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Some conclusions give that |G| = nm

  • #1
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Homework Statement


Let ##G## be a finite group such that ##|G| = nm##.
Suppose ##x\in G## has order ##n## and let ##\sigma_x\in S_G## be the permutation such that ##\sigma_x(g)=xg## for every ##g \in G##. Prove that ##\sigma_x## is a product of ##m## disjoint n-cycles.

Homework Equations




The Attempt at a Solution


Here is my attempt. Note that the cycle decomposition will include every element of ##G##, as if it did not, we would have that for some ##g \in G##, ##xg = g \implies x=1##, which clearly is a contradiction.

Now, suppose that ##a \in G##. Then the cycle that contains ##a## must be ##(1,a,xa,x^2a,\dots, x^{n-1}a)##. Now, consider ##b\in G## s.t. ##b\not = a##. Then either ##b\in \langle x \rangle a## or ##b \not \in \langle x \rangle a##. Suppose the latter, since the former would just result in the same cycle. Then the unique cycle that contains ##b## must be ##(1,b,xb,x^2b,\dots, x^{n-1}b)##. The cycle with ##b## and the cycle with ##a## are disjoint, since if they weren't ##x^ia = x^jb## such that ##i,j\in [0,n)##, and so ##b = x^{j-i}a \implies b\in \langle x \rangle a##, which is a contradiction.

So, every cycle in the cycle decomposition must have n elements and must be disjoint with all other cycles. We also noted that ##\sigma_x## must map all elements, so there must be ##m## n-cycles for this to be the case, since ##|G| = mn## QED

Is this proof correct? I feel like in parts I am hand-wavy, but I don't see how else to argue it.
 
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Answers and Replies

  • #2
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Now, suppose that ##a \in G##. Then the cycle that contains ##a## must be ##(1,a,xa,x^2a,\dots, x^{n-1}a)##.
Which cycle? You don't have any at the start, except that you wrote ##\sigma_x=c_1 \cdot \ldots \cdot c_k## and then show ##k=m## and ##\operatorname{ord}c_i=n##, but you haven't. I don't really understand your notation. ##\langle x \rangle## usually denotes the subgroup generated by ##x##, orbits are written ##G.x =\{\,g.x\,|\,g\in G\,\}## and the stabilizer ##G_x=\{\,g\in G\,|\,x.g=g\,\}##.

I would say you have:
"The orbit of ##a## is ##\langle x \rangle.a = \{\,a,xa,\ldots,x^{n-1}a\,\}##".
How do you get the ##1## into the orbit? I would try to prove it with the orbit-stabilizer theorem, but I haven't checked. The other idea is the one above. I'm a little bit confused, as I don't see which way you followed.
 
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  • #3
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Which cycle? You don't have any at the start, except that you wrote ##\sigma_x=c_1 \cdot \ldots \cdot c_k## and then show ##k=m## and ##\operatorname{ord}c_i=n##, but you haven't. I don't really understand your notation. ##\langle x \rangle## usually denotes the subgroup generated by ##x##, orbits are written ##G.x =\{\,g.x\,|\,g\in G\,\}## and the stabilizer ##G_x=\{\,g\in G\,|\,x.g=g\,\}##.

I would say you have:
"The orbit of ##a## is ##\langle x \rangle.a = \{\,a,xa,\ldots,x^{n-1}a\,\}##".
How do you get the ##1## into the orbit? I would try to prove it with the orbit-stabilizer theorem, but I haven't checked. The other idea is the one above. I'm a little bit confused, as I don't see which way you followed.
Never mind the ##1##, that shouldn't be included in any cycle other than ##(1,x,x^2,\dots, x^{n-1})##. Other than that mistake does the proof look correct?
 
  • #4
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Your proof is correct.

I found some minor issues while "completing" it:
1.) We assume ##n>1## to conclude ##x\neq 1##. The case ##n=1## is trivial, as it implied ##\sigma_x = 1## and ##\sigma_x=\Pi_{g\in G}(g)##, but for sake of completeness, it should be mentioned.
2.) To name the decomposition would be helpful, say ##\sigma_x = c_1\cdot \ldots \cdot c_k##. (I think this is formally needed, see below.)
3,) You didn't mention the case ##b \in \langle x \rangle .a \Longrightarrow \langle x \rangle .b = \langle x \rangle .a##
4.) What you actually have shown is
$$
G = \dot{\bigcup}_{g \in I} \langle x \rangle.g \,\text{ for some set }\, I\subseteq G \text{ and } |\langle x \rangle .a|=n
$$
Since ##|G|=n\cdot m = n \cdot |I|## we have ##|I|=m##.
5.) Now what I formally would add is, that all orbits ##\langle x \rangle .a## are identical to the cycles ##c_i\,##, either at the beginning or at last. It is a simple line: For any ##a\in G## there is a cycle ##c_i## with ##a \in c_i## by the first part of your proof, say ##a=c_{ij}##. Then ##\sigma_x(c_{ij})=c_{ij+1}=x.c_{ij}=x.a## and thus ##\langle x \rangle.a =c_i\; , \;k=|I|=m## and ##c_i \cap c_j = \emptyset ## for ##i \neq j\,.##

Sorry, if this is nitpicking, I just wrote what I added to your proof, so that me could (lazily) read it in one piece.
 
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Likes Mr Davis 97
  • #5
Stephen Tashi
Science Advisor
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\implies x=1##, which clearly is a contradiction.
What is the contradiction? Are you adding the hypothesis that ##n > 1## ?

Edit: Never mind. That's already been discussed.
 

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