# List the elements of the cyclic subgroup of $S_6$

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) ?

• 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. – MJD Feb 25 '15 at 1:02

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.

• 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)]$^6$? – Justin Feb 25 '15 at 1:51
• Also, if you do [(1234)(56)]$^4$, why is (1) rather than (1)(3,5,6) – Justin Feb 25 '15 at 1:55
• 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 $[(1234)(56)], [(1234)(56)]^2, [(1234)(56)]^3, [(1234)(56)]^4, \ldots$ – MarkG Feb 25 '15 at 1:57
• For [(1234)(56)]$^4$, I made a mistake, my fault. I see everything clearer now. thanks a lot! – Justin Feb 25 '15 at 2:13
• Is this a typo: $2$ goes to $3$, and $3$ goes to $4$, so $3 \to 4$.? Should it not read, $2 \to 4$? – Kevin Meredith May 20 '15 at 14:20