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September 10th, 2013, 04:58 AM   #1
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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!
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September 10th, 2013, 10:04 AM   #2
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Re: Question on groups

Quote:
Originally Posted by Leo_J
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!
What are the group axioms?

Can you see one that's certain to fail for M(S)? Once you write down the axioms, this will be fairly obvious.
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September 10th, 2013, 10:20 AM   #3
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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.
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September 10th, 2013, 12:01 PM   #4
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Re: Question on groups

Quote:
Originally Posted by Leo_J
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.
Yes, that is the right intuition!

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.
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September 11th, 2013, 03:52 AM   #5
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Re: Question on groups

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Originally Posted by Maschke
Quote:
Originally Posted by Leo_J
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
Yes, that is the right intuition!

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.
OK so first of all thanks for this it's helping a lot!
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?
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September 11th, 2013, 08:03 AM   #6
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Re: Question on groups

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Originally Posted by Leo_J
\
OK so first of all thanks for this it's helping a lot!
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?
Yes, the collection of all permutations of a set is a group, for the reasons you mention.

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).
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September 11th, 2013, 11:48 AM   #7
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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?
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September 11th, 2013, 11:53 AM   #8
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Re: Question on groups

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Originally Posted by Leo_J
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?
You need to review the definition of "mapping." What does your book say a mapping is? Write out the full definition here. Then show a few examples.
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September 12th, 2013, 03:37 AM   #9
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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 1-1 nor onto so therefore Q o R isn't invertible?

Would that be correct?
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September 12th, 2013, 06:59 AM   #10
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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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