Abstract Algebra Mod 6 Subgroup Computation and Generator Identification

[kathrynag]

Homework Statement

I'm working with a mod 6 addition table.
I want to compute the subgroups <0>,<1>,<2>,<3>,<4>,<5>
I also want to find what elements are generators of the group mod 6.
Then I wnat to use do a subgroup diagram.

Homework Equations

The Attempt at a Solution

I am not sure about how to compute sungroups. Doe sthis incolve something like:
1+1+1 and so on?
I know an element is a generator if <a>=G. My problem is figuring out which elelments are. Is it when 2 subgroups are equal?

 

[kathrynag]

Ok, is there a general way to determine a subgroup from a table. I think if I can figure out one, I'll be fine on the rest.

 

[quantumdude]

The Attempt at a Solution

I am not sure about how to compute sungroups. Doe sthis incolve something like:
1+1+1 and so on?

Yes. And when you start getting repeat elements, you're done.

I know an element is a generator if <a>=G. My problem is figuring out which elelments are. Is it when 2 subgroups are equal?

Just do it! Work out all 6 cyclic subgroups then it will be obvious when two of them are equal.

 

[kathrynag]

Yes. And when you start getting repeat elements, you're done.



Just do it! Work out all 6 cyclic subgroups then it will be obvious when two of them are equal.

What do you mean by repeat elements. So I for
<0> =0+0=0+0=0
<1>=1+1=2+1=3+1=4+1=5+1=0+1=1+1=2=<2,3,4,5,0>
<2>=2+2=4+2=0+2=2+2=4=<4,0,2>
<3>=3+3=0+3=0
<4>=4+4=2+4=0+4=4+4=2=<2,0,4>
<5>=5+5=4+5=3+5=2+5=1+5=0+5=5+5=4=<4,3,2,1,0>
So,<<0> and <3> are generators. Is what I did correct?

 

[quantumdude]

What do you mean by repeat elements.

I mean that when the elements start to repeat, then stop.

<0> =0+0=0+0=0

So for this one, $<0>=\{0\}$. There's nothing more you can get out of it.

<1>=1+1=2+1=3+1=4+1=5+1=0+1=1+1=2=<2,3,4,5,0>

Your notation is very sloppy. You can't just set all of those things equal to each other like that. You've got 1+1=2+1, which is obviously not true. Plus you didn't take it far enough. 1 should be in that subgroup too. Also you shouldn't list the elements of a subgroup in angled brackets. You should use curly braces instead. So in this case, $<1>=\{0,1,2,3,4,5\}$.

<2>=2+2=4+2=0+2=2+2=4=<4,0,2>
<3>=3+3=0+3=0
<4>=4+4=2+4=0+4=4+4=2=<2,0,4>
<5>=5+5=4+5=3+5=2+5=1+5=0+5=5+5=4=<4,3,2,1,0>

Bad notation aside, you got the subgroups $<2>$ and $<4>$ correct. You didn't take $<5>$ far enough, and I can't tell if you know what you're doing on $<3>$ because you didn't write your answer down.

So,<<0> and <3> are generators. Is what I did correct?

No, neither of those are generators. I suggest you look up the definition of "generator of a group".

 

[kathrynag]

Ok, so I want to make sure, I have every possible number in the subgroup.
<5>=5+5=4+5=3+5=265=1+5=0+5=5+5=4
<5>=<4,3,2,1,0,5>

<3>=3+3=0+3=3+3=0
<3>=<0,3>

An element a of a group G generates G and is a generator for G if <a>=G.
Something about the definition doesn't make sense.

My thought is that <1>, <5>, <2>, and <4> are generators because they have the same elements.

 

[quantumdude]

Ok, so I want to make sure, I have every possible number in the subgroup.
<5>=5+5=4+5=3+5=265=1+5=0+5=5+5=4

You really must stop doing this. "5+5=4+5" is simply not true.

<5>=<4,3,2,1,0,5>

Yes, but you should use the curly braces as I said. What you've written here means something else. The notation $<a,b>$ means "the group generated by a and b", not "the set whose elements are a and b."

<3>=3+3=0+3=3+3=0
<3>=<0,3>

Bad notation aside, this is correct.

An element a of a group G generates G and is a generator for G if <a>=G.
Something about the definition doesn't make sense.

It makes perfect sense. Either $<a>=G$ or it doesn't. There's no ambiguity, so it's a fine definition.

My thought is that <1>, <5>, <2>, and <4> are generators because they have the same elements.

Now that doesn't make sense. $<1>$ isn't a generator, it's a subgroup. The fact that $<1>=G$ means that 1 is a generator. 5 is also a generator, but 2 and 4 are not. Do you see why, in light of the definition of "generator"?

 

[kathrynag]

Ok, I think I know where my confusion is I understand <a> is a subgroup, but I don't understand what the group G is.
Oh, so G consists of the elements 0,1,2,3,4,5 and <1> and <5> have all these elements, so they are generators?

 

[quantumdude]

Oh, so G consists of the elements 0,1,2,3,4,5

Yes, that's right.

<1> and <5> have all these elements, so they are generators?

More precisely, <1> and <5> have all these elements, so 1 and 5 are generators. A generator is an element of a group, not a group itself.

 

[kathrynag]

Ok, that makes sense. I have to make an addition table of these subgroubs.

Ok, so <1>+<2>= ?
I'm not sure what to do when the sizes of the groups aren't the same?
Would <2> +<4>=<0>

 

[quantumdude]

<1>+<2> doesn't make any sense. You don't add subgroups, you add elements. What exactly were you asked to do?

 

[kathrynag]

Give the subgroup diagram for the subgroups of mod 6.

I thought this menat creating an addition table.

 

[quantumdude]

No, it doesn't mean that. Isn't there an example of such a thing in your book?

 

[kathrynag]

Not really:
My book doesn't explain very well.
i think I start with 1 and create a diagram connecting with 2, 3, 4, 5, 0?

 

[quantumdude]

What book are you using?

 

[kathrynag]

I'm using Fraleigh.
I did just find a site that genertes a subgroup diagram and am trying to make sense out of it.
http://hobbes.la.asu.edu/courses/site/devel-groups/tabs.html

 

[quantumdude]

That's the book I learned from. It explains subgroup diagrams with examples.

 

[kathrynag]

Ok, so I think I have an idea the subgroup <0> stems out from <3>

 

[kathrynag]

Ok, I think I'll get it eventually. It's just a different way of thinking.

 

[kathrynag]

Ok, I figured it out. I was looking at it all wrong.

 

[quantumdude]

At the top should be the full group. Then below that you put its largest subgroups and connect them with a line segment. Continue mapping out the heirarchy of subgroups until you arrive at $<0>$ at the bottom. So yes, $<0>$ and $<3>$ are connected. But that's not the only subgroup $<0>$ is connected to.