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    (Original post by TheMagicMan)
    There are groups satisfying that property but to my knowledge there are no elementary counterexamples.

    Some research has led me to the fact that the modular group order 16 and C_2 \times C_8 is a counterexample. (http://www.opensourcemath.org/gap/small_groups.html)

    You should never have to resort to such a theorem in A-level though.

    One final thought that occurred to me about order 16 groups. There are 14 non-isomorphic groups order 16, and each of these groups can have at most 10 subgroups (itself, the 5 groups order 8, the 2 groups order 4, the 1 group order 2 and the one group order 1) so of course order 16 will contain a counterexample to your proposition! Indeed there are 51 groups order 32 so there will be an order 32 counterexample too. And so on.
    Are cyclic groups of the same order always isomorphic, if so, why?
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    (Original post by Primus2x)
    Groups A and B have the same order and have the same number of subgroups, are they necessarily isomorphic? If this is true, can I use it in an A-Level MEI FP3 answer without explaining it?
    I came across a past paper question asking me to explain why all groups of order 5 are always isomorphic to eachother. The first thing I wrote down was "5 is a prime number so by Lagrange's Theorem, the only subgroups it can have is itself and a group only containing its identity"

    So the second part of my answer is "So all order 5 groups have the same number of subgroups, therefore all of them are isomorphic to eachother". But I am not sure if this is correct.
    It's because they're cyclic. Let G, H be cyclic of order n, and let g, h be generators for them. You can check that \phi: G \to H by g^i \mapsto h^i is a group isomorphism.
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    (Original post by Primus2x)
    Are cyclic groups of the same order always isomorphic, if so, why?
    Yes of course. Write the groups as A=<a> and B=<b> and then define \phi : A \to B by \phi(a^j)=b^j  \forall j
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    (Original post by Smaug123)
    It's because they're cyclic. Let G, H be cyclic of order n, and let g, h be generators for them. You can check that \phi: G \to H by g^i \mapsto h^i is a group isomorphism.
    I showed that one order five group was cyclic in the first party the question. I was trying a different approach to show that all of order five groups are cyclic so they are all isomorphic, but I am not sure how.
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    (Original post by Renzhi10122)
    From doing n=m=0, you should be able to just compete the question. You should get
    Spoiler:
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     f(f(0))=f(0)+1 and then use the first condition
    may you please post your solution to the possible solutions to f(2001)?
    would be appreciated
    thanks
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    (Original post by demigawdz)
    may you please post your solution to the possible solutions to f(2001)?
    would be appreciated
    thanks
    Spoiler:
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    I get one solution of  f(x)=x+1 , which works, so  f(2001)=2002 is the only solution
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    (Original post by Primus2x)
    I showed that one order five group was cyclic in the first party the question. I was trying a different approach to show that all of order five groups are cyclic so they are all isomorphic, but I am not sure how.
    take a group G of order p (=5). pick an arbitrary non-trivial element and consider the group generated by that element.

    Then write down a homomorphism from G to the cyclic group order p (=5) which maps ______ to ________ and show that it is an isomorphism.
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    (Original post by TheMagicMan)
    Yes of course. Write the groups as A=<a> and B=<b> and then define \phi : A \to B by \phi(a^j)=b^j  \forall j
    You want \langle and \rangle (\langle, \rangle) rather than <, >
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    (Original post by Primus2x)
    I showed that one order five group was cyclic in the first party the question. I was trying a different approach to show that all of order five groups are cyclic so they are all isomorphic, but I am not sure how.
    You could also use Lagrange: the order of any element is either 1 or 5. There's only one element of order 1 - the identity - so the other four must be of order 5. In particular, there is an element of order 5, so that element must generate the group.
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    (Original post by demigawdz)
    may you please post your solution to the possible solutions to f(2001)?
    would be appreciated
    thanks
    Woops, you probably wanted a full solution:

    Spoiler:
    Show

    Let n=0
     f(f(m))=f(m)+1

    Let  f(m)=a , then  f(a)=a+1 is the only possible function. We check this works by a substitution into condition 2 to give  f(m+n+1)=m=n+2 so this function satisfies both conditions, so the only possible value of  f(2001)=2002
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    (Original post by Renzhi10122)
    Woops, you probably wanted a full solution:

    Spoiler:
    Show

    Let n=0
     f(f(m))=f(m)+1

    Let  f(m)=a , then  f(a)=a+1 is the only possible function. We check this works by a substitution into condition 2 to give  f(m+n+1)=m=n+2 so this function satisfies both conditions, so the only possible value of  f(2001)=2002
    I don't think this solution is rigorous. In particular f is not necessarily surjective.
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    (Original post by TheMagicMan)
    I don't think this solution is rigorous. In particular f is not necessarily surjective.
    Ah woops, I'll have another go then.
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    (Original post by TheMagicMan)
    I don't think this solution is rigorous. In particular f is not necessarily surjective.
    Actually, why is my solution not rigorous? I showed that if there is a solution, then it is what I said, and then showed that that solution worked.

    Edit: Nevermind, I see where I went wrong.
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    (Original post by TheMagicMan)
    I don't think this solution is rigorous. In particular f is not necessarily surjective.
    Spoiler:
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    Let m=n=0
     f(f(0))=f(0)+1 \Rightarrow f(0)=1 by condition 1
    Now let m=0
     f(n+f(0))=f(0)+n+1 \Rightarrow f(n+1)=n+2 and the rest follows as before
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    (Original post by Renzhi10122)
    Spoiler:
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    Let m=n=0
     f(f(0))=f(0)+1 \Rightarrow f(0)=1 by condition 1
    Now let m=0
     f(n+f(0))=f(0)+n+1 \Rightarrow f(n+1)=n+2 and the rest follows as before
    Why does f(0) have to be 1? Why can't it be 3 say?
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    (Original post by TheMagicMan)
    Why does f(0) have to be 1? Why can't it be 3 say?
    Because I'm dumb . Welp ,I've done this wrong
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    (Original post by Renzhi10122)
    Because I'm dumb
    dw i got nowhere lol

    but maybe that's because i've never actually solved one before..
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    (Original post by Renzhi10122)
    Because I'm dumb
    Here's a hint: Suppose f(0)=a. What is f(a)? We have that f(x) \geq f(0)+x. use this to determine an inequality on f(a)


    Here's a solution to this part

    Spoiler:
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    We have ff(m)=f(m)+1.

    let f(0)=a.

    Then f(a)=a+1 setting m=n=0 in condition 2

    Now f(x) \geq f(0)+x = x+a

    so for x&lt;a have f(a) \geq f(x) + a -x \geq x+a+a-x = 2a

    so  2a \leq a+1

    so a=0,1

    If a=0, then  f(0)=1 contradiction.

    So a=1 and by the first line f(n)=n+1

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    (Original post by TheMagicMan)
    Here's a hint: Suppose f(0)=a. What is f(a)? We have that f(x) \geq f(0)+x. use this to determine an inequality on f(a)


    Here's a solution to this part
    Spoiler:
    Show


    We have ff(m)=f(m)+1.

    let f(0)=a.

    Then f(a)=a+1 setting m=n=0 in condition 2

    Now f(x) \geq f(0)+x = x+a

    so for x&lt;a have f(a) \geq f(x) + a -x \geq x+a+a-x = 2a

    so  2a \leq a+1

    so a=0,1

    If a=0, then  f(0)=1 contradiction.

    So a=1 and by the first line f(n)=n+1

    Ah thanks, got it after the hint. I'm feeling real dumb right now haha.
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    (Original post by Renzhi10122)
    Ah thanks, got it after the hint. I'm feeling real dumb right now haha.
    You shouldn't feel dumb. It's the last question on a BMO. If you find it easy at 18/19 you're part of a very small group.
 
 
 
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