OCR (non mei) FP3 Friday 17th June 2016

Someone please explain how non-commutative multiplication works? Besides matrix multiplication I thought all multiplication was commutative.

Grade boundary predictions?
I'm going with:
Full ums: 65
A*: 58
A: 51
B: 44

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

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 = a^n-1
Then b²=e=a^2n-2=a^4k for some k
So 2n-2 = 4k
=> n = 2k+1
=> b = a^2k = 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 = a^{2 + 4k} for any integer k
So ba = a^{1 + 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 a^mbⁿ, 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 😂

Predictions for grade boundaries?

Same as last year?

(52 A, 57 90%, 62 100)

Do you think it was as hard as last year?

Last year's boundaries are seriously low, I'll be surprised if OCR keeps them there.