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A Summer of Maths (ASoM) 2016

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    (Original post by Gregorius)
    Yup.
    hmm im not sure what example to be looking for, cus is the definition of an isomorphism the same for all algebraic structures? Cus I've heard of ring isomorphisms but not much about them. Does this counterexample need to hold for all the isomorphisms different to the usual  \phi(xy)= \phi(x) \phi(y) one? (assuming they are different).
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    EnglishMuon Here's another question that you might like... (it's quite straightforward but part (b) draws on knowledge of other groups)

    Consider the group G = ( \mathbb{Z}_{2}, +)
    (a) Construct the composition table for the external product G \times G.
    (b) Can you think of a group that you may have already studied that is isomorphic to G \times G?
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    (Original post by A Slice of Pi)
    EnglishMuon Here's another question that you might like... (it's quite straightforward but part (b) draws on knowledge of other groups)

    Consider the group G = ( \mathbb{Z}_{2}, +)
    (a) Construct the composition table for the external product G \times G.
    (b) Can you think of a group that you may have already studied that is isomorphic to G \times G?
    Assuming the addition is mod2:
     \mathbb{Z}_{2} = \{ 0,1 \} so  \mathbb{Z}_{2} \times \mathbb{Z}_{2} = \{ (0,0) , (1,0), (0,1) , (1,1) \} .

    This is isomorphic to the symmetries of the rectangle,

     D_{4}= \{ x^{i}y^{j} : i,j =0,1\ xy=y^{-1}x(=yx) \} by the isomorphism defined by  \phi (x^{i}y^{j} ) = (i,j) . e.g. as  x^{i}y^{j} x^{a}y^{b} = x^{i+a}y^{j+b} as abelian (hopefully)
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    (Original post by EnglishMuon)
    Assuming the addition is mod2:
     \mathbb{Z}_{2} = \{ 0,1 \} so  \mathbb{Z}_{2} \times \mathbb{Z}_{2} = \{ (0,0) , (1,0), (0,1) , (1,1) \} .

    This is isomorphic to the symmetries of the square,

     D_{4}= \{ x^{i}y^{j} : i,j =0,1\ xy=y^{-1}x(=yx) \} by the isomorphism defined by  \phi (x^{i}y^{j} ) = (i,j) . e.g. as  x^{i}y^{j} x^{a}y^{b} = x^{i+a}y^{j+b} as abelian (hopefully)
    This is what I did...
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    (Original post by A Slice of Pi)
    This is what I did...
    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
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    (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
     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.
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    (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).
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    (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.
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    (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
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    (Original post by EnglishMuon)
    maybe, but I've been used to the other as the books I used on gt happened to use it
    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.
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    (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 Im pretty sure this is just 1st year gt, on the beginning of the 1st year lecture notes i read anyway.
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    (Original post by EnglishMuon)
    ah ok, will do 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.
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    (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!
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    (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?
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    (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.
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    Could someone please explain the small o and big O notation?
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    (Original post by l1lvink)
    Could someone please explain the small o and big O notation?
    is there a context?
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    (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.
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    (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 ?
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    (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.
 
 
 
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