Order of the conjugacy class generated by a permutation.

Question

Let $\sigma=(12345)\in S_5$ what is $|Cl_{S_5}(<\sigma>)|$ i.e. what is the order of the conjugacy class of subgroup generated by $\sigma$?

What I've done is to write down this subgroup:

$$<\sigma>=\{e,(12345),(13524),(14254),(15432)\}. $$

The definition of a conjugacy class gives $Cl_G(H)=\{xHx^{-1}|x\in G\}$. All 5-cycles are conjugate. And I know that $|Cl(\sigma)|=\dfrac{5!}{5}=24$ but I don't see how to put these facts together.

Answer 1

I think you may have got a little mixed up in your notation.

The conjugacy class of a single element $h$ in a group $G$ is defined to be $\{xhx^{-1}: x\in G\}$. This is usually denoted by $C_{h}$. It is a subset of $G$ and with the exception of when $h$ is the identity element it is NOT a subgroup of $G$.

Conjugation is a group action on the group itself and so the conjugacy classes are the orbits of this action and as such partition $G$. Thus two conjugacy classes are either equal or disjoint.

Now there is a result that says that in $S_{n}$ two permutations are conjugate if and only if they have the same cycle type (when decomposed into disjoint cycles). Thus the conjugacy class of a 5-cycle in $S_{5}$ is the set of all the 5-cycles in $S_{5}$.Thus for your example $C_{(12345)}$ is the set of all the 5-cycles of $S_{5}$. I think this is possibly what you want to find.

However in case what you want to find is the object you defined, $Cl_{S_{5}}(<\sigma>)$, proceed as follows. So as you run through each $h\in <\sigma> $, to get the set you want to find, $Cl_{S_{5}}(<\sigma>)$, you get the set that is the union of each of the conjugacy classes of each of your elements of $<\sigma>$. Thus all you have to do is evaluate $<\sigma>$ and then see which cycle types this subgroup contains. In general your answer will NOT be a subgroup and the union of conjugacy classes is not in general a subgroup (though sometimes is).

Are you sure this is the set you wanted to find?

To answer your follow-up comments.

Let $(a_{1} a_{2} a_{3} a_{4} a_{5})$ be a 5-cycle in $S_{5}$. You have 5 choices for$a_{1}$, 4 for $a_{2}$ and so on. So that is 5! possible cycles of length 5. However the same cycle can be written in 5 different ways (as $(a_{1} a_{2} a_{3} a_{4} a_{5}) = (a_{2} a_{3} a_{4} a_{5} a_{1})$ and so on). So the number of (distinct) 5-cycles in $S_{5}$ is 5!/5 = 4! = 24.

$<\sigma>$ means the subgroup generated by $\sigma$. This is the subgroup that consists or all powers of $\sigma$ (assuming using multiplicative notation for the B.O of the group, which is the case for $S_{5}$). Now you know that your 5-cycle (12345) has order 5 (so $(12345)^{5}=(1)$ where (1) denoted the identity permutation). Thus $<\sigma> = \{ (1),(12345), (12345)^{2}, (12345)^{3}, (12345)^{4} \} $. Now $(12345)^{2} = (12345)(12345)=(13524)$. Similarly you can find the other elements of$<\sigma>$.

There are some other important sets associated with groups which are

The centralizer of (the subset) $S$ of a group $G$, often denoted by $C_{G}(S)$ which is the set $\{ g\in G: gs=sg \mbox{ for all } s\in S\}$

The normalizer of $S$ in $G$ often denoted by $N_{G}(S)$ which is the set $\{ g\in G: gS=Sg \}$ which is not (in general) the same thing as the centralizer.

The centre of $G$ usually denoted bt $Z(G)$ which is the set $\{ z\in G: zg=gz \mbox{ for all } g\in G\}$.

Finally two subgroups $H,K$ of $G$ are called conjugate if there exists a $g\in G$ such that $H =gKg^{-1}$

Not sure what you have been told $N<\sigma>$ means.

I hope this helps.

This extension of the answer is to try and answer you latest comment. If by "the conjugacy class of $<\sigma>$ in $S_{5}$" you/your lecturer means the set of all subgroups that are conjugate to $<\sigma>$ (and I do not know if that is what is meant since it is not a standard phrase) then you can proceed as follows. The order of $<\sigma>$ is 5 and the order of $S_{5}$ is 5! and so $<\sigma>$ is a Sylow 5-subgroup of$S_{5}$ (since 5!=5(4!) and 5 does not divide 4! and 5 is prime). Thus using Sylow's third theorem the number of Sylow 5-subgroups $k$ say (which is the number you are looking for) satisfies $5\mid k-1$ and $k\mid 4.3.2.1$ so $k\mid 24$. Thus you get $k=6$.

I hope this helps

Answer 2

Hint: The conjugacy class of a cycle $\sigma$ contains only the cycles that you can write as compositon of cycles that have the same lenght of $\sigma$.

Answer 3

Given a subgroup $H$ of $S_5$ then the elements $g$ act on $H$ by creating new subgroups $g^{-1}Hg$. The set of these subgroups is called the conjugacy 0class of $H$. To count them one can use the orbit-stabilizer theorem. So one has to find out which sugroup of $G$ leaves $H = \langle \sigma \rangle$ unchanged (this subgroup is also called the normalizer of $H$ in $G$). For this we have to find out which elements of $S_5$ map $H$ into $H$, but these are the elements of $H$ itself, and the elements that map the generator $\sigma$ to another generator $\sigma^3$ so the stabilizer has order $20$ and the orbit has $\frac{5!}{20} = 6$ elements.