September 10th, 2013, 04:58 AM  #1 
Newbie Joined: Sep 2013 Posts: 7 Thanks: 0  Question on groups
Question from Modern Algebra an Intro: Q. if S>1, then M(S), the set of all mappings from S to S, is not a group with respect to composition, Why? I can't figure it out! 
September 10th, 2013, 10:04 AM  #2  
Senior Member Joined: Aug 2012 Posts: 2,413 Thanks: 755  Re: Question on groups Quote:
Can you see one that's certain to fail for M(S)? Once you write down the axioms, this will be fairly obvious.  
September 10th, 2013, 10:20 AM  #3 
Newbie Joined: Sep 2013 Posts: 7 Thanks: 0  Re: Question on groups
I'm thinking that it's the fact that not all elements of M(S) have an inverse? I know it probably should be obvious, but I just can't get my head around it.

September 10th, 2013, 12:01 PM  #4  
Senior Member Joined: Aug 2012 Posts: 2,413 Thanks: 755  Re: Question on groups Quote:
Now your task is to prove it. Can you give an example of some set S and write down a few elements of M(S)? Any old set S will do. As a way to get started, choose some set S and write down a few elements of M(S) that do have inverses.  
September 11th, 2013, 03:52 AM  #5  
Newbie Joined: Sep 2013 Posts: 7 Thanks: 0  Re: Question on groups Quote:
Second, I think I'm getting there. I wrote out a Cayley table for S = set of permutations of order 3. That worked out well and it satisfies all the group axioms. Is it true that all permutations are groups in M(S)? due to the fact that they bijective and therefore all have inverses? I'm finding it hard to think of a set S in M(S) that has elements without an inverse. Can you let S = the set of all integers or something like that?  
September 11th, 2013, 08:03 AM  #6  
Senior Member Joined: Aug 2012 Posts: 2,413 Thanks: 755  Re: Question on groups Quote:
But you're overthinking this. Let S = any old set at all. Can you write down a function that's not invertible? Let S = {1,2,3}. Write down all the elements of M(S).  
September 11th, 2013, 11:48 AM  #7 
Newbie Joined: Sep 2013 Posts: 7 Thanks: 0  Re: Question on groups
if S = {1,2,3} then the elements of M(S) would be the mappings 1>2, 1>3, 2>1, 2>3, 3>1, 3>2 is that right?

September 11th, 2013, 11:53 AM  #8  
Senior Member Joined: Aug 2012 Posts: 2,413 Thanks: 755  Re: Question on groups Quote:
 
September 12th, 2013, 03:37 AM  #9 
Newbie Joined: Sep 2013 Posts: 7 Thanks: 0  Re: Question on groups
OK so my book says: A mapping from a set S to a set T is a relationship that assigns to each element in S a uniquely determined element in T. So if I'm looking at mappings in M(S), S={1,2,3}, then two examples would be ... Q: 1>2, 2>3, 3>2 R: 1>3, 2>1, 3>3 So Q,R are elements of M(S) but Q and R don't form a group with respect to composition, because both of them are neither 11 nor onto so therefore Q o R isn't invertible? Would that be correct? 
September 12th, 2013, 06:59 AM  #10 
Senior Member Joined: Aug 2012 Posts: 2,413 Thanks: 755  Re: Question on groups
Do you know what a function is? Do you know that a function is the same thing as a mapping? They're just two words for the same thing. What about a function that maps everything to the same element? Perhaps "uniquely determined" is a confusing phrase. It means that if 1 > 2 then 1 can't also go to 3. But 1 could go to 2 and 2 could also go to 2. 

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