Klein and Cyclic Groups: Definitions and Subgroups

[nille40]

Hi everybody!

Question #1
What is the definition of a Klein group? The $$K_4$$ group has a table that looks like this:
$$\begin{array}{c|cccc}<br /> *&e&a&b&c \\\hline<br /> e&e&a&b&c\\<br /> a&a&e&c&b\\<br /> b&b&c&e&a\\<br /> c&c&b&a&e<br /> \end{array}$$

What is the strict definition of a Klein group? That every element generates the entire group? Is $$\langle \lbrace 1, 3, 5, 7 \rbrace, +\rangle$$ a Klein group?

Question #2
All subgroups of the cyclic group $$C_{24}$$ are cyclic groups $$C_n$$ where $$n \mid 24$$. So to find all subgroups, one can locate all divisors for 24. Correct?

This is what I do not understand: The subgroups of $$C_24$$ with the order 24 is $$c, c^5, c^7, c^{11}, c^{13}, c^{17}, c^{19}, c^{23}$$. What does $$c^n$$ mean? Does $$c^n$$ generate $$\frac{24}{\gcd(24, n)}$$ elements?

I would really, really appreciate some help with this!

Thanks in advance,
Nille

 

[Lonewolf]

Hint for question 1:

The Klein group is a symmetry group for a geometrical object. The fact the group is of order 4 is significant...

Btw, are you sure that every element generates the group?

 

[nille40]

"Btw, are you sure that every element generates the group?"

Nope. I am pretty much fumbling in the dark... The source of the table in the previous post is my teacher and some document(s) on the internet, so the table should be correct. In the table every element generates the entire group, hence the conclusion.

I am not sure about anything, in the case of groups. For instance, the klein group is defined as $$\langle \lbrace e, a, b, c \rbrace, *\rangle$$, where $$*$$ represents some operation. Is this correct?

And the elements in a cyclic group can be written as $$C_n = \langle \lbrace e, c^1, c^2, \ldots c^n \rbrace, * \rangle$$. Is this correct?

To quote someone famous: I'm going crazy.

 

[Lonewolf]

Hmm, ok. Well, if we take the element a and mulitply a by itself, we arrive at the identity. So, the group generated by a is <a> = {e,a}, which means a has order 2, but the Klein group is of order 4. Your table is correct, however.

And the elements in a cyclic group can be written as $$C_n = \langle \lbrace e, c^1, c^2, \ldots c^n \rbrace, * \rangle$$

Yes, that's correct. One interesting feature of the Klein group is that it is the smallest non-cyclic group. (Every group of prime order is cyclic).

You may find this site interesting http://www.wordiq.com/cgi-bin/knowledge/lookup.cgi?title=Klein_four-group

Edit: Used QUOTE tags instead of URL. Whoops.

 

[nille40]

So a group $$G_4 = \langle \lbrace e, a, b, c \rbrace, * \rangle$$ is a Klein group iff it is not cyclic, ie it has a "multiplication table" identic to the Klein table?

Thanks again!
Nille

 

[HallsofIvy]

There are only two groups of order two ("up to isomorphism"), the cylic group of order 4 and the Klein group. So one way of defining the Klein group is to say "it is the group of order 4 that is not cyclic"!

Since any group is defined (again, "up to isomorphism") by its multiplication table, yes, you could say the Klein group is any group that has that multiplication table.

 

[nille40]

Thank you both very much!

I'll post yet another question, in case you don't have anything better to do...

2 groups $$\langle M_1 \mid *\rangle$$ and $$\langle M_2 \mid \diamond\rangle$$ are isomorphic if there exists a bijection $$f:M_1 \rightarrow M_2$$ such that

$$f(x * y) = f(x) \diamond f(y)$$

Correct? Say we have the groups $$\langle \mathbb{Z}_6, + \rangle$$ and $$\langle \mathbb{Z}_7\backslash\lbrace 0 \rbrace, \times \rangle$$. How can one prove that these are isomorphic?

We should locate a bijection $$f:\mathbb{Z}_6 \rightarrow \mathbb{Z}_7\backslash\lbrace 0 \rbrace$$, with the property

$$f(x+y) = f(x) \times f(y)$$

How do you do that? Do you explicitly express mappings between elements in the group, like this:
$$f(0) = 1,<br /> f(1) = 3,<br /> f(2) = 2,<br /> f(3) = 6,<br /> f(4) = 4,<br /> f(5) = 5$$.

How can you locate such a bijection?

Thanks again,
Nille

 

[Lonewolf]

2 groups...Correct?

Yes, that's correct. The map $f$ that satisfies the property $f(x * y) = f(x) \diamond f(y)$ for a group is called a group homomorphism. You'll probably see the term used over and over again.

How do you do that? Do you explicitly express mappings between elements in the group, like this

Yes, you can do it that way. There's usually an 'obvious' bijection, but in this case, I don't think there is. There's no general method to find a bijection between two groups.

All you have to do to prove that the two groups are isomorphic is to find just one bijection between them. I haven't checked yours, but it does appear to be a bijective homomorphism, so the two groups are indeed isomorphic.

 

[nille40]

Aha...
I always enjoy the "there is no good way to do that" answer :)
But that was pretty much what I figured...

I tried a number of ways:
1. If you organize the order of each element in a list you can easily find a bijection between the groups. But this becomes a bit complex when you're dealing with large groups.
2. If you draw a directed graph, where nodes represent elements and edges represent generation ($$a \rightarrow b \Rightarrow b \in \langle a \rangle$$). But this wasn't a very efficient approach either...

Thanks for all your help!
Nille

 

[matt grime]

Look at the order of the elements of the group

an isomorphism will send an element of order n to an element of order n.

if it sends x to y it sends x*x to y@y so that narrows down the bijections you need to examine

* and @ are my notations for the group composition.

express things as generators and relations and simpler objects

the Klein 4 group is C_2xC_2 for instance

learn some more theory: it is trivial to decide when finite abelian groups are isomorphic because of the structture theorem.

as for the example you gave. the integers mod 6 under addtion are cyclic of order 6. (1 generates the group.

to show if the units mod 7 are cyclic of order 6 then find an element that does not square or cube to the identity; its order must then be 6

hint try 3 as the generator.

matt