a divides n so n/a=k EZ(Original post by drandy76)
Always get this mixed up, is that a divides n, or n divides a?
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 15072016 20:01
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 15072016 20:02
(Original post by drandy76)
Always get this mixed up, is that a divides n, or n divides a?
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 15072016 20:06
Dunno how to double quote so cheers lads
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 15072016 20:30
Was looking at this and noticed something odd when I looked at the middle two/one, with the outer sides you can 'follow' the trail from top to bottom with no problems, but if you try the same for either of the middle pillar thingies it can't be done, does anyone know how they managed this?
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 15072016 21:32
(Original post by drandy76)
Was looking at this and noticed something odd when I looked at the middle two/one, with the outer sides you can 'follow' the trail from top to bottom with no problems, but if you try the same for either of the middle pillar thingies it can't be done, does anyone know how they managed this?
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Proper weird though.
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 15072016 21:46
(Original post by physicsmaths)
They joined gaps on one end and respectively opposite on the other side the extra comes from gap between 4 is 3 spaces etc.
Proper weird though.
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 15072016 22:21
(Original post by drandy76)
Yeah it was giving me a headache trying to trace it lol, thanks geometry nerd
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 15072016 22:27
(Original post by physicsmaths)
No geometry in this one lad. Just using those eyes.
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 16072016 16:18

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 16072016 16:26
(Original post by EnglishMuon)
If for p prime, why are there p1 choices for a,b such that (mod p) ?
edit: lol didn't notice the +/ tho
I presume it means we care either only about pairs such that ab = 1 (modp) or only about pairs such that ab = 1 (modp), or else there seem to be counterexamples. In this case proving the former is also fairly simple; multiplication is commutative and we can just multiply one of the elements in the inverse pair by p1 (or 1 if you prefer) to get the corresponding pair that multiply to 1Last edited by 13 1 20 8 42; 16072016 at 16:35. 
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 16072016 16:36
(Original post by 13 1 20 8 42)
for each of 1,...,p1 there is a multiplicative inverse to pair off with (e.g. for p = 3 you have (1,1), (2,2), for p = 5 you have (1,1), (2,3), (3,2), (4,4) )
edit: lol didn't notice the +/ tho 
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 16072016 16:40
(Original post by EnglishMuon)
Yep, thought so for the +1 case. Was the 1 case I wasnt sure how to explain though. thanks 
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 16072016 17:02
(Original post by 13 1 20 8 42)
dunno if you've seen the edit. But it's probably not worded amazingly; basically the point is that given each pair (a,b) for the +1 case, you can multiply b by 1 to find a pair for the 1 case (as you get a(b) = (ab) = 1), i.e. you can sort of set up a correspondence 
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 22072016 17:59
Any hints on how to prove this? :
If is a group and are subgroups of G of finite index in G, prove that is of finite index in .
Some initial thoughts:Spoiler:ShowIts straightforwards to show is also a subgroup of G.
The problem is trivial if G is finite so we assume it is infinite (order).
Let the index of H be m, the index of K be n.
Then there are only m distinct elements such that is distinct
and n distinct elements such that is distinct.
Now Itd be nice to say that these constitute to all elements g in G such that are distinct, but I dont think thats obviously the case imagine forming where for some . Although the elements in are the same as in couldnt the elements in be different to still be different ? (i.e. we have permuted H)

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 22072016 18:23
(Original post by EnglishMuon)
Any hints on how to prove this? :
If is a group and are subgroups of G of finite index in G, prove that is of finite index in .
Some initial thoughts:Spoiler:ShowIts straightforwards to show is also a subgroup of G.
The problem is trivial if G is finite so we assume it is infinite (order).
Let the index of H be m, the index of K be n.
Then there are only m distinct elements such that is distinct
and n distinct elements such that is distinct.
Now Itd be nice to say that these constitute to all elements g in G such that are distinct, but I dont think thats obviously the case imagine forming where for some . Although the elements in are the same as in couldnt the elements in be different to still be different ? (i.e. we have permuted H)
Since you have proven that the intersection of H and K is a group, you are done! From the proof of Lagrange, this subgroup has a particular (finite) index.
G being infinite. A good example of this that I can think of is G = Z with the + operation and subgroups H,K being the {3n  n in Z} and {4n  n in Z}. These have finite indexes in G. (Eg, the left cosets of H are {3n  n in Z}, {3n+1  n in Z}, and {3n+2  n in Z}.) The intersection of H and K is {12n  n in Z} which has index 12 in G  clearly finite.
The only problem is that it would seem difficult to generalise this example to other infinite groups. Here I would turn to contradiction.
Suppose that H and K have finite indexes in G but their intersection has an infinite index. The intersection of H and K is a subgroup of H. At this point, I'm going to point out that K is irrelevant  K is only here to give us a subset of H (intersection of H,K), which is actually a group (so that we can talk about the index of this subset). If a subset (subgroup) of H has infinite index... How can H possibly have a finite index?
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 22072016 21:30
(Original post by Ecasx)
G being finite.
Since you have proven that the intersection of H and K is a group, you are done! From the proof of Lagrange, this subgroup has a particular (finite) index.
G being infinite. A good example of this that I can think of is G = Z with the + operation and subgroups H,K being the {3n  n in Z} and {4n  n in Z}. These have finite indexes in G. (Eg, the left cosets of H are {3n  n in Z}, {3n+1  n in Z}, and {3n+2  n in Z}.) The intersection of H and K is {12n  n in Z} which has index 12 in G  clearly finite.
The only problem is that it would seem difficult to generalise this example to other infinite groups. Here I would turn to contradiction.
Suppose that H and K have finite indexes in G but their intersection has an infinite index. The intersection of H and K is a subgroup of H. At this point, I'm going to point out that K is irrelevant  K is only here to give us a subset of H (intersection of H,K), which is actually a group (so that we can talk about the index of this subset). If a subset (subgroup) of H has infinite index... How can H possibly have a finite index?
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I have the following argument so far:
Let be the elements that account for all distinct
and
be the elements that account for all distinct .
As any two subgroups are either equal or disjoint,
But is a rightcoset so we can choose our xi, yj such that wlog . Still cant see the final argument from this though...
p.s apologies for this disgusting formatting, if someone knows how to 'displaysyle' the intersections and have 2 lines of subscript one under the other, id appreciate it.Last edited by EnglishMuon; 22072016 at 21:35. 
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 22072016 21:56

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 22072016 22:40
(Original post by EnglishMuon)
p.s apologies for this disgusting formatting, if someone knows how to 'displaysyle' the intersections and have 2 lines of subscript one under the other, id appreciate it. 
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 23072016 00:25
(Original post by EnglishMuon)
Ah I might have it:
Basically I think you can show if isnt the empty set. So
hence
for i.e. there are less than or equal to mn right cosets of H n K in G:
(as may equal some )
I would suggest looking for a contradiction. We have H and K being finiteindexed subgroups. We have T = (H intersection K) being a subgroup of H. If T has infinite index (infinitely many different cosets), then surely H must have infinite index too  why?
The benefit of contradiction here is that there is no detailed considerations needing to be made, in contrast to m, n being the number of cosets of H, K in your attempt, for example.
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 23072016 10:08
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Updated: September 28, 2016
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