3
$\begingroup$

I am working though an Introduction to the theory of Groups. I have come the following exercise:

"Let $\alpha = \begin{pmatrix} i_1 & i_2 & \cdots & i_r \end{pmatrix}$ and $\beta = \begin{pmatrix} j_1 & j_2 & \cdots & j_s \end{pmatrix}$. Prove that $\alpha\text{ and } \beta$ are disjoint if and only if $\{ i_1 , i_2 ,\ldots,i_r\} \cap \{ j_1 , j_2 ,\ldots,j_s\} = \varnothing$"

The Definition of disjoint cycles: $\alpha$ and $\beta$ are disjoint if for every $x \in X $ if $\alpha(x) \neq x $ then $ \beta(x) =x$ and if $\beta(y) \neq y $ then $\alpha(y) =y $

Reading other books on group theory, I have come across the notion of a Support:

The Support of $\alpha$ denoted $supp(\alpha) = \{ i_1 , i_2 ,\ldots,i_r\}$

I have thought of using this in my proof of the above. However I am somewhat weary about this .

Can anyone suggest a proof without Supports?

Update:

Is this okay proof for the "if" part of the proof.

Assume that $\{i_i , i_2 ,\ldots ,i_r\} \cap \{ j_1 , j_2 ,\ldots ,j_s\} = \emptyset $

then for all $x \in \{i_i , i_2 ,\ldots ,i_r\} $ we have that $\alpha(x) \ne x $ and since $ x \notin \{ j_1 , j_2 ,\ldots ,j_s\} $ we have that $\beta (x) =x $

Similarly for all $y \in \{ j_1 , j_2 ,\ldots ,j_s\}$ we have that $\beta(y) \ne y $ and since $ y \notin \{i_i , i_2 ,\ldots ,i_r\} $ we have that $\alpha(y) = y $

$\endgroup$
3
$\begingroup$

For example, let's go over the "only if" part of the proof.

First note that if $x\in \{i_1,...,i_r \}$ then $\alpha(x) \neq x$, and if $x\in \{j_1,...,j_s \}$ then $\beta(x) \neq x$ (why?).

Assume that the cycles are disjoint, and assume for contradiction that $\{i_1,...,i_r \} \cap \{j_1,...,j_s \} \neq \emptyset$. Thus there exists some $x\in \{i_1,...,i_r \} \cap \{j_1,...,j_s \}$. For this $x$ we have both $\alpha(x) \neq x$ and $\beta(x) \neq x$, and that contradicts the definition of disjoint cycles.

$\endgroup$
  • $\begingroup$ I am right in saying if $x\in \{i_1,...,i_r \}$ then $\alpha(x) \neq x$, because $x\in \{i_1,...,i_r \}$ is the set of elements that move under $\alpha$? $\endgroup$ – Mathman May 5 '14 at 19:40
  • $\begingroup$ Is that okay proof for the "if" part of the proof. $\endgroup$ – Mathman May 8 '14 at 5:00
  • 1
    $\begingroup$ It is almost correct. Note that the definition of disjoint cycles begins with "for every $x\in X$...". You only covered the cases in which the $x$ comes from the supports. You should say something about the other $x$'s too. $\endgroup$ – Ludolila May 8 '14 at 12:02
1
$\begingroup$

The cycle $\alpha$ only "does something" with the numbers $i_1,\ldots, i_r$. This means that if $\alpha(x)\neq x$, then x has to be one of these numbers. Bearing this in mind you can pretty much follow the definitions and your proof comes rolling out.

$\endgroup$
  • $\begingroup$ It seems too simple. $\endgroup$ – Mathman May 5 '14 at 17:36
  • $\begingroup$ @Matthew But it really is that simple. You could have suspected as much when people started wordering if you mixed up your definitions, because usually one would use the fact you are trying to prove as a definition for disjoint cycles and then prove your definition is equivalent. $\endgroup$ – gebruiker May 5 '14 at 17:40
  • $\begingroup$ I tend to overthink it when it is this simple $\endgroup$ – Mathman May 5 '14 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.