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    "Let G be a set with just one element. Define a binary operation * on G and show that (G,*) is a group"

    This question confuses me in so many ways. For one, you can use any binary operation tbf, (+,-,x,divide etc) so would I just say that for an element say, b, and that it can defined in anyway(+,-, x etc) hence that is a binary operation defined? for showing that it's a group, well it will be associative, the inverse will be itself and identity is itself as well?

    Thanks
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    You can't use, say, + on it without defining what + is; the element b needn't be a number, it could be a yellow stripey kitten, and you can't really define addition on those, can you? Similarly for multiplication. You're given that your binary operation is called *; forget what it might mean in terms of numbers, and just think in terms of this abstract concept that is a group.

    A binary operation acts on pairs of elements of the group; that is, it takes two elements and spews out another element which is in the group. Since your group only has two elements, this is fairly simple to do. There are four things you need to define work out e*e, e*b, b*e, b*b. Fortunately because these have to satisfy the group axioms, you're very limited as to what these can be. Draw out a multiplication table and put in what the values have to be.
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    (Original post by nuodai)
    You can't use, say, + on it without defining what + is; the element b needn't be a number, it could be a yellow stripey kitten, and you can't really define addition on those, can you? Similarly for multiplication. You're given that your binary operation is called *; forget what it might mean in terms of numbers, and just think in terms of this abstract concept that is a group.

    A binary operation acts on pairs of elements of the group; that is, it takes two elements and spews out another element which is in the group. Since your group only has two elements, this is fairly simple to do. There are four things you need to define work out e*e, e*b, b*e, b*b. Fortunately because these have to satisfy the group axioms, you're very limited as to what these can be. Draw out a multiplication table and put in what the values have to be.
    The group has only 1 element though, do we assume that the identity element will be in there as well?
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    (Original post by boromir9111)
    The group has only 1 element though, do we assume that the identity element will be in there as well?
    Oh sorry, for some reason I read "one" as "two". In that case your group is just {b}. The group must have an identity, so what must the identity be? So what is b*b? Once you've defined b*b you're done, since there is no other combination of elements. The axioms are trivial to check.
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    (Original post by nuodai)
    Oh sorry, for some reason I read "one" as "two". In that case your group is just {b}. The group must have an identity, so what must the identity be? So what is b*b? Once you've defined b*b you're done, since there is no other combination of elements. The axioms are trivial to check.
    b*b = identity? there b is an inverse of itself?
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    (Original post by boromir9111)
    b*b = identity? there b is an inverse of itself?
    Well yes, but you're missing an even simpler point. The group has to contain the identity and b is the only element of the group, so what is b?

    To define * you just have to say what b*b is. Once you've done that, you just have to show that (G,*) behaves as you'd expect, which requires you to know what b is... but if you know what b is then it makes defining * a lot easier.
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    (Original post by nuodai)
    Well yes, but you're missing an even simpler point. The group has to contain the identity and b is the only element of the group, so what is b?

    To define * you just have to say what b*b is. Once you've done that, you just have to show that (G,*) behaves as you'd expect, which requires you to know what b is... but if you know what b is then it makes defining * a lot easier.
    Oh, I see.....b is the identity element?
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    (Original post by boromir9111)
    Oh, I see.....b is the identity element?
    Yup!
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    (Original post by nuodai)
    Yup!
    Sorry to ask you of this but I was hoping you could explain what an Homomorphism and Isomorphism is, please?
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    (Original post by boromir9111)
    Sorry to ask you of this but I was hoping you could explain what an Homomorphism and Isomorphism is, please?
    Well say you have two groups (G, *) and (H, \circ), and a function \varphi which maps elements of G to elements of H. That is, for any g \in G, \varphi(g) \in H. Then we say that \varphi is a homomorphism if for any two elements x,y \in G, we have that \varphi (x * y) = \varphi(x) \circ \varphi(y).

    That is, it doesn't matter whether you 'multiply' x and y in G and then apply the map, or apply the map and then multiply the resulting elements of H. So in some sense, it preserves the group structure.

    An example of this is with odd and even numbers. Take one group to be (\mathbb{Z}, +) (integers under addition) and the other group to be (\{ \text{even} , \text{odd} \}, +) (i.e. where even + even = odd, even + odd = odd, and so on). Then if for n \in \mathbb{Z} we say that \varphi(n) is "even" if n is even and "odd" if n is odd, the then map \varphi is a homomorphism. In other words, if you add two numbers together and then check if it's even or odd, it's the same as checking if the two numbers are even or odd and then applying the "even + odd = odd" rule.

    An isomorphism is a bijective homomorphism. That is, if a map is a homomorphism which has an inverse, then it's an isomorphism. For finite groups, this basically means that G and H have the same size, and every element of H is 'hit' by some element of G under \varphi.

    Hope this helps.

    I really should stop procrastinating and get round to doing my own maths at some point...
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    (Original post by nuodai)
    Well say you have two groups (G, *) and (H, \circ), and a function \varphi which maps elements of G to elements of H. That is, for any g \in G, \varphi(g) \in H. Then we say that \varphi is a homomorphism if for any two elements x,y \in G, we have that \varphi (x * y) = \varphi(x) \circ \varphi(y).

    That is, it doesn't matter whether you 'multiply' x and y in G and then apply the map, or apply the map and then multiply the resulting elements of H. So in some sense, it preserves the group structure.

    An example of this is with odd and even numbers. Take one group to be (\mathbb{Z}, +) (integers under addition) and the other group to be (\{ \text{even} , \text{odd} \}, +) (i.e. where even + even = odd, even + odd = odd, and so on). Then if for n \in \mathbb{Z} we say that \varphi(n) is "even" if n is even and "odd" if n is odd, the then map \varphi is a homomorphism. In other words, if you add two numbers together and then check if it's even or odd, it's the same as checking if the two numbers are even or odd and then applying the "even + odd = odd" rule.

    An isomorphism is a bijective homomorphism. That is, if a map is a homomorphism which has an inverse, then it's an isomorphism. For finite groups, this basically means that G and H have the same size, and every element of H is 'hit' by some element of G under \varphi.

    Hope this helps.

    I really should stop procrastinating and get round to doing my own maths at some point...
    Perfect explanation. Cheers!
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    Just a quick question, the normal subgroups for symm 'square' are 2 which is the identity and group itself, right?
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    i'm doing some coursework on this atm, really helpful man thanks
 
 
 
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