4
$\begingroup$

List the elements of the cyclic subgroup of $S_6$ generated by:

\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix}

My textbook doesn't mention how to do this so I decided to research online and I've found this answer:

[(1234)(56)]$^2$ = (1234)(56) * (1234)(56) = (13)(24)

[(1234)(56)]$^3$ = (13)(24) * (1234)(56) = (1432) (56)

[(1234)(56)]$^4$ = [(13)(24)]$^2$ = (1).

So, <(1234)(56)> = {(1234)(56), (13)(24), (1432)(56), (1)}.

So i'm trying to make sense of this and I've come up with this:

For [(1234)(56)]$^2$: \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \\ 3 & 4 & 5 & 6 & 1 & 2 \end{smallmatrix}

but it seems unusual because for the 4th column you would think it would be 2 and the last column it would be 6.

So now my problem is: [(1234)(56)]$^2$ = (1234)(56) * (1234)(56) = (13)(24) how did the person arrive at (13)(24)

Is it because you have:

1-3 3-5 5-1

and since we're back at 1, then we only include (13)

as for (24) 2-4 4-6 6-2

since we arrive at 2, then we only have (24)

If so, then i'm pretty fine with that but then I get completely lose here:

[(1234)(56)]$^3$ = (13)(24) * (1234)(56) = (1432) (56)

I get that you can rewrite:

[(1234)(56)]$^2$ [(1234)(56)]=(13)(24) * (1234)(56)

but how do we get (1432) (56)?

So, I decided to come up with the permutation:

For [(1234)(56)]$^3$: \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \\ 3 & 4 & 5 & 6 & 1 & 2 \\ 4 & 1 & 6 & 5 & 2 & 3 \end{smallmatrix}

which doesn't help me unless I did it wrong. So how did the person get [(1234)(56)]$^3$=(1432) (56) ?

$\endgroup$
  • 1
    $\begingroup$ Do you understand what the $(1\,2\,3\,4)(5\,6)$ notation means? I guess not, but I'm not sure. Wikipedia has an explanation. $\endgroup$ – MJD Feb 25 '15 at 1:02
3
$\begingroup$

Let's do $ [(1234)(56)]$ first:

You have:

$$\begin{matrix} 1& 2& 3& 4& 5& 6\\ 2& 3& 4& 1& 6& 5\\ \end{matrix} $$

Now for $ [(1234)(56)]^2$:

$1$ goes to $2$, and then $2$ goes to $3$. So $1 \to 3$.

$2$ goes to $3$, and $3$ goes to $4$, so $2 \to 4$.

$3$ goes to $4$, and $4$ goes to $1$, so $3 \to 1$.

$4$ goes to $1$, and $4$ goes to $2$, so $4 \to 2$.

$5$ goes to $6$, and $6$ goes to $5$, so $5 \to 5$.

$6$ goes to $5$, and $5$ goes to $6$, so $6 \to 6$.

Putting this together gives $[(13)(24)]$, since $5$ and $6$ don't change.

$\endgroup$
  • $\begingroup$ Alright, that helped a lot! Do you know why it ends at [(1234)(56)]$^4$? Like why the person did not do [(1234)(56)]$^5$ and [(1234)(56)]$^6$? $\endgroup$ – Justin Feb 25 '15 at 1:51
  • $\begingroup$ Also, if you do [(1234)(56)]$^4$, why is (1) rather than (1)(3,5,6) $\endgroup$ – Justin Feb 25 '15 at 1:55
  • $\begingroup$ Because, for example, you have $1 \to 2 \to 3 \to 4 \to 1$. Similar for all the other elements. After 4 iterations, you end up with the identity map. As for $[(1234)(56)]^5$, the trick is that $[(1234)(56)]^5 = [(1234)(56)][(1234)(56)]^4 = [(1234)(56)] \cdot \text{id}$. So there are really only 3 different permutations in the list $[(1234)(56)], [(1234)(56)]^2, [(1234)(56)]^3, [(1234)(56)]^4, \ldots$ $\endgroup$ – MarkG Feb 25 '15 at 1:57
  • $\begingroup$ For [(1234)(56)]$^4$, I made a mistake, my fault. I see everything clearer now. thanks a lot! $\endgroup$ – Justin Feb 25 '15 at 2:13
  • $\begingroup$ Is this a typo: $2$ goes to $3$, and $3$ goes to $4$, so $3 \to 4$.? Should it not read, $2 \to 4$? $\endgroup$ – Kevin Meredith May 20 '15 at 14:20

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.