Extract 9.2
Context: This is the beginning of the tutorial with student Camille.
The Episode:
Camille says she has been confused with 'what the lecturer referred to as something like a~b when f(a)=f(b)'. The tutor replies that the lecturer might have been trying to prove one of the Isomorphism Theorems for Groups:
if f:G®G, where G is a group, then there is a bijection between the equivalence classes of the equivalence relation defined by 'a~b when f(a)=f(b)' and Imf.
The tutor draws fig.2a to illustrate that if b and b' are different then f-1(b) and f-1(b') are different. Camille looks at fig.2a and asks:
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C1: Why are they all straight lines? |
T1: Because I drew them this way. It doesn't mean anything. It's only the way I drew it. And you can draw little 'squigges' if you like (fig.2b). |
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C2: And these are the equivalence classes? |
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The tutor nods in agreement: the set of all elements equivalent to b is the equivalence class of b and mapping b to its equivalence class establishes a correspondence between the elements of the group and their equivalence classes. Camille seems to be sceptical and says:
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C3: Yeah... but the lecturer does it in a more complicated way... he defines an equivalence class like this and then he defines g of ea equals f(a). It's very confusing because we don't know what g is. |
T2: [pointing at her correspondence] That's his definition of g, is it? |
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C4: When he proves it I don't see at all a bijection. |
T3: [reassuringly] What he's done is equivalent to what I have done except that I've phrased it slightly differently. |
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C5: You've shown us a bijection and he didn't really show that... he just had it onto by definition and he was looking at the 1-1. |
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The tutor then says that it has to be proved that to define an equivalence class it doesn't matter which representative you take. So, she continues, in ea=ea' Û a~a' Û f(a)=f(a') one direction is about proving that f is well defined and the other one is to prove it is 1-1. Camille looks convinced and moves on to her next question: can the tutor prove that 'there is a 1-1 correspondence between the conjugates of x and x?'. The tutor asks Camille to define C(x) and Camille defines the centraliser of x as
C(x) = {hÎG/ xh=hx},
the conjugate of x as
'g-1xg for some g'.
and also points out that
if gÎC(x) then g-1xg=x.
So, says the tutor, the size of C(x) tells you 'how many repetitions you get if you try to write x as a conjugate'. So, she continues, if two elements g1 and g2 give the same conjugate they belong to the same coset of C(x). And vice versa. This happens only when g2g1-1 belongs to C(x). Therefore there is a 1-1 relation between cosets of the centraliser and conjugates of x and these have the same number of elements. Camille is quiet and looks sceptical. Then she asks:
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C6: What are cosets materially? |
T4: What do you mean by that? |
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C7: If we have a group G and a subgroup H why do we bother to find the cosets? |
T5: Because of results like this. They turn up naturally. |
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C8: Cosets are a group multiplied by an element in the big group. |
T6: [hesitantly] Yes...it's a set... |
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C9: Cosets are just a moving... |
T7: That's right. That's one way... you can look at it as translates of a subgroup... sort of multiplying g with everything in H and it shifts it... |
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C10: [after a pause] So if we have a square of size one and then the group G is like this...(fig2c) |
T8: You have to be slightly careful... It is slightly... don't think about in... you're not thinking of applying it on squares, are you? |
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C11: So then it would have four cosets? |
T9: Mmm... if the subgroup was a quarter of the size of the whole thing, yes... it would have four cosets... that's right. |
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C12: And the cosets are always the same size as the original. |
T10: That's right. As we know they partition the group. |