Write down the cosets (right/left) of this:

[RJLiberator]

Homework Statement

Q: In Qu, write down the elements in all the right cosets and all the left cosets of <j>.

Homework Equations

Let H ≤ G. A coset of H is a subset of G of the form Hg for some g ∈ G.
where ≤ denotes subgroup.

The Attempt at a Solution

I need help understanding what all this means. Can someone give me an example of what I should be looking for / doing here?

I don't understand what they mean by the group Qu. Is this some group u multiplied by Q? Or just the group Qu.

etc.

 

[micromass]

$Qu$ is just the name of the group (also known as the quaternion group). First, can you write out al the elements in $<i>$?

 

[RJLiberator]

Thanks for the help. I am at work now/tonight, so I'll have to get back to this problem then.
for my reference: https://en.wikipedia.org/wiki/Quaternion_group

To answer your question: Write out all the elements in < i >:
{1, -1, i, -i}
or {e, i, -i}

I think?

I think it is {1, -1, i, -i} based on the quick reading of the Wiki page.

 

[micromass]

Thanks for the help. I am at work now/tonight, so I'll have to get back to this problem then.
for my reference: https://en.wikipedia.org/wiki/Quaternion_group

To answer your question: Write out all the elements in < i >:
{1, -1, i, -i}
or {e, i, -i}

I think?

I think it is {1, -1, i, -i} based on the quick reading of the Wiki page.

There isn't even an element $e$ in your group??

Anyway, can you now compute $g<i>$ for each element in the group? I'll do one for you if $g=i$:
$$i<i>=i\{1,-1,i,-i\} = \{i\cdot 1, i\cdot (-1), i\cdot i, i\cdot (-i)\} = \{i, -i, -1, 1\} = <i>$$
What for the other $g$'s in the group?

What about $<i>g$?

 

[Dick]

There isn't even an element $e$ in your group??

Anyway, can you now compute $g<i>$ for each element in the group? I'll do one for you if $g=i$:
$$i<i>=i\{1,-1,i,-i\} = \{i\cdot 1, i\cdot (-1), i\cdot i, i\cdot (-i)\} = \{i, -i, -1, 1\} = <i>$$
What for the other $g$'s in the group?

What about $<i>g$?

Just wanted to point out that the original question seems to be asking for the cosets of $<j>$, not $<i>$.

 

[RJLiberator]

Thanks for the help so far. I'm on my lunch break, but here's some of my (very quick) thoughts:

<i>g where g = i is the following:

<1,-1,i,-i> => <1*i,-1*i,i*i, -i*i> = <i, -i, -1, 1> = <i>

so the left coset and the right coset for this particular item came out to be the same.

Is my analysis correct here?
Then I would continue this process of right and left operation with the other members, 1, -1, i, -i.

 

[micromass]

Thanks for the help so far. I'm on my lunch break, but here's some of my (very quick) thoughts:

<i>g where g = i is the following:

<1,-1,i,-i> => <1*i,-1*i,i*i, -i*i> = <i, -i, -1, 1> = <i>

so the left coset and the right coset for this particular item came out to be the same.

Is my analysis correct here?
Then I would continue this process of right and left operation with the other members, 1, -1, i, -i.

Be careful with your brackets. $<i>$ means the group generated by $i$. This is right. But you clearly mean $\{i,-i,-1,1\}$ as a set, and so it shouldn't be written as $<i, -i,1, -1>$.

In any case, this is correct. Try to do the other ones.

Side question, it turns out that $i<i> = <i>$ and that $<i> i = <i>$. Should this surprise you?

 

[RJLiberator]

Why is it that < i > is just {1, -1, i, -i} ?
Is < j > just {j, -j, ij, -ij} ? If so, how in the world was I supposed to know this based on the question information?

The fact that i< i > = < i > = < i >i suprises me a bit. I don't really understand the operation of i or what it is doing.

So to determine the right and left cosets of j all I have to do is the following

1< j >
-1< j >
i< j >
-i< j >
j< j >
-j< j >
ij< j >
-ij< j >

For the left cosets, where < j > = { j, -j, ij, -ij }

and then the same for the right cosets.

And then this problem is solved.

But how was I supposed to know that < j > = { j, -j, ij, -ij } ?

 

[micromass]

No, your $<j>$ is wrong. Can you tell us what the definition of $<j>$ is?

 

[RJLiberator]

Damn. I have no idea what the definition of < j > is. All I have is this table.
If < i > = {1, -1, i, -i}

Perhaps j is < j > = {1, -1, j, -j} ?

 

[micromass]

Damn. I have no idea what the definition of < j > is. All I have is this table.
If < i > = {1, -1, i, -i}

Perhaps j is < j > = {1, -1, j, -j} ?

Stop guessing. Look up the definition of $<...>$.

 

[RJLiberator]

Definition of what?
Is there a wikipedia link you can post to me?

I will continue to look through my notes, tho.

 

[micromass]

Definition of what?
Is there a wikipedia link you can post to me?

I will continue to look through my notes, tho.

I don't get it. You're trying to solve this problem about $<j>$ without first looking up what $<j>$ even means? Who gave you this problem? He should have explained you these things.

 

[RJLiberator]

Well, we have 2 lectures of 50 minutes per week and then a homework assignment over the weekend. The homework assignment kind of molds/forms everything. I have never in my life seen this Qu quaternion group before, but I often learn a lot in these homeworks.

Now, with the help you've given, if I look up the Qu wikipedia page:
https://en.wikipedia.org/wiki/Quaternion_group

< j > = { j, k, -j, -k } ?

 

[micromass]

I don't see why your wasting your time. Don't you have a book or notes? Look up what $<j>$means instead of guessing.

 

[RJLiberator]

Oi.

< j > = {j, -j, ij, -ij}

 

[micromass]

I'm not helping any more until you post the definition of $<g>$ where $g\in G$.

 

[RJLiberator]

Aha! Now I have the words/hints that I needed.

For any group G and any g ∈ G, <g> is a subgroup of G.
< g> is called the cyclic subgroup generated by g.

Note: A group G is called cyclic if there exists a ∈ G such that < a > = G.

Note2: If g ∈ G, then <g> is the smallest subgroup containing the element g. Precisely, if g ∈ H ≤ G, then < g > ≤ H.

Definition: Let G be a group and H ⊆ G. Then H is a subgroup if
1) e ∈ H
2) For all x,y∈ H, xy ∈ G.
3) For all x ∈ H, x^(-1) ∈ H

So from here, < j > = { 1, -1, j, -j}
As you need the identity element and j and the inverses ?

 

[micromass]

Yes, that is correct.

 

[RJLiberator]

Whew. Thanks for helping me with that. I'm not sure why, but with this being new material for me, the switching of simple letters really messes me up. I will try to keep that in mind as I go through this.

But now, it should be pretty easy for me to figure out. Simply as stated above,

1< j >
-1< j >
i< j >
-i< j >
j< j >
-j< j >
ij< j >
-ij< j >

for the left coset and then the opposite side for the right coset.

 

[micromass]

Yep, the Cayley table in the OP should help you how to do that.

 

[RJLiberator]

Just completed it! Thank you for your help, this problem is solved.