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How do I find the conjugacy classes in $ \langle (12345), (2354) \rangle $? I really have no idea how to do this nicely. I know that two elements will be conjugate only if they have the same cycle type, but that doesn't help me much

Any help would be appreciated!

Thanks

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There might be a nice slick solution to this problem...but long term you might want to learn to use GAP. It's not hard to use and it'll crank out conjugacy classes or whatever else you need. –  Bill Cook Mar 23 '12 at 23:28
    
@BillCook Thanks for your feedback. This is something I'd like to be able to do swiftly in an exam situation, so I don't think GAP would be much help... –  Matt Mar 23 '12 at 23:33
    
@Mat Does $\langle \sigma_1, \sigma_2,\ldots \rangle$ denote a group generated by $\sigma_1, \sigma_2, \ldots$ ? –  Sasha Mar 23 '12 at 23:47
    
@Sasha Yes, it does –  Matt Mar 23 '12 at 23:55
    
@GeoffRobinson I said "only if"! –  Matt Mar 24 '12 at 0:52

4 Answers 4

up vote 2 down vote accepted

Your subgroup has an element of order $5$. There are only a handful of subgroups of $S_5$ like this. So if you are familiar with $S_5$'s structure, that might help on an exam.

  • Your subgroup cannot be $A_5$, since it has an odd permutation.
  • Your group is clearly larger than a copy of $C_5$.
  • Your group is not isomorphic to the dihedral group of order $10$, since it has an element of order $4$.

This only leaves two options: it's all of $S_5$, or it's isomorphic to $C_5\rtimes C_4$, which has (only?) $20$ elements. Stressed for time on exam, I might resort to making a multiplication table (don't laugh!). Either I end up with something closed that is $20\times20$, or we have $S_5$. Of course, at any step in the production of the table, if we find a transposition, a $3$-cycle, or an element of order $6$, we can rule out $C_5\rtimes C_4$ and know that we have $S_5$.

EDIT: Forget the table! At this point, I would suspect that we in fact have $C_5\rtimes C_4$. Does the $4$-cycle that is provided actually act on the $5$-cycle in the way we would expect for $C_5\rtimes C_4$? Yes! If I call your elements $a$ and $b$, we check that $a^b=a^2$. So these elements are canonical generators for $C_5\rtimes C_4$.

To finally answer what was asked of you, list all of the conjugacy classes of your copy of $C_5\rtimes C_4$. In an abstract $C_5\rtimes C_4$, generated by $a$ and $b$ with $b^{-1}ab=a^2$, $a^5=e=b^4$, you'd have the $20$ elements of the form $b^ja^i$ in the group. (Any words with $ab$ should be replaced with $ba^2$.) There is only one subgroup of order $5$, and all 4 of its order-$5$ elements are conjugate via $b^{-1}ab=a^2$. So that gives two classes: $$\begin{align} \left\{e\right\} && \left\{a,a^2,a^3,a^4\right\} \end{align}$$ There are $15$ elements that remain, coming from copies of $C_4$. There is one copy given by each of $\langle a^{-i}ba^i\rangle$ for $i=0,1,2,3,4$. That makes for $15$ non-identity elements, and they are all distinct since each $a^{-i}ba^i$ omits a unique number from $1$ through $5$ from its cycle notation. That gives potentially three more classes: $$\begin{align} \left\{b,b^a,b^{a^2},b^{a^3},b^{a^4}\right\} && \left\{b^2,(b^2)^a,(b^2)^{a^2},(b^2)^{a^3},(b^2)^{a^4}\right\} && \left\{b^3,(b^3)^a,(b^3)^{a^2},(b^3)^{a^3},(b^3)^{a^4}\right\} \end{align}$$

The middle class here consists of order-$2$ elements while the others consist of order-$4$ elements, so the middle class is truly a conjugacy class. Can the first and third be merged? No, and here's why. I've implicitly used this fact several times already, but one choice for the cycle notation of $x^y$ is the same as for $x$, with $y$ having acted on each number. For example, $$(12345)^{(2354)}=(2453)(12345)(2354)=(13524)=(1^{(2354)}2^{(2354)}3^{(2354)}4^{(2354)}5^{(2354)})$$ Now if $b^3$ were conjugate to $b$, then $(2354)^x=(2453)$. Because of the way $x$ affects $b$'s cycle notation, we would have to have an $x$ that omitted $1$ from its cycle notation. That is, $x$ would have to already be in $\langle b\rangle$. But of course that would make $b^x=b$, not $b^3$.

So the five classes that have been listed are the conjugacy classes for your group. This whole discussion relied heavily on a familiarity with $S_5$. I wouldn't trust any approach like this for a similar problem with an ambient $S_6$, say. But I would like to say that $S_5$ is such an important group in so many contexts that it warrants more familiarity than $S_{n}$ does in general.

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Thanks for your response. I think perhaps I didn't make my question clear enough - my main issue is listing the conjugacy classes of this group. How many are there, for example? How do I find them? –  Matt Mar 24 '12 at 1:26
    
@Matt I've added some more to get at an actual list of conjugacy classes. –  alex.jordan Mar 24 '12 at 2:43

This is not an answer, just following BillCook's advice and trying out GAP from Sage

sage: g = gap.Group(['(1,2,3,4,5)','(2,3,5,4)']); g
Group( [ (1,2,3,4,5), (2,3,5,4) ] )

sage: g.ConjugacyClasses()
[ ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), () ), 
  ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,3,5,4) ), 
  ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,4,5,3) ), 
  ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,5)(3,4) ), 
  ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (1,2,3,4,5) ) ]

and Sage as well

sage: G = PermutationGroup(['(1,2,3,4,5)','(2,3,5,4)']); G
Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]

sage: g = G.gens()[0]; g
(2,3,5,4)

sage: g = G.gens()[1]; g
(1,2,3,4,5)

sage: G.order()
20

G.conjugacy_classes_subgroups()
[Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [()], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,5)(3,4)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,3,5,4)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by (1,2,3,4,5)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,5)(3,4), (1,5,4,3,2)], Subgroup of (Permutation group with generators [(2,3,5,4),(1,2,3,4,5)]) generated by [(2,3,5,4), (2,5)(3,4), (1,5,4,3,2)]]

# this is an ugly hack:
for x in G.conjugacy_classes_subgroups():
s=str(x).replace('Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by ','')
print s.replace('[','<').replace(']','>')
<()>
<(2,5)(3,4)>
<(2,3,5,4)>
<(1,2,3,4,5)>
<(2,5)(3,4), (1,5,4,3,2)>
<(2,3,5,4), (2,5)(3,4), (1,5,4,3,2)>

to show the results. I think there is a slick trick to know but I don't remember it.

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Let's do it more theoretically without a computer. Conjugation by the $4$-cycle sends the $5$-cycle to its square. The subgroup you are looking at therefore has a normal Sylow $5$-subgroup, and has order $20.$ The Sylow 2-subgroup is cyclic of order $4$, and since is unique element of order $2$ conjugates the $5$-cycle to its inverse, there are no elements of order $10$ or $20$ in the subgroup. Hence all elements have order $1,2,4$ or $5$. All elements of order $5$ are conjugate, since all lie in the only Sylow $5$-subgroup, and the $5$-cycle is conjugate to its square, fourth power, and cube (its cube also being its eight power). Every element of $2$-power order is conjugate to an element of the Sylow $2$-subgroup, which is cyclic, and only contains on element of order $2$. Hence there is one conjugacy class of elements of order $2$. There are two elements of order $4$ in the Sylow $2$-subgroup. There are several ways to check that they are not conjugate (for example, if they were conjugate, their images would be conjugate in the homomorphic image obtained by factoring out the normal Sylow $5$-subgroup. That factor group is cyclic of order $4$ and the iemages of the two elements of order $4$ are different, so are not conjugate in the Abelian factor group. Or you could count elements if you haven't seen factor groups. In any case, the conjugacy classes are as follows ( I omit the arguments for calculating their sizes: the class of the identity, (size 1), one class of elements of order $5$, size $4$. One class of elements of order $2$, size $5$, two classes of elements of order $4$, each of size $5.$ The representatives of the non-identity classes are respectively: $(12345), (25)(34), (2354)$ and $(4532)$.

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Since this is too long for a comment, here goes...

Thanks to @bgins for providing some code. This is what I got from GAP:

>g := Group([(1,2,3,4,5),(2,3,5,4)]);
Group([ (1,2,3,4,5), (2,3,5,4) ])

>Order(g);
20

>c := ConjugacyClasses(g);
[ ()^G, (2,3,5,4)^G, (2,4,5,3)^G, (2,5)(3,4)^G, (1,2,3,4,5)^G ]

>Size(c[1]);
1

>Size(c[2]);
5

>Size(c[3]);
5

>Size(c[4]);
5

>Size(c[5]);
4

>Elements(c[1]);
[ () ]

>Elements(c[2]);
[ (2,3,5,4), (1,2,4,3), (1,3,2,5), (1,4,5,2), (1,5,3,4) ]

>Elements(c[3]);
[ (2,4,5,3), (1,2,5,4), (1,3,4,2), (1,4,3,5), (1,5,2,3) ]

>Elements(c[4]);
[ (2,5)(3,4), (1,2)(3,5), (1,3)(4,5), (1,4)(2,3), (1,5)(2,4) ]

>Elements(c[5]);
[ (1,2,3,4,5), (1,3,5,2,4), (1,4,2,5,3), (1,5,4,3,2) ]

So the class equation is $20=1+5+5+5+4$. The conjugacy classes turn out (for this subgroup) to be: the identity, pairs of 2 cycles, 5 cycles, and then the 4 cycles split into 2 classes.

As I mentioned in the comments, there may be some "slick" solution to this problem, but determining conjugacy classes for a random subgroup of some permutation group is quite a difficult thing to do. As in @alex.jordan's post, you can determine quite a bit without direct computation, but unless you're quite familiar with groups of order 20 (I think there are only 5 up to isomorphism) there's not a great way to tackle this without getting your hands dirty.

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