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I am doing some exercises in the textbook Group Theory in a Nutshell for Physicists. However, there is an exercise in which I do not even understand what they ask. Could I kindly ask anybody to clarify this, please?

The question is in the chapter I.2, exercise 15

Map a finite group $G$ with $n$ elements into $S_n$ in a la Cayley. The map selects $n$ permutations, known as "regular permutations", with various special properties, out of the $n!$ possible permutations of $n$ objects.

(a) Show that no regular permutation besides the identity leaves an object untouched.

(b) Show that each of the regular permutations takes object 1 (say) to a different object.

(c) Show that when a regular permutation is resolved into cycles, the cycles all have the same length.

Well, I do not know what the "regular permutation" is (I tried to google it but couldn't find anything useful), so I just leave that word and do the exercise. The questions (a) and (b) seem to be quite obvious (and I have no idea what I need to prove). However, for the question (c), if I map $Z_4=\{1,i,-1,-i\}$ into a subgroup of $S_4$, we will get that the subgroup of $S_4$ which is isomorphic to $Z_4$ is $\{(),(1432),(13)(24),(1234) \}$ which all the cycles don't have the same length. Therefore, I misunderstood somehow.

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  • $\begingroup$ This question seems to suit Math.SE better. Nevertheless, note that (c) means that in each of the permutations you've found, all the cycles are of the same length. For example, in Id all the cycles are of length 1, in (1432) all cycles are of length 4 and in (13)(24) all cycles are of length 2. $\endgroup$ – eranreches Nov 11 '17 at 17:19
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    $\begingroup$ The definition of 'regular' permutation is easy enough to find. Leaving it out is a recipe for disaster - it's caused you to completely misunderstand the question. $\endgroup$ – E.P. Nov 11 '17 at 17:19
  • $\begingroup$ I understand it's a physics-oriented book you're using, but might Mathematics be better suited for this math question? $\endgroup$ – Kyle Kanos Nov 11 '17 at 18:10
  • $\begingroup$ The statement of the problem tells you what "regular" means in this context. A permutation is regular if it is in the image of the Cayley map from $G$ to $S_n$. $\endgroup$ – WillO Nov 11 '17 at 18:46
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The question is asking you to write down the map $\tau : G \to A(G)$, where $|G| = n$, given by Cayley's theorem. This mapping given by Cayley's theorem maps elements $g \in G$ to $\sigma \in A(G) \cong S_n$ where $\sigma$ is a so-called "regular permutation". The idea is to show that this object you get from the mapping from Cayley's theorem satisfies (a), (b) and (c).

So, it seems important to know and understand what Cayley's theorem states. Cayley's theorem (restricted to finite groups) says that every finite group $G$, $|G| = n$, is isomorphic to a subgroup of $S_n$; this map is given by $\tau : G \to A(G)$, $\tau(g) = \tau_g$, where $\tau_g : G \to G$, $\tau_g(x) = g x$ and $A(G)$ is the group of all permutations $\tau_g$ of $G$.

$G$ has $n$ elements, and $\tau$ selects from each of these $n$ elements an element $\tau_g \in A(G)$, and this $\tau_g$ is called a regular permutation. This is what the question states.

Now, let $g \in G$, $g \neq 1_G$, and consider the element $\tau(g) = \tau_g$. For an arbitrary $x \in G$, $\tau_g(x) = g x$. If $g = 1$, we can see that any object $x$ remains "untouched". This is (a). (b) follows from (a).

This group $A(G)$ is isomorphic to $S_n$, so we can identify each element $\tau_g$ with an element $\sigma$. This makes it slightly easier to work with, for now we can write $\tau_g$ as a permutation of the elements of the set $I_n = \{ 1, \ldots, n \}$. Hopefully you have the theorem which states that every non identity permutation can be written as a product of disjoint cycles.

Consider an element $g \in G$ of order $k$; then the permutation $\tau_g$ is a $k$-cycle (and has order $k$ as well). If $k = 1$, $\tau_g$ is the identity permutation; $k = 2$, $\tau_g$ is just a 2-cycle.

At this point, I have hopefully helped lead you in the right direction.

EDIT: (1) I incorrectly stated some things about Cayley's theorem. Hopefully things are correct now. (2) Attempted to elaborate on the hint given by @eranreches in a comment below

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  • $\begingroup$ I wanted to add an hint for (c) in addition to this answer: you might want to consider the order of the element $g\in G$ and see how it is connected to the length of cycles in $\tau_{g}$. $\endgroup$ – eranreches Nov 11 '17 at 18:50
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    $\begingroup$ @eranreches ah thank you. Great point. I included a brief remark about what you said $\endgroup$ – phoko Nov 11 '17 at 19:28
  • $\begingroup$ There are still some point that is not very clear to me. One is that $\tau$ is not a bijective map from $G$ to $A(G)$ (since there are more elements in $A(g)$ than $G$). What is the element in $A(G)$ that $\tau$ send to. For example, if $G=\{1,2,3 \}$ then $A(G)$ might be $\{(123),(231),(312),(132),(321),(213) \}$. So what is the value of $\tau(1)$ is this case? $\endgroup$ – Kimari Nov 11 '17 at 23:40
  • $\begingroup$ You're right about $\tau$ not being a bijective map; it is, however, a monomorphism. Thus, $G$ is isomorphic to its image $\tau(G)$, which is a subgroup of $A(G)$, and $A(G)$ can be shown to be isomorphic to $S_n$, at least when $|G| < \infty$. Also, $A(G)$ is defined to be the group of all permutations of $G$, hence $A(G)$ will always have an identity. Consider your example, where $G = \{ 1, 2, 3 \}$. Then $\tau(1) = \tau_1$, which is equivalent to the identity permutation. $\endgroup$ – phoko Nov 12 '17 at 2:00
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The question is asking you specifically about regular permutations instead of arbitrary ones, and if you drop that qualifier then the statements in the question are patently false. (If that's not clear, then it's a good exercise to construct counter-examples to each of the parts of your question from non-regular permutations.) As such, any progress you might have made on (a) and (b) assuming an arbitrary permutation will be resting on hidden assumptions and will thus be wrong.

The definition of a regular permutation is easy enough to find, though - on my machine, the top two google results are this and this threads on Math Stack Exchange. The second thread provides a clear definition in the form

Define a permutation $\alpha\in S_n $ to be regular if either $\alpha$ has no fixed points and it is the product of disjoint cycles of the same length, or $\alpha=(1)$,

for which all three parts of your question are essentially trivial (i.e. a red flag that this isn't the definition you should use).

That then tells you that as far as your exercise is concerned, you should use the definition of 'regular' permutations as given by the problem: that is, the image $\tau(G)\subset S_n$ of the group $G$ under the mapping $\tau:G\to S_n$ provided by Cayley's theorem. You can then take that definition and show that it is equivalent to the one in the MSE thread quoted above, from whence the rest of the question is easy to solve, or you can work more directly - take Cayley's mapping and show that the permutations in its image have the properties you're asked about.

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