1

I'm currently in first year math and got given this question, but I have absolutely no idea how to go about it.

After asking me to write the permutations given by $f$ as a product of disjoint cycles, it asked me to find the least number $m\in\mathbb{N}$ such that $f^m=\mathrm{id}$, where id denotes the identity function.

I tried composing it again and again so I saw what was obtained for $m=2,m=3,...$, but it seems like it's going nowhere: I cannot spot any patterns in the lengths of the cycles corresponding to the value of $m$. I was wondering if I could get a hint that could take me in the correct direction?

Thank you!

user356
  • 467
  • What is this $f$? – Angina Seng Mar 03 '19 at 08:11
  • 3
    If $f$ is the product of some disjoint cycles $f=c_1 \cdots c_n$, then the least number for which $f^m=id$ is $m=\mathrm{lcm}(l_1, \cdots , l_n)$ where $l_i$ denotes the length of $c_i$. – Crostul Mar 03 '19 at 08:14
  • @Crostul Thank you! May I ask why? – user356 Mar 03 '19 at 08:44
  • What is $f^4$? $f^6$? – Angina Seng Mar 03 '19 at 08:49
  • @LordSharktheUnknown $f^4 = (1 4 2)(3 8 9)(5)(10 6)(7)$ and $f^6 = (1)(9)(4)(3)(2)(8)(5 7)(6)(10)$? – user356 Mar 03 '19 at 08:56
  • I'd check $f^4$ again.... – Angina Seng Mar 03 '19 at 09:01
  • 1
    Both your $f^4$ and your $f^6$ are wrong. Try again. Be careful. Also, $f$ isn't a function on ${,1,2,3,4,5,6,7,}$. – Gerry Myerson Mar 03 '19 at 09:01
  • @LordSharktheUnknown $f^4$=(1 4 2)(3 8 9)(5)(10)(7)(6) and $f^6$=(1)(9)(4)(3)(2)(8)(5 7)(6 10) – user356 Mar 03 '19 at 09:16
  • @GerryMyerson sorry $f$ is a function on {1,...,10}! – user356 Mar 03 '19 at 09:16
  • @LordSharktheUnknown do you count the first composition of $f$ with itself as $f^1$ or $f^2$? I've been doing the latter but if I were to do the former, then I'd see where you're getting at with the $f^4$ and $f^6$... In my case, $f^5$ and $f^7$ are the more significant ones in that they are the same as $f$. How does this then link in with the LCM idea? – user356 Mar 03 '19 at 09:28
  • 1
    The order of a group is its number of elements. The order of an element $x $ is the order of the subgroup it generates, and equals the smallest positive $k$ such that $x^k=1$. See the connection? – darij grinberg Mar 03 '19 at 10:06
  • For an element $x$ in a group one defines its order as the least positive integer $m$ such that $x^m=e$, where $e$ is the identity element (which for permutations is the identity map from ${1,\dots,n}$ to itself). See the connection with your question? – user26857 Mar 03 '19 at 10:19
  • @darijgrinberg So in this case each permutation is a subgroup, and the order of each permutation is the number of disjoint cycles the permutation is expressed as? How do we get that the order of each permutation equals the smallest positive k such that $x^k=1$? I see that $x^k=1$ links in with $f^k$=id but I don't see how it links with the LCM. – user356 Mar 03 '19 at 10:21
  • No, the order of each permutation is the least common multiple of the lenghts of its cycles. – user26857 Mar 03 '19 at 10:23
  • 1
    No, a permutation is not a subgroup. He said "the subgroup it generates". – user26857 Mar 03 '19 at 10:24
  • You can consider the order just as the smallest $k>0$ such that $x^k=1$. @darijgrinberg wanted to show you the connection between the order of a group and the order of an element. – user26857 Mar 03 '19 at 10:26
  • @user26857 where do we get that the order is the least positive k such that $x^k=1$? (sorry I think I'm in a right old muddle now!) – user356 Mar 03 '19 at 10:26
  • https://en.wikipedia.org/wiki/Order_(group_theory) – user26857 Mar 03 '19 at 10:26
  • 2
    @user26857 Ah I see, so it's just a universal definition! (I wonder why I've never come across this before in lectures) Do you know where LordSharktheUnknown was getting at with the $f^4$ and $f^6$? I know that 4 and 6 are the lengths of the disjoint cycles of $f$, but I don't understand how I'd use that to get to the conclusion that the LCM of the lengths of the cycles gives the order of each permutation. – user356 Mar 03 '19 at 10:29
  • Yes, this can be considered as a definition of the order. Later on can show the connection with the order of a group by proving that the order of an element $x$ equals the order of the (sub)group generated by $x$, usually denoted by $\left< x\right>$. – user26857 Mar 03 '19 at 10:31
  • In this case the order is $12$. After finding $f^4$ can see that you arrived at a few disjoint $3$-cycles which raised to the power $3$ give the identity, so $(f^4)^3=e$. If compute $f^6$ then you get few disjoint transpositions which raised to the power $2$ give the identity, so $(f^6)^2=e$. – user26857 Mar 03 '19 at 10:35
  • Btw, I hope you know that $(c_1\cdots c_t)^m=c_1^m\cdots c_t^m$ if $c_1,\dots,c_t$ are disjoint cycles. – user26857 Mar 03 '19 at 10:38
  • 1
    @user26857 Thank you so much! Yes they must be disjoint, otherwise they wouldn't be commutative and none of this would work – user356 Mar 03 '19 at 10:42
  • $f$ composed with itself is $f^2$. Neither $f^5$ nor $f^7$ is equal to $f$. – Gerry Myerson Mar 03 '19 at 11:20

1 Answers1

1

The answer has been given in the comments, with many helpful remarks from other users. Having said that, the way I think about this problem is in the following way:

You now have your permutation written down as a product of two disjoint cycles:

$$f=(5\ 6\ 7\ 10)(1\ 8\ 2\ 3\ 4\ 9).$$

Lets look at these two cycles individually. Lets denote $f_1=(5\ 6\ 7\ 10)$ and $f_2=(1\ 8\ 2\ 3\ 4\ 9)$ so that $f=f_1 \circ f_2$. (WARNING: Do not get subscripts confused with superscripts. Here I am just labelling things with subscripts. Superscripts (used later) are for multiplication.)

You either know, or will be able to figure out, that if you have a cycle $\alpha$ of length $n$, then $\alpha^n = Id$. Lets be clear about what I mean: If you have a cycle of length $n$, and you do it $n$ times, then that is the same as the identity permutation. In plain english: doing a cycle of length $n$ $n$ times is the same as doing nothing at all. Everything goes back to where it started.

Now, that means that if you do $f_1$ four times you get the identity (i.e. $(f_1)^4 = Id$), and similarly $(f_2)^6 = Id$.

Initially, that may seem like it's not important and it doesn't help. After all, we are doing $f_1$ and $f_2$ "at the same time" when we do $f$. The key thing to realise here is that if, for example, $(f_1)^4=Id$, then for any multiple of four, doing $f_1$ that many times is the same as doing the identity. The same goes for $f_2$: doing $f_2$ any number of times which is a multiple of $6$ results in doing the identity.

What does that all add up to? Well, when we do $f$, we are doing $f_1$ and $f_2$ at the same time. The conclusion is that if you do $f$ a number of times which is both a multiple of four and six, that will be the same as doing nothing, because that is the same as doing nothing for both of the cycles which we put together to make $f$.

For example, $f^{24} = Id$.

Now, if you check the definition of what the lowest common multiple of two numbers is vs the above logic, perhaps that will bring the insight which you're after to understand Crostul's answer in the comments.

This exercise actually demonstrates some of the motivation for why we like to write permutations as products of disjoint cycles.

Matt
  • 3,296