Prove/disprove these subgroups

[RJLiberator]

Homework Statement

Prove or disprove each of the following:

a) {f ∈ S_n | f(1) = 1} is a subgroup of Sn
b) {e}∪{f∈D_n | f is a rotation} is a subgroup of D_n
c) {e} ∪ { f∈D_n | f is a reflection} is a subgroup of D_n

Homework Equations

e = identity element

Subgroup if:
i) e exists (identity)
ii) if x, y exist in the set then x*y exists
iii) for all x existing in the set, there is an inverse existing.

The Attempt at a Solution

a) we can prove that this is indeed a subgroup by following the subgroup definition
1. the identity is 1 and that exists in f(1) = 1.
2. For everything that exists in f(1)=1, 1*1 = 1. That exists still.
3. Since 1 is the only thing that exists in this set, 1*1 = 1 which is also the inverse element.

So yes, this is a subgroup of Sn.

b) 1. The identity clearly is part of this set.
2. since all elements are rotations, then they all exist when you multiply each of them. rot1*rot2 = just some other rotation?
3. Wouldn't the inverse be the opposite rotation?

c) 1. The identity clearly is part of this set.
2. the reflections all exist, if you multiply them, theyll just be reflections that still exist.
3. Wouldn't the inverse be the opposite reflection?Is my thinking correct?

Thank you.

 

[mfb]

1. the identity is 1 and that exists in f(1) = 1.

"f(1) = 1" is not a set, you cannot have something that exists in it.
{f ∈ S_n | f(1) = 1} has more than the identity element. For n=3, it has (f(1)=1, f(2)=3, f(3)=2) for example.

What exactly is a rotation? A cyclic permutation? Same question for reflection.

rot1*rot2 = just some other rotation?

That's what you have to find out.

3. Wouldn't the inverse be the opposite rotation?

Sure.

 

[geoffrey159]

I understand $S_n$ is the symmetric group. What is $D_n$ ?

 

[RJLiberator]

"f(1) = 1" is not a set, you cannot have something that exists in it.
{f ∈ S_n | f(1) = 1} has more than the identity element. For n=3, it has (f(1)=1, f(2)=3, f(3)=2) for example.

I see, so now I am unclear on what a) is trying to say. They say f exists in S_n such that f(1) = 1. But this implies that f(2) = 3 somehow? How is that possible?@geoffrey159 I am unsrue what D_n signifies here. It wasn't brought up to us in our notes yet. I may have to do some digging.

 

[mfb]

But this implies that f(2) = 3 somehow?

It does not. I made an example.

 

[RJLiberator]

Oh, I see now, the n dictates it.

So, if n = 4
then we can safely say that f(2) = 2, f(3) = 3, f(4) = 4
correct?

 

[Samy_A]

$D_n$ could be the dihedral group.

That is at least consistent with the mention of rotations and reflections.

https://en.wikipedia.org/wiki/Dihedral_group

 

[mfb]

Oh, I see now, the n dictates it.

So, if n = 4
then we can safely say that f(2) = 2, f(3) = 3, f(4) = 4
correct?

That is one of your permutations in the set. It is not the only one.
f(1)=1, f(2)=2, f(3)=4, f(4)=3 is another one in the set, and there are more.

Dihedral groups - okay, then reflections and rotations are clear.

2. the reflections all exist, if you multiply them, theyll just be reflections that still exist.

You'll have to prove that.

 

[RJLiberator]

Okay so let's start with a then. I thought I initially had it, but clearly I was a bit off.

Subgroup if:
i) e exists (identity)
ii) if x, y exist in the set then x*y exists
iii) for all x existing in the set, there is an inverse existing.

a) {f ∈ S_n | f(1) = 1} is a subgroup of Sn

i) e exists as e exists in a symmetrical group?
ii) if x,y are elements in this set that is defined by the question, then x*y exists as f(x) = x and f(y) = y so f(xy) = xy.
iii) Identity exists as it must in a symmetric group?

Meh, this is not going well.

 

[Samy_A]

Okay so let's start with a then. I thought I initially had it, but clearly I was a bit off.

Subgroup if:
i) e exists (identity)
ii) if x, y exist in the set then x*y exists
iii) for all x existing in the set, there is an inverse existing.

This is somewhat unclear.
Basically, a subgroup is a subset of a group that is also a group under the same binary operation as the group.
To paraphrase your definition:
A ⊆ S_n is a subgroup if:
i) e ∈ A (identity)
ii) if x, y ∈ A, x*y ∈ A
iii) if x ∈ A, x^-1 ∈ A

a) {f ∈ S_n | f(1) = 1} is a subgroup of Sn

i) e exists as e exists in a symmetrical group?
ii) if x,y are elements in this set that is defined by the question, then x*y exists as f(x) = x and f(y) = y so f(xy) = xy.
iii) Identity exists as it must in a symmetric group?

Meh, this is not going well.

For a), you have to prove that A = {f ∈ S_n | f(1) = 1} is a subgroup of S_n.
Referring to the definition of a subgroup, you have answer the following questions:
i) is the identity permutation in A?
ii) if f,g ∈ A, meaning f(1)=1, g(1)=1, is f*g (where the * stands for the group operation of S_n) also in A?
iii) if f ∈ A, meaning f(1)=1, is f^-1 ∈ A ?

If the answer is Yes to all of them, A is a subgroup.

 

[RJLiberator]

Here's what I am a bit unclear of. We talk of * being the group operation, but how is the group operation defined in the problem?
Is this something about S_n that I am not knowing/using.

I see the identity permutation as being f(1)=1. done!
I see, ii) being simply done as f*g = f(1)g(1) = 1*1 = 1 done!
I see iii) as f(1)=1=e=so yes it works.

Perhaps my proof is written hastly and unclear, but i hope you can see what I mean. While I am struggling with the correct notation and wording, I think i have the general idea of this part done. I would say that part a is definitely a subgroup.

The key distinction for me is what is the operation here and what is the identity. If the identity is simple "1" then yes it all works out for me.

 

[Samy_A]

I see, ii) being simply done as f*g = f(1)g(1) = 1*1 = 1 done!

Let's start with this one.
S_n is by definition the group of all permutations of a finite set of n elements.
Traditionally, one takes as set S={1,2,3,...,n}, and S_n is the set of all permutations of S.
But, the numbers in S are not important. You could, instead of using {1,2,3,4} for the definition of S₄, use the sets S'={a,b,c,d}, S"={cat, dog, lion, tiger}, ... Any set consisting of exactly 4 elements.

So, when you have to check something about f*g for f,g ∈ S_n, f*g is the composition of two permutations. First you apply g, then f. The product 1*1, as you wrote, is irrelevant. If we had used the set S", how would you have calculated cat*cat?

The key distinction for me is what is the operation here and what is the identity. If the identity is simple "1" then yes it all works out for me.

The group operation is composition of functions (here permutations). The identity is not the number 1, but the identity permutation, the permutation that maps each element to itself.
If we denote the identity element of S₄ by e, then:
e(1)=1
e(2)=2
e(3)=3
e(4)=4

Here I used the usual set {1,2,3,4}.

If I want to be a little silly and use S"={cat, dog, lion, tiger}, then e, the identity element of S₄ would be the following permutation:
e(cat)=cat
e(dog)=dog
e(lion)=lion
e(tiger)=tiger
In this case, of course, part a) of the exercise would be: prove that {f ∈ S_n | f(cat)=cat} is a subgroup of S_n.

Let us not be silly, and stick with permutations of {1,2,3,4}. Just remember that these numbers are symbols, and that multiplying them as numbers (1*1,2*3) is irrelevant when discussing S₄.

 

[mfb]

I think the problem is the set definition. A = {f ∈ Sn | f(1) = 1} is the set of all permutations where the number 1 is mapped onto the number 1. The other elements can be mapped in any way. Now you have to show:
(i) the identity permutation is a permutation where 1 is mapped onto 1 (well, that is trivial)
(ii) if you perform two permutations which both map 1 onto 1, then the resulting permutation is also one where 1 is mapped onto 1
(iii) the inverse permutation to any permutation that maps 1 onto 1 is again a permutation that maps 1 onto 1.

 

[RJLiberator]

Okay, thank you both for the words. Samy_A I think you did much to improve my understanding of this problem.

With mfb's i, ii, and iii, words, let me try this:

(i) the identity permutation is a permutation where 1 is mapped onto 1 (well, that is trivial)

Could we say, by the definition of the statement, we have the permutation. Since the statement was f(1)=1, this IS the identity permutation.

Here, I am a touch confused, as we noted earlier and as you said now:

The other elements can be mapped in any way.

So you are saying that the other elements could be mapped as f(2) = 3. But this is NOT the identity permutation.
How do I deal with this? Is it enough to simply say, by the definition we see f(1)=1 so the identity permutation exists?

(ii) if you perform two permutations which both map 1 onto 1, then the resulting permutation is also one where 1 is mapped onto 1

Is this as simple as saying (f°f)(1) = 1 = f(1)f(1)@Samy_A

If we had used the set S", how would you have calculated cat*cat?

I just don't understand how you could. From the looks of it, we don't have a defined operation. We say that it is a permutation. So cat*cat would be a permutation on cat with a permutation on another cat.

 

[mfb]

Since the statement was f(1)=1, this IS the identity permutation.

That does not make sense. f(1)=1 is not a permutation. It is an equation, telling you something (but not everything) about f. Don't take that out of context. Your set of permutations is A = {f ∈ Sn | f(1) = 1}. It is not "f(1)=1".

Is this as simple as saying (f°f)(1) = 1 = f(1)f(1)

What is f(1)f(1)? There is no multiplication involved at all. The group operation means (f°f)(1) = f(f(1)).

I just don't understand how you could.

You cannot, that is the point. Yet you try to do so with "1*1". Which has some reasonable definition with real numbers but that is pure coincidence.

 

[RJLiberator]

What is f(1)f(1)? There is no multiplication involved at all. The group operation means (f°f)(1) = f(f(1)).

Precisely. I suppose that it was I meant.

That does not make sense. f(1)=1 is not a permutation. It is an equation, telling you something (but not everything) about f. Don't take that out of context. Your set of permutations is A = {f ∈ Sn | f(1) = 1}. It is not "f(1)=1"

Ok. So I've been reading on wiki about permutations and so forth. I feel like, what I have is the following:

[ 1 2 3 ... n ]
and this symmetric map maps to
[ 1 x x ... x ]

Where I don't have any further knowledge on what x is other than the fact that this is a symmetric group.

Considering I don't know what x is in the symmetric group, all I feel I know (based on my lofty definition of symmetric groups) is that x is unique and one to one, and all x's possible get mapped to some number.

In theory, let's say we have
[1 2 3 4] maps to (map f)
[1 3 2 4]Could I say then, well the identity is available as f(f(2)) = 2 (this is just an example for understanding not a full fledged proof yet)

 

[Samy_A]

Precisely. I suppose that it was I meant.
Ok. So I've been reading on wiki about permutations and so forth. I feel like, what I have is the following:

[ 1 2 3 ... n ]
and this symmetric map maps to
[ 1 x x ... x ]

Where I don't have any further knowledge on what x is other than the fact that this is a symmetric group.

Considering I don't know what x is in the symmetric group, all I feel I know (based on my lofty definition of symmetric groups) is that x is unique and one to one, and all x's possible get mapped to some number.

In theory, let's say we have
[1 2 3 4] maps to (map f)
[1 3 2 4]Could I say then, well the identity is available as f(f(2)) = 2 (this is just an example for understanding not a full fledged proof yet)

Yes. The identity element e of S₄ maps 1→1, 2→2, 3→3, 4→4. So certainly it satisfies e(1)=1, and thus is an element of A.
Now you still have to prove point iii) as mentioned by @mfb , and that will conclude part a) of the exercise.

 

[RJLiberator]

So, to summarize.

i) e is an element of A as e(1)=1.
ii) We have associativity as f(1)f(1)=1=(f*f)(1) where * is the composition of the two maps.
iii) The inverse is part of A as f(1)*f(1) = 1 = e

Does that wrap up part a correctly?

 

[Samy_A]

So, to summarize.

i) e is an element of A as e(1)=1.
ii) We have associativity as f(1)f(1)=1=(f*f)(1) where * is the composition of the two maps.
iii) The inverse is part of A as f(1)*f(1) = 1 = e

Does that wrap up part a correctly?

Not really.
i) is correct.
ii) f(1)f(1) doesn't make any sense. You have to prove that for f,g ∈ A, f°g ∈ A.
iii) Again, f(1)*f(1) doesn't make any sense. You have to prove that for f ∈ A, f^-1 ∈ A.

Remember that by definition of A, f ∈ A means that f(1)=1.

 

[RJLiberator]

but there is no g and that is my confusion, perhaps you guys have told me over and over in this thread and its not hitting me.

Let me think.

If we let g(1)=1 and f(1)=1 then we have a g, I guess. So f°g(1) = f(g(1)) = f(1) = 1.
Now that seems more correct, I think.

for iii), for f(1)=1, f^-1 exists as f°g(1)=e. Letting g(1)= 1 as prescribed in the defintion statement we see g(1) = f^(-1)

 

[Samy_A]

but there is no g and that is my confusion, perhaps you guys have told me over and over in this thread and its not hitting me.

You have to prove that A is a subgroup. That means (among other things) that the group product of two elements of A is also an element of A.
To do that, you have to take two arbitrary elements of A, give them a name (I used f and g), and see if the product is in A. You did this correctly, see below.

Let me think.

If we let g(1)=1 and f(1)=1 then we have a g, I guess. So f°g(1) = f(g(1)) = f(1) = 1.
Now that seems more correct, I think.

Yes, that is all there is to it for ii).

for iii), for f(1)=1, f^-1 exists as f°g(1)=e. Letting g(1)= 1 as prescribed in the defintion statement we see g(1) = f^(-1)

Well, this is somewhat unclear. I assume that g=f^-1. You don't have to "let g(1)=1",you have to prove that this the case for g, the inverse of f ∈ A.

 

[RJLiberator]

Excellent. I am starting to learn this and i'll try to apply it to parts b and c next ( either now or tonight).

Yes g=f^-1 and so (f°g)(1) = e implies that g is the inverse.
f(g(1)) where g(1) = 1 so then we have f(1) = 1 = e.

 

[Samy_A]

Excellent. I am starting to learn this and i'll try to apply it to parts b and c next ( either now or tonight).

Yes g=f^-1 and so (f°g)(1) = e implies that g is the inverse.
f(g(1)) where g(1) = 1 so then we have f(1) = 1 = e.

I think you have things backward here. You know that f ∈ A, meaning f(1)=1.
You have to prove that g(1)=1.
Hint: since g=f^-1, g°f=e (where e is the identity element of S₄, the permutation that sends each element to itself).

 

[mfb]

(f°g)(1) = e

That is not true. The left side is a number, the right side is a permutation, they cannot be equal.
(f°g) = e - combining both gives the identity element (which is what you want)
(f°g)(1) = e(1)=1 - here we look where 1 gets mapped to. This equation follows from the one above.

Now you have to prove g(1)=1.