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 December 17th, 2017, 11:19 AM #1 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 group = function Can I define some functions as groups?
December 17th, 2017, 03:20 PM   #2
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Quote:
 Originally Posted by policer Can I define some functions as groups?
The set of continuous functions from the reals to the reals with the operation of pointwise addition is an Abelian group. Is that the kind of thing you have in mind?

Last edited by Maschke; December 17th, 2017 at 03:40 PM.

 December 18th, 2017, 12:01 AM #3 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 Can somebody give me, (please!!!), more examples?
 December 18th, 2017, 01:21 AM #4 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 Thanks. I don't think that there is more examples. You so nice, that you answer me. Thanks.
December 18th, 2017, 08:57 AM   #5
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Quote:
 Originally Posted by policer Can somebody give me, (please!!!), more examples?
Same as above with differentiable functions. Or with arbitrary functions. You can make up a lot of different examples based on this general idea.

 December 18th, 2017, 11:02 AM #6 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 Can you give an example(s) of arbitrary function that you notice? It is very nice from you to answer me.
 December 18th, 2017, 11:06 AM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152 I should point out at this point that OP is getting a different answer over on mathhelpforum.com, or rather an answer to a different question. Maschke is giving examples of sets of functions that via their properties form a group. On the other board the question "is a function a group" is being answered no. I just want to keep things clear. Thanks from Maschke
December 18th, 2017, 11:53 AM   #8
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Quote:
 Originally Posted by policer Can you give an example(s) of arbitrary function that you notice? It is very nice from you to answer me.
The set of bijections from any set to itself is a group under composition of functions. In the case of a finite set of $n$ elements, the set of bijections is the symmetric group on $n$ letters, denoted $S_n$.

Can you say more about why you're asking these questions? There are many sets of functions that are groups under various operations.

Last edited by Maschke; December 18th, 2017 at 11:59 AM.

 December 18th, 2017, 09:56 PM #9 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 I don't understand a function (or some functions) is a group(s) or not?!
 December 19th, 2017, 05:19 AM #10 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Your question, frankly, doesn't make much sense. Do you know what a "group" is? A group is a set of things, together with a "binary operation" such that: For any two objects in the set, combining with the binary operation, a*b, is an object in the set. (called "closure"- we don't get outside the set.) We can "combine" three objects because a*(b*c)= (a*b)*c. (The operation is "associative".) There is an "identity"- a single member "i" such that, for any member of the set, a, a*i= i*a= a. Every member of the set has an "inverse"- for every a in the set there exist b such that a*b= b*a= i. I don't know what you could mean by a single function being a group. Given a single function, f, we could always define the operation f*f= f so that we have a group containing only f- but that is not very interesting! The responses above have shown that there are many sets of functions that are groups using the "composition of functions" as operation. And since the real numbers form a group using "addition" as operation, there exist many sets of functions that are groups using addition of functions as the operation. Thanks from greg1313

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