Groups and Subgroups: Clarifying Questions

  • Thread starter Shackleford
  • Start date
  • Tags
    Groups
In summary, the conversation discussed questions about sets and groups, specifically in the context of Z18 and Z4. The concept of generated sets and the use of <> notation were also mentioned. The group (Z,+) and the concept of a closed subset were also discussed, with an example of finding a closed subset that is not a group provided. The conversation also highlighted the importance of understanding the reasoning behind solutions and not just the final answer.
  • #1
Shackleford
1,656
2
Well, apparently, I'm not too clear on a few things.

For #4, what is <[8]> in the group Z18? What does the <> mean around the congruence class? Is my work correct?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161645.jpg?t=1311456158

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161721.jpg?t=1311456173 For #8, I understand making a closed subset of Z. It appears H = {[0], [1], [3]} is closed under addition in Z4. What exactly is the additive group Z?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161701.jpg?t=1311456186

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161733.jpg?t=1311456197
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
An additive group is one where the function is addition (a group is a set and a function, remember).
Z is the standard notation for the integers.

I recall that we used <> to denote the generated set. For instance <[8]> in Z_18 is the set of all numbers, modulo 18, that can be generated by x*8.

For instance, [8] is a member, as is [16], as is [6] (16+8=24=[6]).
 
  • #3
Disconnected said:
An additive group is one where the function is addition (a group is a set and a function, remember).
Z is the standard notation for the integers.

I recall that we used <> to denote the generated set. For instance <[8]> in Z_18 is the set of all numbers, modulo 18, that can be generated by x*8.

For instance, [8] is a member, as is [16], as is [6] (16+8=24=[6]).

Did you take a look at the problems and my work?
 
  • #4
Yeah, I sure did. Was there something that made you think I didn't?

I wasn't wanting to go way in depth.

To be honest, I couldn't really follow the work for the first question.

As for question 2, it asked about Z, the integers. You seemed to instead work in Z_4, which is a different set. I was trying to hint that this was not correct.
 
  • #5
Disconnected said:
Yeah, I sure did. Was there something that made you think I didn't?

I wasn't wanting to go way in depth.

To be honest, I couldn't really follow the work for the first question.

As for question 2, it asked about Z, the integers. You seemed to instead work in Z_4, which is a different set. I was trying to hint that this was not correct.

For #4, I said <[8]> = <[2]> in Z18. Is that correct?

Then, I generated a set in Z18 from <[2]>: {[16], [14], [12], [10], [8], [6], [4], [2], [0]}.

The order of this set is 9.

For #8, I know what a closed subset of Z is. A subset H of a group G has to be nonempty and if a, b are contained in H, then ab-1 is in H. They're asking for a subset of Z. Do they want a set of congruence classes in Z?
 
  • #6
I am not sure if <[8]>=<[2]> in Z_18 or not, I don't recall that being a rule, though.

I believe that for question 8 they are asking for a set that is closed under addition but is not a subset of Z. They are not asking for a modulo.

(Z,+) is a group, and is also a sub-group of (R,+).

There are more axioms then the two you stated.
if G=(S,*) is a group, then
S contains the identity I under *
for all x in S there exists x^-1 s.t. x*x^-1=I
for all x,y in S x*y is in S.
a*(b*c)=(a*b)*c

So I guess you got to find a set that is closed under addition but doesn't meet one of those criteria.
 
  • #7
Disconnected said:
I am not sure if <[8]>=<[2]> in Z_18 or not, I don't recall that being a rule, though.

I believe that for question 8 they are asking for a set that is closed under addition but is not a subset of Z. They are not asking for a modulo.

(Z,+) is a group, and is also a sub-group of (R,+).

There are more axioms then the two you stated.
if G=(S,*) is a group, then
S contains the identity I under *
for all x in S there exists x^-1 s.t. x*x^-1=I
for all x,y in S x*y is in S.
a*(b*c)=(a*b)*c

So I guess you got to find a set that is closed under addition but doesn't meet one of those criteria.

Those two iff conditions are a later theorem. The other properties were listed earlier. I suppose the earlier properties would be easier to produce a counterexample with a particular set.

The additive group is (Z,+). Now, I just need a closed subset that is not a group of the additive group (Z,+).
 
  • #8
"The additive group is (Z,+). Now, I just need a closed subset that is not a group of the additive group (Z,+). "
Yep, do any come to mind?
With this kind of thing I always find it easier to decide which axiom I am going to not meet and then decide on the set.
 
  • #9
Disconnected said:
"The additive group is (Z,+). Now, I just need a closed subset that is not a group of the additive group (Z,+). "
Yep, do any come to mind?
With this kind of thing I always find it easier to decide which axiom I am going to not meet and then decide on the set.

Well, the first thing that comes to mind is the set of positive even integers. There's no identity or inverse.
 
  • #10
Shackleford said:
Well, the first thing that comes to mind is the set of positive even integers. There's no identity or inverse.
Perfect. Though, the positive integers alone would also have worked. So, did the thread clear up what you were wondering about?
 
  • #11
Disconnected said:
Perfect. Though, the positive integers alone would also have worked. So, did the thread clear up what you were wondering about?

Yeah, that's true. The identity exists, but an inverse doesn't exist.

Does my work for #4 make sense?
 
  • #12
I couldn't really follow your working for question 4.
What made it so you can say <[2]>=<[8]>? I didn't catch that part.
 
  • #13
Shackleford said:
For #4, I said <[8]> = <[2]> in Z18. Is that correct?

Then, I generated a set in Z18 from <[2]>: {[16], [14], [12], [10], [8], [6], [4], [2], [0]}.

The order of this set is 9.
Hi, I looked at your solution. Once you have <[8]> = <[2]>, your enumeration of the elements of <[2]> is fine.

However, you didn't say how you know that <[8]> = <[2]>. It's true, but how do you know it's true? You either have to calculate it out by enumerating the elements of <[8]>; or else you have to have some other argument.
 
  • #14
Shackleford said:
For #8, I know what a closed subset of Z is. A subset H of a group G has to be nonempty and if a, b are contained in H, then ab-1 is in H. They're asking for a subset of Z. Do they want a set of congruence classes in Z?

No, they asked for a subset, not a congruence class. You're overthinking this. A subset is just a subset. What's a subset of Z? How could a subset be closed under addition but not be a subgroup of Z when Z is regarded as an additive group?
 
  • #15
SteveL27 said:
Hi, I looked at your solution. Once you have <[8]> = <[2]>, your enumeration of the elements of <[2]> is fine.

However, you didn't say how you know that <[8]> = <[2]>. It's true, but how do you know it's true? You either have to calculate it out by enumerating the elements of <[8]>; or else you have to have some other argument.

Well, in Z18, 13[2] = [26] = [8]. Is that how I correctly show they're equivalent?
 
Last edited:
  • #16
Shackleford said:
Well, in Z16, 12[2] = [24] = [8]. Is that how I correctly show they're equivalent?

Didn't we start out in Z18?
 
  • #17
SteveL27 said:
Didn't we start out in Z18?

Oops. Yes.
 
  • #18
Is it because (8,18) = 2? Is that how I found a similar subgroup?
 
  • #19
I don't understand what you're asking in that last post.

As for the other thing,
You said
"Well, in Z18, 13[2] = [26] = [8]. Is that how I correctly show they're equivalent? "

But isn't 26[1]=[26]=[8]? But <[1]>\=<[8]> Unless I am forgetting some theorem here, I don't think that your method is a full proof that <[2]>=<[8]>. I may be wrong, though.
Personally, to be safe, I would just find <[8]> directly.
 
  • #20
Disconnected said:
I don't understand what you're asking in that last post.

As for the other thing,
You said
"Well, in Z18, 13[2] = [26] = [8]. Is that how I correctly show they're equivalent? "

But isn't 26[1]=[26]=[8]? But <[1]>\=<[8]> Unless I am forgetting some theorem here, I don't think that your method is a full proof that <[2]>=<[8]>. I may be wrong, though.
Personally, to be safe, I would just find <[8]> directly.

gcd(8, 18) = 2. That's why I said they're equivalent.
 

Related to Groups and Subgroups: Clarifying Questions

1. What is the difference between a group and a subgroup?

A group is a collection of objects or elements that follow a set of rules or operations. A subgroup is a subset of a group that also follows those same rules or operations. Essentially, a subgroup is a smaller group within a larger group.

2. How are groups and subgroups related?

Groups and subgroups are related in that subgroups are a part of a larger group. Subgroups inherit the properties and operations of the larger group, but may also have additional properties and operations specific to the subgroup.

3. What are some examples of groups and subgroups?

Some examples of groups include the set of integers with addition as the operation, or the set of rotations of a square. Examples of subgroups include the set of even integers within the set of integers, or the set of diagonal reflections within the set of all rotations of a square.

4. Can a subgroup also be a group?

Yes, a subgroup can also be a group. To be a group, a subgroup must satisfy the same requirements as a group, such as having closure, associativity, identity, and inverse elements.

5. How do you determine if a subset is a subgroup?

To determine if a subset is a subgroup, you must check if it satisfies the requirements of a group. This includes checking for closure, associativity, identity, and inverse elements. If all of these requirements are met, then the subset can be considered a subgroup.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
933
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
36
Views
10K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
Back
Top