A Summer of Maths (ASoM) 2016

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    (Original post by drandy76)
    Always get this mixed up, is that a divides n, or n divides a?


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    a divides n so n/a=k EZ


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    (Original post by drandy76)
    Always get this mixed up, is that a divides n, or n divides a?


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    a divides n
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    Dunno how to double quote so cheers lads


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    Name:  ImageUploadedByStudent Room1468610947.215133.jpg
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    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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    (Original post by drandy76)
    Name:  ImageUploadedByStudent Room1468610947.215133.jpg
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    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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    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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    (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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    Yeah it was giving me a headache trying to trace it lol, thanks geometry nerd


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    (Original post by drandy76)
    Yeah it was giving me a headache trying to trace it lol, thanks geometry nerd


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    No geometry in this one lad. Just using those eyes.


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    (Original post by physicsmaths)
    No geometry in this one lad. Just using those eyes.


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    I have 1 1 vision bro


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    If  a,b \in \mathbb{Z}_{p} for p prime, why are there p-1 choices for a,b such that  ab = \pm 1 (mod p) ?
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    (Original post by EnglishMuon)
    If  a,b \in \mathbb{Z}_{p} for p prime, why are there p-1 choices for a,b such that  ab = \pm 1 (mod p) ?
    for each of 1,...,p-1 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
    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 p-1 (or -1 if you prefer) to get the corresponding pair that multiply to -1
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    (Original post by 13 1 20 8 42)
    for each of 1,...,p-1 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
    Yep, thought so for the +1 case. Was the -1 case I wasnt sure how to explain though. thanks
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    (Original post by EnglishMuon)
    Yep, thought so for the +1 case. Was the -1 case I wasnt sure how to explain though. thanks
    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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    (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
    thanks yea that makes sense. My mistake was that after multiplying one of a or b but p-1 stupidly I forgot that that is congruent to one of the other elements in Zp.
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    Any hints on how to prove this? :

    If  G is a group and  H, K are subgroups of G of finite index in G, prove that  H \cap K is of finite index in  G .

    Some initial thoughts:
    Spoiler:
    Show
    Its straightforwards to show  H \cap K 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  {a_{1}, a_{2}, ... a_{m}} \in G such that  Ha is distinct

    and n distinct elements  {b_{1}, b_{2}, ... b_{n}} \in G such that  Kb is distinct.

    Now Itd be nice to say that these  {a_{1}, a_{2}, ... a_{m}},  {b_{1}, b_{2}, ... b_{n}} \in G constitute to all elements g in G such that  (H \cap K ) g are distinct, but I dont think thats obviously the case- imagine forming  Ha where  Ha=Hb for some  a \not= b . Although the elements in  Ha are the same as in  Hb couldnt the elements in  (H \cap K)a be different to  (H \cap K)b still be different ? (i.e. we have permuted H)


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    (Original post by EnglishMuon)
    Any hints on how to prove this? :

    If  G is a group and  H, K are subgroups of G of finite index in G, prove that  H \cap K is of finite index in  G .

    Some initial thoughts:
    Spoiler:
    Show
    Its straightforwards to show  H \cap K 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  {a_{1}, a_{2}, ... a_{m}} \in G such that  Ha is distinct

    and n distinct elements  {b_{1}, b_{2}, ... b_{n}} \in G such that  Kb is distinct.

    Now Itd be nice to say that these  {a_{1}, a_{2}, ... a_{m}},  {b_{1}, b_{2}, ... b_{n}} \in G constitute to all elements g in G such that  (H \cap K ) g are distinct, but I dont think thats obviously the case- imagine forming  Ha where  Ha=Hb for some  a \not= b . Although the elements in  Ha are the same as in  Hb couldnt the elements in  (H \cap K)a be different to  (H \cap K)b still be different ? (i.e. we have permuted H)


    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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    (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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    Thanks, and Indeed the integers was the example I was drawn to to help get a feel for whats happening!

    I have the following argument so far:

    Let  x_{1},...,x_{m} be the elements that account for all distinct  Hx
    and
     y_{1},...,y_{n} be the elements that account for all distinct  Ky .
    As any two subgroups are either equal or disjoint,
     G= G \cap G= ( \cup ^{m} _{i=1} Hx_{i} ) \cap ( \cup ^{n} _{i=1} Ky_{i}) 

= \displaystyle\cup _{ 1 \leq i \leq m ,1 \leq j \leq n} (Hx_{i} \cap Ky_{j})

    But  He=H is a rightcoset so we can choose our xi, yj such that wlog  x_{1} = y_{1}=e . 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.
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    Ah I might have it:

    Basically I think you can show  Hx_{i}= Ky_{j} = Hg , g \in G if  Hx_{i} \cap Ky_{j} isnt the empty set. So  Hx_{i} \cap Ky_{j} = Hg \cap Hg = (H \cap K) g
    hence
     G= \cup ^{nm} _{i=1} (H \cap K) g_{i} for  g_{i} \in G i.e. there are less than or equal to mn right cosets of H n K in G:

     i_{G} (H \cap K) \leq i_{G} (H) i_{G} (K) (as  (H \cap K) g_{i} may equal some  (H \cap K) g_{j}, g_{i} \not= g_{j} )
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    (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.
    \bigcup: \displaystyle \bigcup_{1 \leq i \leq m, 1 \leq j \leq n}
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    (Original post by EnglishMuon)
    Ah I might have it:

    Basically I think you can show  Hx_{i}= Ky_{j} = Hg , g \in G if  Hx_{i} \cap Ky_{j} isnt the empty set. So  Hx_{i} \cap Ky_{j} = Hg \cap Hg = (H \cap K) g
    hence
     G= \cup ^{nm} _{i=1} (H \cap K) g_{i} for  g_{i} \in G i.e. there are less than or equal to mn right cosets of H n K in G:

     i_{G} (H \cap K) \leq i_{G} (H) i_{G} (K) (as  (H \cap K) g_{i} may equal some  (H \cap K) g_{j}, g_{i} \not= g_{j} )
    I don't quite see what you are trying to do. I don't think your first sentence is true: 'Basically I think you can show...'. If a coset of H, say Hx, and a coset of K, say Ky, are to be equal, then H and K must be of equal size, which is not guaranteed in the question.

    I would suggest looking for a contradiction. We have H and K being finite-indexed 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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    Ok. whats wrong with the first sentence? We know that any two right cosets in a group are either equal or disjoint. So if we are looking at the intersection of 1 right coset  Hx_{i} and another  Ky_{i} and this set is not empty, they must both equal some  Hg for some  g \in G . There was a typo actually in the 2nd line as I meant to say  Hx_{i} \cap Ky_{j} = Hg \cap Kg = (H \cap K) g but I don't see why the rest of argument doesn't work.
 
 
 
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