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The Proof is Trivial!

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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 C2×C8C_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?
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,HG, H be cyclic of order nn, and let g,hg, h be generators for them. You can check that ϕ:G→H\phi: G \to H by gi↦hig^i \mapsto h^i is a group isomorphism.
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>A=<a> and B=<b>B=<b> and then define ϕ:A→B\phi : A \to B by ϕ(aj)=bj∀j\phi(a^j)=b^j \forall j
(edited 8 years ago)
Original post by Smaug123
It's because they're cyclic. Let G,HG, H be cyclic of order nn, and let g,hg, h be generators for them. You can check that ϕ:G→H\phi: G \to H by gi↦hig^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.
Original post by Renzhi10122
From doing n=m=0, you should be able to just compete the question. You should get

Spoiler



may you please post your solution to the possible solutions to f(2001)?
would be appreciated
thanks
Original post by demigawdz
may you please post your solution to the possible solutions to f(2001)?
would be appreciated
thanks


Spoiler

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.
(edited 8 years ago)
Original post by TheMagicMan
Yes of course. Write the groups as A=<a>A=<a> and B=<b>B=<b> and then define ϕ:A→B\phi : A \to B by ϕ(aj)=bj∀j\phi(a^j)=b^j \forall j


You want \langle and \rangle (⟨,⟩\langle, \rangle) rather than <, > :smile:
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.
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

Original post by Renzhi10122
Woops, you probably wanted a full solution:

Spoiler



I don't think this solution is rigorous. In particular f is not necessarily surjective.
(edited 8 years ago)
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.
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.
(edited 8 years ago)
Original post by TheMagicMan
I don't think this solution is rigorous. In particular f is not necessarily surjective.

Spoiler

Original post by Renzhi10122

Spoiler



Why does f(0)f(0) have to be 1? Why can't it be 3 say?
Original post by TheMagicMan
Why does f(0)f(0) have to be 1? Why can't it be 3 say?


Because I'm dumb :frown: . Welp ,I've done this wrong
(edited 8 years ago)
Original post by Renzhi10122
Because I'm dumb :frown:


dw i got nowhere lol

but maybe that's because i've never actually solved one before..
Original post by Renzhi10122
Because I'm dumb :frown:


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


Here's a solution to this part

Spoiler

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


Here's a solution to this part

Spoiler



Ah thanks, got it after the hint. I'm feeling real dumb right now haha.
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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