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I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$

This is my attempt: \begin{align} a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\ a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\ a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\ a^5 &= (1\ 3\ 8\ 2\ 10)(7\ 6) \\ a^6 &= (1\ 8\ 10)(4\ 6\ 9) \\ a^7 &= (1\ 3\ 10)(4\ 7\ 9\ 6) \\ a^8 &= (1\ 10)(4\ 9\ 7) \\ a^9 &= (1\ 3\ 8\ 2\ 5\ 10)(7\ 6) \\ a^{10} &= (1\ 8\ 5)(4\ 6\ 9) \\ a^{11} &= (1\ 3\ 5\ 10)(4\ 7\ 9) \\ a^{12} &= (1\ 5)(4\ 9\ 7\ 6) \end{align}

I suspect I already went wrong somewhere. I understand I need to get to $e = (1)$ at some point. Is there a way to check and make sure there are no mistakes when you calculate this?

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It would help if you used $0$ for the tenth element permuted by the group, rather than $10$, or use a comma or separator, or letters. Your element seems to be the product of a $6$-cycle and a disjoint $4$-cycle, so should have order lcm$(4,6)=12$. Using $0$ for the tenth element, note that your powers are wrong, and if $a=(138250)(4769)$ then $a^2=(185)(320)(46)(79)$, for example. So take care. –  Mark Bennet Mar 29 at 21:54
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You're calculation of $a^2$ is wrong. It should be $a^2=(1:8:5)(3:2:10)(4:6)(7:9)$. I put a colon between elements of $\{1, ..., 10\}$ because it can be confusing when you get two digit numbers –  Owen Sizemore Mar 29 at 21:55
    
@owen \, also seems to work nicely for separating numerals. –  MJD Mar 30 at 20:40
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2 Answers 2

up vote 3 down vote accepted

If you have $$a = (1\,3\,8\,2\,5\,10)(4\,7\,6\,9)$$

that means that $a$ is the permutation that takes 1 to 3, 3 to 8, 8 to 2, and so on.

The permutation $a^2$ is obtained by applying $a$ twice. Since $a$ takes 1 to 3, and then 3 to 8, $a^2$ takes 1 to 8. $$\begin{array}{ccc} a^0 & a^1 & a^2 \\ \hline 1 & 3 & 8 \\ 2 & 5 & 10 \\ 3 & 8 & 2 \\ 4 & 7 & 6 \\ 5 & 10 & 1 \\ 6 & 9 & 4 \\ 7 & 6 & 9 \\ 8 & 2 & 5 \\ 9 & 4 & 7 \\ 10 & 1 & 3 \end{array}$$

Reading off the first and last column of the first row, we have that $a^2$ takes 1 to 8, so it begins $a^2 = (1\,8\ldots)\ldots$. Reading the first and last column of the 8th row, we see that $a^2$ takes 8 to 5, so $a^2 = (1\,8\,5\ldots)\ldots$. Reading off the rest of the rows similarly, we get: $$a^2 = (1\,8\,5)(2\,10\,3)(4\,6)(7\,9).$$

Perhaps you can take it from here.

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I like the approach of first writing out $a^k(i)$ for fixed $i$ and all $k$ instead of finding $a^2, a^3, \dots$ successively, that's pretty efficient! –  Christoph Mar 29 at 22:09
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You can of course find $a^3$ from the same table, without writing anything else out. –  MJD Mar 29 at 22:11
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Your calculations look wrong. Keep in mind that $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$ is the map $a:\{1,2,\dots,10\}\to\{1,2,\dots,10\}$ given by $$ a : 1\mapsto 3, 3\mapsto 8, 8 \mapsto 2, 2\mapsto5, 5\mapsto 10, 10\mapsto 1, 4\mapsto 7, 7\mapsto 6, 6 \mapsto 9, 9\mapsto 4. $$

Applying this map twice yields $$ a^2 : 1\mapsto 8, 8\mapsto 5, 5\mapsto 1, 3\mapsto 2, 2\mapsto 10, 10\mapsto 3, 4\mapsto 6, 6\mapsto 4, 7\mapsto 9, 9 \mapsto 7, $$ which is written in cycle notation $a^2 = (1\ 8\ 5)(3\ 2\ 10)(4\ 6)(7\ 9)$.

For $a^3$ you have to apply $a$ three times and get $a^3 = (1\ 2)(3\ 5)(8\ 10)(4\ 9\ 6\ 7)$.

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I feel a little lost here, I see how $a^2$ starts with (185), how do you know to start the next one with $3$? –  Bob Mar 29 at 22:06
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@Bob It doesn't matter really, take any number you didn't map already. –  Christoph Mar 29 at 22:10
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@Bob You can start it with anything other than $1, 8$ or $5$ since you know what happens to them. It is just convenient to go until you've finished what happens in the $6$-cycle before choosing something permuted by the $4$-cycle. It helps to be systematic about these things, since that helps to avoid making mistakes. Another systematic way to proceed is to choose the lowest number you haven't yet encountered. The sub-cycles may appear in a different order, but they will be the same, and completely equivalent, whatever proper system you use. –  Mark Bennet Mar 29 at 22:11
    
OK, I think I got it now, every other one would be then: $a^2$ = $(185)(2103)(46)(79)$, $a^4$ = $(158)(2310)(4)(6)(7)$, $a^6$ = $(1)(2)(3)(46)(5)(79)(10)$, $a^8$ = $(185)(2103)(4)(6)(7)$, $a^{10}$ = $(158)(2310)(46)(79)$ –  Bob Mar 30 at 0:05
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