# Some conclusions give that |G| = nm

## 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.

## 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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fresh_42
Mentor
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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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?

fresh_42
Mentor

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.

• Mr Davis 97
Stephen Tashi
\implies x=1$, which clearly is a contradiction. What is the contradiction? Are you adding the hypothesis that$n > 1## ?