Extract 9.6
Context: In Tutorial
1: Eleanor
2: Cary and Beth
3: Cleo and Patricia
4: Abidul and Frances
the tutor suggests they prove the First Isomorphism Theorem for Groups:
Let G, G' be groups and f:G®G' a homomorphism. If K=kerf then: G/K ~ Imf.
Proof (fig.6): since KÑG, G/K can be defined. Then
y
: G/K ® Imfy
(Kg)=f(g)is an isomorphism, namely y is a well-defined, 1-1 and onto homomorphism.
In the following, I present extracts from the proving process in the four tutorials.
The Episode:
In the four tutorials the conversation starts with the tutor's request towards the students to state the theorem.
I note that by (*) I mean the expression
Kg1=Kg2 Û f(g1)=f(g2) (*)
Tutorial 1. Eleanor listens to the tutor silently and throughout the presentation of the proof she only answers the following very closed questions:
• The tutor repeatedly requests the student to recall the theorem. Finally Eleanor hesitantly says that she has associated the theorem with a normal subgroup. The tutor insists that it is a particular normal subgroup and Eleanor remembers kerf. She responds affirmatively to whether kerf <G and kerfÑG. She also dictates the calculation to why gK=Kg.
• Once the tutor has illustrated verbally that the number of cosets in G/K is the same as the number of elements in Imf, she asks Eleanor to see the 'obvious map' between G/K and Imf. Eleanor whispers 'Kg' and the tutor introduces y.
• While proving that y is an isomorphism between G/K and Imf, the tutor asks what the Ü of (*) proves. Eleanor replies 'onto' and, after the tutor explains to her that 'onto' is obvious from the definition, she changes her mind to '1-1'.
• Finally, while Eleanor dictates correctly the rule for the product of cosets (Kgh=KgKh) she remains silent when the tutor asks her to use it in proving that y is a homomorphism. Eventually, under the tutor's firm direction she does.
Tutorial 2. I note the following:
• Beth states the theorem. The tutor corrects: f can be between any two groups G and G' and not necessarily on G itself.
• Similarly to Eleanor, in Tutorial 1, Beth also suggests mapping an element of Imf to its coset.
• In proving (*) Beth recalls (from the group tutorial in this college the day before) that two cosets Kg1 and Kg2 are equal iff 'g1g2-1 is in the kernel'. Also both students dictate the necessary calculations until f(g1)=f(g2) is reached.
• Beth knows that the reverse direction of (*) proves that y is 1-1.
• Once well-definedness, 1-1 and onto is proved the tutor asks for the last thing to prove and discussion is as follows:
B1: That it is closed.
T1: What do you mean by "closed"? If we are talking about an isomorphism of groups and we've proved it's a bijection...?
B2 and C: It's an isomorphism.
T2: Right. So it is a homomorphism because..? What do you want to check?
B3: f of g1g2 is f of g1 times f of g2.
T3: Ah, we are talking about y.
B4: Oh, then y of these...
T4: Er,...y is...
B5: Oh, Kg1 then and Kg2.
Beth then dictates the calculations.
Tutorial 3. I note the following:
• Patricia says that she can only recall that the theorem is related to cosets and a mapping which she says she thinks is an isomorphism. The tutor corrects: it is a homomorphism. She then asks for the conclusion of the theorem. Patricia replies:
P1: G... divided by... kernel... er,...f...Does this [~] actually mean equals?
T1: Isomorphic. It means isomorphic
• Similarly to tutorials 1 nad 2, Cleo suggests, after listening to the tutor's explanations, mapping Kg to g. The tutor asks her to reverse that ('it is the other way around') to g to Kg.
• The tutor continues the presentation of the proof without any participation until she asks what is the meaning of g1g2-1ÎK while proving (*). The students are silent She returns to the 'basic notion' of a kernel and asks what a kernel is:
C1: It maps the elements at zero.
T2: Not zero!
C2: Identity.
T3. [writes down the definition 'properly': kerf={gÎG /. She leaves it unfinished and asks] Such that?
C3: Maps them to the identity.
T4: Which means something is equal to the identity.
C4: [after a pause] g does.
T5: When you say something goes to the identity, is equal to the identity, what do you mean by that?
C5: f does.
• The tutor then returns to the proof of (*). Starting from g1K=g2K she asks the students to prove that f(g1)=f(g2). The students are silent. The tutor points out that two cosets Kg1 and Kg2 are equal when g1g2-1 is in K. The students are still silent. She returns to the basic notion of a kernel (see conversation above). In the end of that discussion Patricia points out that f(g1g2-1) will be the identity. With gradual prompting by the tutor the students arrive at f(g1)=f(g2). The tutor points out that Þ of (*) means y is well- defined. What about Ü? she asks. The students are silent. Then they both mumble something about f(g1)=f(g2) saying something about f. The tutor stresses that their answer must be related to y, not f. The students look lost. The tutor then repeats that they have to prove that y is an isomorphism. The students are still silent and the tutor asks what an isomorphism is.
P2: It is a homomorphism..
T6: Yes? And...?
P3: It has to be 1-1...
T7: And...?
P4: A bijection...
T8: Yes...
P5: Onto.
T9: Yes and in other words which bit have we proved here? You've got a choice: you can say it's a homomorphism, it's 1-1 and it's onto [laughter] Which bit have we proved? [pause] Is it the bit that says it is a homomorphism?
P6 and C6: [after a pause] No.
T10: Is it the bit that says it's 1-1? [pause] What do you need to prove y is 1-1?
C7: Prove... er,... oh, it's this direction that proves it's a well-defined operation... so it's the other one that proves it's 1-1.
y
is obviously onto, says the tutor and suggests that they prove y is a homomorphism. Patricia hesitantly dictates the equality they have to prove: I note that only with the tutor's prompting she recalls and uses the rule for the product of cosets. The tutor concludes that the key point about y is that the homomorphic property works ('The point here is that this thing works').Tutorial 4. The students cannot recall the theorem and remain silent. The tutor then tells them off: 'Number one thing is you learn the theorems. Have you learned what the theorem says? Specially when it's something that's got a name attached to it'. She then states the theorem.
• Similarly to previous tutorials Abidul suggests mapping g to Kg.
• Similarly to previous tutorials Abidul recalls that two cosets Kg1 and Kg2 are equal 'When g1g2-1 is in K' which is true iff 'f( g1g2-1) is one'. Frances uses the homomorphic property to manipulate f(g1g2-1) but 'gets stuck' with the -1. Abidul helps her by pointing out that ' It [f] doesn't really matter, does it?' that is f(g2-1)=f(g2)-1. Thus the well definedness is proved. Frances points out that the Ü direction of (*) shows that y is 1-1 and the tutor explains why y is obviously onto.
• The discussion for proving that y is an isomorphism is as follows:
A1: y of Kg1g2 is y of Kg1....
T1: Say that again.
A2: y of Kg1, Kg2...
T2: Right. And what did you say at the beginning?
A3: y Kg1g2...
T3: I mean you have to show that y of the product is...
Frances finishes the tutor's sentence (applies the definition of multiplying products) and the proof is completed. The tutor closes by calling this 'a very standard method of proof involving quotients'. The problem that they have to be concerned is well-definedness. She also asks them to learn and remember the theorem because it is very important.
Return to Appendices for Chapter 9.