on a maths course at university.

maybe, but I've been used to the other as the books I used on gt happened to use it :tongue:

Reply 545

Original post by EnglishMuon

maybe, but I've been used to the other as the books I used on gt happened to use it :tongue:

Yeah I've seen it a couple of times as well. It's no big deal, and I really don't know why there is two versions, but if your professors next year use a different definition to you just put a little note next to the working out to let them know what you mean and I'm sure it'll be fine. I don't know what year this will be studied in though.

Reply 546

Original post by A Slice of Pi

Yeah I've seen it a couple of times as well. It's no big deal, and I really don't know why there is two versions, but if your professors next year use a different definition to you just put a little note next to the working out to let them know what you mean and I'm sure it'll be fine. I don't know what year this will be studied in though.

ah ok, will do :smile:

Im pretty sure this is just 1st year gt, on the beginning of the 1st year lecture notes i read anyway.

Reply 547

Original post by EnglishMuon

ah ok, will do :smile:

Im pretty sure this is just 1st year gt, on the beginning of the 1st year lecture notes i read anyway.

Have you looked at any other of the 1st year topics? I did a lot of reading on new topics last year and was surprised to learn some of the stuff is 2nd and 3rd year, because there wasn't an obvious increase in difficulty.

Reply 548

Original post by A Slice of Pi

Have you looked at any other of the 1st year topics? I did a lot of reading on new topics last year and was surprised to learn some of the stuff is 2nd and 3rd year, because there wasn't an obvious increase in difficulty.

yeah Ive done some vector spaces and nt (+some other few things but not in much detail). pretty engrossed in gt currently tbh!

Reply 549

Original post by EnglishMuon

yeah Ive done some vector spaces and nt (+some other few things but not in much detail). pretty engrossed in gt currently tbh!

Have you looked at rings, fields and quotient rings?

Reply 550

Original post by A Slice of Pi

Have you looked at rings, fields and quotient rings?

not yet, I've been recommended to cover vector spaces and carry on gt first. I've looked some basic principles with fields but not enough to say I've studied them properly.

Reply 551

l1lvink

11

Could someone please explain the small o and big O notation?

Reply 552

Original post by l1lvink

Could someone please explain the small o and big O notation?

is there a context? :tongue:

Reply 553

InOrbit

9

Original post by l1lvink

Could someone please explain the small o and big O notation?

If there exists a constant

M > 0

, for which there exists

N > 0

, such that

n > N \quad \Rightarrow \quad |f(x)| < M|g(x)|

,
then we write

f(x) = O(g(x))

.

If for every

\varepsilon > 0

, there exists

N > 0

, such that

n > N \quad \Rightarrow \quad |f(x)| < \varepsilon|g(x)|

,
then we write

f(x) = o(g(x))

.

The analogue to big-O and little-o is very similar to less than and strictly less than. Big-O gives an upper bound to the growth, but the function can still approach its Big-O function asymptotically. Little-o is much more strict.

There's other things like Omega, omega and Theta notation. A kinda rough intuition of them could be:

o: f < g.

O: f \leq g.

\Theta: f = g.

\Omega: f \geq g.

\omega: f > g.

(edited 7 years ago)

Reply 554

Original post by Alex:

If there exists a constant

M > 0

, for which there exists

N > 0

, such that

n > N \quad \Rightarrow \quad |f(x)| < M|g(x)|

,
then we write

f(x) = O(g(x))

.

If for every

\varepsilon > 0

, there exists

N > 0

, such that

n > N \quad \Rightarrow \quad |f(x)| < \varepsilon|g(x)|

,
then we write

f(x) = o(g(x))

.

I think O(g(x)) and o(g(x)) are sets, so shouldn't the notation instead be

f(x) \in o(g(x))

etc ?

Reply 555

InOrbit

9

Original post by A Slice of Pi

I think O(g(x)) and o(g(x)) are sets, so shouldn't the notation instead be

f(x) \in o(g(x))

etc ?

You're correct, the relation is not symmetric at all and you can blame the computer scientists for this notation.

The worst notation I've found was the Fourier transform by engineers:

S(\xi) = \int_{-\infty}^{\infty} g(x) \exp{(-2\pi j x\xi)}\,dx

.
Or should I say the Sourier transform on a gunction using jmaginary numbers.

Reply 556

Original post by Alex:

Or should I say the Sourier transform on a gunction using jmaginary numbers.

Nice one xD

Reply 557

Could someone give me a hint for how to solve this question without the fundamental theorem for finite abelian groups?

Let

G

be a finite abelian group such that it contains a subgroup

H_{0}

which is contained in every subgroup

H \not = (e)

. Prove that

G

is cyclic.

This is easy if I'm allowed to use the fact that G is the direct product of cyclic groups (with trivial intersections), as if we have more than 1 cyclic group in the product, we must have H0=(e) so G must be cyclic. But I am not allowed to use this theorem for the question. Prove a couple facts about the orders of G, H0 but not sure where to go from there...

Reply 558

Original post by EnglishMuon

Could someone give me a hint for how to solve this question without the fundamental theorem for finite abelian groups?

Let

G

be a finite abelian group such that it contains a subgroup

H_{0}

which is contained in every subgroup

H \not = (e)

. Prove that

G

is cyclic.

This is easy if I'm allowed to use the fact that G is the direct product of cyclic groups (with trivial intersections), as if we have more than 1 cyclic group in the product, we must have H0=(e) so G must be cyclic. But I am not allowed to use this theorem for the question. Prove a couple facts about the orders of G, H0 but not sure where to go from there...

Hint:

Spoiler

Reply 559