# Math Help - cycle decomposition and group actions.

1. ## cycle decomposition and group actions.

Question 1) If $\tau=$(1 2)(3 4 5), determine whether there is an n-cycle with $\tau=\sigma^k$ for some integer k.

Question 2)Assume that G is an abelian, transitive subgroup of $S_A$. Let G act on A.
Show that $\sigma (a)\neq a$ for all $\sigma \in$ G-{1} and all $a \in A$. Deduce that |G|=|A|.
(there was a hint given to this question saying: use $\sigma G_a \sigma^{-1}=G_{\sigma (a)}$. using this hint the question is very easy to solve. can someone please solve it without using this hint.)

2. along with the above two questions please also answer this simple one.

Let there be a set A. Any subgroup of $S_A$ acts faithfully on A. true or false??
I think its true. if its false then please explain.

3. Originally Posted by abhishekkgp
Question 1) If $\tau=$(1 2)(3 4 5), determine whether there is an n-cycle with $\tau=\sigma^k$ for some integer k.

Question 2)Assume that G is an abelian, transitive subgroup of $S_A$. Let G act on A.
Show that $\sigma (a)\neq a$ for all $\sigma \in$ G-{1} and all $a \in A$. Deduce that |G|=|A|.
(there was a hint given to this question saying: use $\sigma G_a \sigma^{-1}=G_{\sigma (a)}$. using this hint the question is very easy to solve. can someone please solve it without using this hint.)
Question 1 can be done by inspection - the elements in question, $\tau$ has order 6. So find the other 4 non-trivial powers (what is $n$, by the way? Is it just 5?).

Question 2 is more subtle. The hint is the neatest way of doing it - and in fact, it isn't toooooo hard to prove. Basically, it comes down to proving that $(a \text{ } b \ldots c)^{\sigma} = (a\sigma \text{ } b\sigma \ldots c\sigma)$. This is a very useful fact, and so is worth proving then committing to memory! So, I would advise you to at least attempt to prove it.

I have had a quick attempt to prove the result without the hint. I then realised you can use the orbit-stabliser theorem! Are you familiar with this theorem? It works here because the stabiliser of every element of $A$ is trivial.

For your third question, you are correct. $G$ acts transitively on $A$ for all $G \leq S_A$. However, it needn't acts transitively on subsets of $A$.

4. Originally Posted by Swlabr
Question 1 can be done by inspection - the elements in question, $\tau$ has order 6. So find the other 4 non-trivial powers (what is $n$, by the way? Is it just 5?).
HI Swlabr! nice to see you again.
In the book its written that $5 \leq n$.
you can see this link too: http://www.mathhelpforum.com/math-he...lp-168487.html

I dont understand the meaning of $(a \text{ } b \ldots c)^{\sigma}$....
You have raised an element in A to the power $\sigma$. Can you please explain it if its not so much trouble. This must be some standard notation but i have not yet come across it.

I have started reading the chapter on group actions just today so i dont know orbit-stabilizer theorem yet. I will read it and then try the question again.

Thanks for comment on my third question.

5. One more thing. your avatar is the rubik's cube pic.
I started reading group theory to figure out my own algorithm to solve the cube... nevertheless group theory in itself turned out to be the most wonderful topic of maths that i have ever read.
All i could do is that now i can find order of any operation on the rubik's cube.
What other math do i need to read to be able to write an algo?? Graph theory??
Although an algo is easy to find on internet, i, however, want to be able to PROVE that the algorithm will definitely work.. etc.. and want to develop an algo by a mathematical approach not just by observation. I hope you understand.
Thanks for help.

6. Originally Posted by abhishekkgp
HI Swlabr! nice to see you again.
In the book its written that $5 \leq n$.
you can see this link too: http://www.mathhelpforum.com/math-he...lp-168487.html

I dont understand the meaning of $(a \text{ } b \ldots c)^{\sigma}$....
You have raised an element in A to the power $\sigma$. Can you please explain it if its not so much trouble. This must be some standard notation but i have not yet come across it.

I have started reading the chapter on group actions just today so i dont know orbit-stabilizer theorem yet. I will read it and then try the question again.

Thanks for comment on my third question.
Oh, sorry - I missread your first question. To solve your question, you need to think about how long cycles square. Basically, they will either square to give another long cycle (if they have odd length, say 5) or split in two and give two m-cycles, of they have length 2m (e.g. $(1 \text{ }2\text{ } 3\text{ } 4\text{ } 5\text{ } 6)^2 = (1\text{ }3\text{ }5)(2\text{ }4\text{ }6)$). Can you generalise? What happens when you take a long cycle of length $n$ to a power of $p$ where $p|n$ and when $p \nmid n$? Basically, you should see that the power of a long cycle will always consist of a number of short cycles all of the same length. The permutation you are looking at is a product of a two-cycle and a three-cycle, so this can't happen.

$(a \text{ } b \ldots c)^{\sigma} = \sigma^{-1}(a \text{ } b\ldots c)\sigma$. It means conjugating the cycle by sigma. Sorry - it is standard, I just forget that not everyone knows such things...

7. Originally Posted by abhishekkgp
One more thing. your avatar is the rubik's cube pic.
I started reading group theory to figure out my own algorithm to solve the cube... nevertheless group theory in itself turned out to be the most wonderful topic of maths that i have ever read.
All i could do is that now i can find order of any operation on the rubik's cube.
What other math do i need to read to be able to write an algo?? Graph theory??
Although an algo is easy to find on internet, i, however, want to be able to PROVE that the algorithm will definitely work.. etc.. and want to develop an algo by a mathematical approach not just by observation. I hope you understand.
Thanks for help.
Hmm...this is a good question. Have you read this page before? Gap is a very powerful tool for studying algebra.

Now, as the article I linked to above says, you can think of the Rubik's cube as being generated by some permutations. If your algorithm works, the moves involved in it should generate the whole of the Rubik's group. So try working out what these moves are as permutations, and check using Gap that they generate the group.

Of course, this doesn't prove that your algorithm works, it just proves that the individual parts of your algorithm can be put together in such a way that they will work...

The Rubik's cube is a wonderful way of applying group theory. For example, you may hope that there is a single move such that you can apply this move some number of times to any position to solve the cube. In group theoretic terms, this would mean that the group was cyclic. This is clearly not so, as the group is non-abelian! (rotate the right face once, and then the top face once. This is different from rotating the top face then the right face.)

What is also nice is that if you choose a random set of moves and apply these moves a finite number of times to a solved cube you will return you to the solved cube! (Because every element in your group has finite order).

8. Originally Posted by Swlabr
Oh, sorry - I missread your first question. To solve your question, you need to think about how long cycles square. Basically, they will either square to give another long cycle (if they have odd length, say 5) or split in two and give two m-cycles, of they have length 2m (e.g. $(1 \text{ }2\text{ } 3\text{ } 4\text{ } 5\text{ } 6)^2 = (1\text{ }3\text{ }5)(2\text{ }4\text{ }6)$). Can you generalise? What happens when you take a long cycle of length $n$ to a power of $p$ where $p|n$ and when $p \not\div n$? Basically, you should see that the power of a long cycle will always consist of a number of short cycles all of the same length. The permutation you are looking at is a product of a two-cycle and a three-cycle, so this can't happen.

$(a \text{ } b \ldots c)^{\simga} = \sigma^{-1}(a \text{ } b\ldots c)\sigma$. It means conjugating the cycle by sigma. Sorry - it is standard, I just forget that not everyone knows such things...
I generalized it in the case when $p|n$ but did not do it for $p \not\div n$. Thanks i will try this one.
In the LHS of $(a \text{ } b \ldots c)^{\simga} = \sigma^{-1}(a \text{ } b\ldots c)\sigma$ you probably forgot to raise it to the power $\sigma$.

thanks a lot for help.

9. Originally Posted by Swlabr
What is also nice is that if you choose a random set of moves and apply these moves a finite number of times to a solved cube you will return you to the solved cube! (Because every element in your group has finite order).
I guess you meant the following:
We start with a solved cube. Then any crazy set of moves a finite number of times gives the solved cube back( in general we arrive at the initial configuration back).

I am currently working on the 2x2x2 cube. It comes out to be a subgroup of $S_{24}$ symmetric group.
Right-slice-clockwise-rotaion( this is my terminology- non standard of course but i guess conveys the idea ) followed by top-slice-clockwise rotaion has order 15 in 2x2x2 cube.
I can show that six basic rotations will generate all members of this 2x2x2 cube group but here comes the main problem...
suppose you give me an element of the 2x2x2 cube group and ask me to express it in terms of its generators then i am completely lost.
is this problem solved??

10. deleted post. sry.