Groups and Subgroups: Clarifying Questions

[Shackleford]

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

 

[Disconnected]

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]).

 

[Shackleford]

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?

 

[Disconnected]

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.

 

[Shackleford]

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?

 

[Disconnected]

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.

 

[Shackleford]

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,+).

 

[Disconnected]

"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.

 

[Shackleford]

"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.

 

[Disconnected]

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?

 

[Shackleford]

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?

 

[Disconnected]

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.

 

[SteveL27]

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.

 

[SteveL27]

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?

 

[Shackleford]

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?

 

[SteveL27]

Well, in Z₁₆, 12[2] = [24] = [8]. Is that how I correctly show they're equivalent?

Didn't we start out in Z₁₈?

 

[Shackleford]

Didn't we start out in Z₁₈?

Oops. Yes.

 

[Shackleford]

Is it because (8,18) = 2? Is that how I found a similar subgroup?

 

[Disconnected]

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.

 

[Shackleford]

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.