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# OCR (non mei) FP3 Friday 17th June 2016

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1. Can someone fully go to town and explain/ run through question 8, as I got 100% on 1-7 but 8 was a dik
2. Someone please explain how non-commutative multiplication works? Besides matrix multiplication I thought all multiplication was commutative.
3. (Original post by Ying watson)
Someone please explain how non-commutative multiplication works? Besides matrix multiplication I thought all multiplication was commutative.
I was thinking the same. Maybe it was secretly matrix?
I'm going with:
Full ums: 65
A*: 58
A: 51
B: 44
5. (Original post by TheNicholas)
I'm going with:
Full ums: 65
A*: 58
A: 51
B: 44
I bloody hope it's something like that, or I can wave goodbye to an A* in further.
Last year was 57 A*, 52 A, 47B, and I think it's probably comparable (so I'd expect the B boundary to be a bit higher than 44 unfortunately)
6. (Original post by afang02)
finished question 1-7 in half an hour and spent entire hour doing the last one...groups and proof are always my least favourite
FFS
oh my god, me too - I looked up at the clock after Q6 and was like wow I have an hour to go, this is going so well!
famous last words
7. (Original post by Mandos)
Can someone fully go to town and explain/ run through question 8, as I got 100% on 1-7 but 8 was a dik
Okay, so I'm not the best on this, but until somebody else comes along I'll run through what I did. Correct me if I'm wrong, anybody

i)
So if ba = an for some n, b = an-1
Then b2=e=a2n-2=a4k for some k
So 2n-2 = 4k
=> n = 2k+1
=> b = a2k = either a2 or e, depending on (k mod 2)
but e, b, and a2 are all given independent elements of G, so b =/= e and b =/= a2
Therefore, ba =/= an for any integer n

ii)
Suppose ba = a2b
Then bba = b2a = ea = a = ba2b = ba ab = a2b ab = a2 ba b = a2 a2 b b = a4b2 = e
So a = e
This obviously cannot be true, hence by contradiction ba =/= a2b

Suppose ba = ab
Then (ba)2 = (ab)2 = abab = aabb = a2 = a2 + 4k for any integer k
So ba = a1 + 2k for some integer(s) k
But by (i), ba =/= an for any n, so this cannot be true, hence by contradiction ba =/= ab

Now, by the way the set is defined for us, we can see that any element of it can be expressed as ambn, where m ∈ { 0, 1, 2, 3 }, n ∈ { 0, 1 }
By the fact that a group is closed, ba must be in the set, so it must be possible to express it in this form.
Part i showed that n =/= 0, so n = 1
Part ii showed that, given n = 1, m =/= 1 and m =/= 2
If m = 0, then ba = b, so a = e, which obviously is not true
The only remaining possibility is m=3, or ba = a3b

iii)
ba2 = ba a = a3 ba = a3a3b = a6b = a4a2b = a2b

iv)
The subsets are { e, a, a2, a3 }, { e, a2, b, a2b }, { e, a2, ab, a3b }, and a little fiddling with the given equivalences leads to all the products of elements.
Edit explaining how to get the subsets:First note that powers of 'a' generate a cyclic subgroup of order 4. This means two things. First, that the powers of a are one of our subgroups.
Second, that neither a nor a3 appears in any other subgroup of order 4 (a2 is not excluded because it is of order 2).Now, any rearrangement of powers of 'a' and 'b' keeps the same number of 'b's, even as the 'a's change (ba = a3b, for instance). Therefore, any two terms containing a 'b', when multiplied together, will give a term containing no 'b's.
Therefore, by considering that every row and column of the Latin square must contain each element exactly once, we see that two of our terms will need to have a 'b', and two will not.The identity element 'e' does not contain a 'b', so that's one down - the other will be a pure power of 'a', and therefore, as we have shown, must be a2.
Now we have {e, a2, ...}, where the missing pair of elements must each contain a 'b'.Trying 'b' yields one subgroup including a2b, and trying 'ab' yields the other including a3b.
8. I think its a dick move to make you do proofs before you properly learn how to construct them in university.

Oh well, I think the rest of the paper was pretty decent so I should still manage to scrape 90%. Even if I don't 88% is still enough hopefully that I can easily make up in FP2.
9. Looking at the solutions to question 8 given above. Chances are 95% of people dropped at least 10-12 marks.
10. I'm not sure if my working for question 6ii) is right, but this is what I had:
Plane 3 equation: x+5y-z=1
Line equation: (0,1,1) + t(7,-2,-3) = (7t,1-2t,1-3t)
Sub in to plane equation: (7t) + 5(1-2t) - (1-3t) = 1
Simplifies to 4 = 1 which shows an inconsistency.
I think that this means the line and plane don't meet (in other words they are parallel), but also that the line doesn't lie in the plane.

Also that last group question was vile. Hats off to anyone who got those proofs and subgroups.
11. (Original post by farryharnworth)
Okay, so I'm not the best on this, but until somebody else comes along I'll run through what I did. Correct me if I'm wrong, anybody

i)
So if ba = an for some n, b = an-1
Then b2=e=a2n-2=a4k for some k
So 2n-2 = 4k
=> n = 2k+1
=> b = a2k = either a2 or e, depending on (k mod 2)
but e, b, and a2 are all given independent elements of G, so b =/= e and b =/= a2
Therefore, ba =/= an for any integer n

ii)
Suppose ba = a2b
Then bba = b2a = ea = a = ba2b = ba ab = a2b ab = a2 ba b = a2 a2 b b = a4b2 = e
So a = e
This obviously cannot be true, hence by contradiction ba =/= a2b

Suppose ba = ab
Then (ba)2 = (ab)2 = abab = aabb = a2 = a2 + 4k for any integer k
So ba = a1 + 2k for some integer(s) k
But by (i), ba =/= an for any n, so this cannot be true, hence by contradiction ba =/= ab

Now, by the way the set is defined for us, we can see that any element of it can be expressed as ambn, where m ∈ { 0, 1, 2, 3 }, n ∈ { 0, 1 }
By the fact that a group is closed, ba must be in the set, so it must be possible to express it in this form.
Part i showed that n =/= 0, so n = 1
Part ii showed that, given n = 1, m =/= 1 and m =/= 2
If m = 0, then ba = b, so a = e, which obviously is not true
The only remaining possibility is m=3, or ba = a3b

iii)
ba2 = ba a = a3 ba = a3a3b = a6b = a4a2b = a2b

iv)
The subsets are { e, a, a2, a3 }, { e, a2, b, a2b }, { e, a2, ab, a3b }, and a little fiddling with the given equivalences leads to all the products of elements.
Edit explaining how to get the subsets:First note that powers of 'a' generate a cyclic subgroup of order 4. This means two things. First, that the powers of a are one of our subgroups.
Second, that neither a nor a3 appears in any other subgroup of order 4 (a2 is not excluded because it is of order 2).Now, any rearrangement of powers of 'a' and 'b' keeps the same number of 'b's, even as the 'a's change (ba = a3b, for instance). Therefore, any two terms containing a 'b', when multiplied together, will give a term containing no 'b's.
Therefore, by considering that every row and column of the Latin square must contain each element exactly once, we see that two of our terms will need to have a 'b', and two will not.The identity element 'e' does not contain a 'b', so that's one down - the other will be a pure power of 'a', and therefore, as we have shown, must be a2.
Now we have {e, a2, ...}, where the missing pair of elements must each contain a 'b'.Trying 'b' yields one subgroup including a2b, and trying 'ab' yields the other including a3b.
Decent effort. I struggled with this question. For one of the contradictions of ii, I proved that the orders of two elements were different, and for another I proved their inverses were different. Do you think I'll get any marks for this? Fortunately I got pretty much everything else right so at worse I reckon I've got 54. Hopefully 80 ums so the A is solid if FP2 isn't dreadful 😂
12. (Original post by tobybes)
Decent effort. I struggled with this question. For one of the contradictions of ii, I proved that the orders of two elements were different, and for another I proved their inverses were different. Do you think I'll get any marks for this? Fortunately I got pretty much everything else right so at worse I reckon I've got 54. Hopefully 80 ums so the A is solid if FP2 isn't dreadful 😂
Yeah, me too.
I can get both of those methods to work on ba = a2b, but neither on ba = ab (I may well just be doing it wrong - I had to try a few things before finding something ba = ab didn't work with), but if they worked, I see no reason why they shouldn't get you the marks.

Yeah, I thought the rest of the paper was alright on the whole. Group theory's always weird. Best of luck
13. My anus was not ready for question 8, rest of the paper was a dream. I answered a grand total of zero on the question 8 :'(
14. Remember, last year you could drop 10 marks and get 100UMS. There is hope.
16. (Original post by Inert1a)
Same as last year?

(52 A, 57 90%, 62 100)
17. (Original post by marioman)
Same as last year?

(52 A, 57 90%, 62 100)
Do you think it was as hard as last year?
18. (Original post by TheNicholas)
Do you think it was as hard as last year?
I thought the last question was very hard. The rest of the paper was nice but Q1-Q7 in June 2015 were also quite nice with one hard groups question at the end.
19. Last year's boundaries are seriously low, I'll be surprised if OCR keeps them there.
20. (Original post by 16characterlimit)
Last year's boundaries are seriously low, I'll be surprised if OCR keeps them there.
What do you think they will be?

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