9
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I know that I need to find a $\theta \in S_9$ such that $\theta \sigma \theta^{-1}=(\theta(1),\theta(2),\theta(3))\cdot (\theta(4),\theta(5),\theta(6),\theta(7))=(1,2,3)(4,5,6,7)$.

Kind of stuck here. How do I check which elements of $S_9$ will satisfy it?

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    Can you think of any (other than the identity and itself of course)? – 2017-02-06
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    Hint: Note that $\theta$ just relables the points moved by $\sigma$ so that the elements of the subgroup generated by $\sigma$ are candidates. On the other hand you also have the elements that relable the points not moved by $\sigma$. – 2017-02-06

4 Answers 4

8

Consider the action of $ S_9 $ on itself, given by conjugation. The stabilizer of an element $ x \in S_9 $ is its centralizer, and the orbit-stabilizer theorem gives

$$ 9! = |S_9| = |C(x)| \cdot |S| $$

where $ S $ denotes the set of conjugates of $ x $, and $ C(x) $ is the centralizer. Conjugate elements in $ S_n $ are precisely those which have the same cycle type, therefore, $ S $ is the set of all $ 3, 4 $ cycles when $ x = (123)(4567) $. By elementary combinatorics, there are exactly

$$ \binom{9}{3} \cdot 2! \cdot \binom{6}{4} \cdot 3! = 15120 $$

such cycles. Thus,

$$ C((123)(4567)) = \frac{9!}{15120} = 24 $$

elements of $ S_9 $ commute with $ (123)(4567) $.

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    Nice answer! Knowing the count is 24, we can even list them: $(89), (123), (4567)$ all commute with this element and with each other, and generate a subgroup of order $2 \cdot 3 \cdot 4$, so that subgroup is the whole centralizer. – 2017-02-06
1

Hint $ O(Class(a))=\frac{O(S_n)}{O(C(a))}$

$C(a)$ is the centralizer of $a$

You need $C(a)$ as your answer. Can you proceed? In general if you have to find out the $O(C(a))$ of any element $a$ then first express the element into a product of disjoint cycles .say this representation has cycles of length $n_1,n_2,......n_n$ with occurence $r_1,r_2,.....r_n $ times respectively. Then $C(a)$=$(r_1×(n_1)^{r_1}).(r_2×(n_2)^{r_1})......$

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    What do you mean by the normalizer of an element? – 2017-02-06
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    $N(a)=${$x \in G : ax=xa$} why the downvote – 2017-02-06
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    Agree. Not sure why people are downvoting. This looks right to me. – 2017-02-06
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    The downvotes are probably for confusingly nonstandard notation. What you have called $N(a)$ is usually called $C(a)$ (for centralizer). – 2017-02-06
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    have edited now should the downvoted by removed – 2017-02-06
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    @Upstart This is now correct, but it just rephrases the original question, but it still doesn't answer the question it only reformulates it. – 2017-02-06
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    @MarcBogaerts It was correct before (the normalizer is the centralizer in this case cf. https://proofwiki.org/wiki/Size_of_Conjugacy_Class_is_Index_of_Normalizer) and it does answer the question... it's just Starfall's answer in hint form. Reduces it to computing the size of a conjugacy class. – 2017-02-06
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The centralizer (also called the normalizer by some) of $\sigma$ is the intersection of the centralizer of $(1,2,3)$ and the centralizer of $(4,5,6,7)$. The centralizer of a cycle in general is given by the conjugation by the permutations that relabels the cycle (e.g. for $(1,2,3)$ giving $(2,3,1)$ and $(3,1,2)$). For a given cycle these permutations form the group generated by the cycle itself and all permutations of the elements that are not moved by the cycle (for $(1,2,3)$ this is $S_{\{4,5,6,7,8,9\}}$). So this intersection consists of the group generated by $(1,2,3),(4,5,6,7)$ and $(8,9)$ giving the group $C_3 \times C_4 \times C_2$ of order 24.

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    Just a note that there are two types of element in the group. There are those which come from powers of the individual cycles, and there are those which come from permuting cycles of the same type (here $C_2$ which swaps the two $1-$ cycles, or fixed points). If there is more than one cycle of the same type, these components also have to be taken into account. – 2017-02-06
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    @mark This example is rather chosen to be simple. A more complex case would be $\theta = (1,2,3)(4,5)$. In this case the centralizer would consist of the group generated by the cycles of $\theta$ and the permutations of the set ${6,7,8,9}$ resulting in the non abelian group $C_3 \times C_2 \times S_4$ of order $144$. – 2017-02-06
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Now the thing you have computed states that the three length cycle is only possible to be $(1,2,3)$, the four length cycle is $(4,5,6,7)$ in a similar manner. The only possiblity is $(8,9)$ is possible. So the permutations that shall commute are $\{(1,2,3)(4,5,6,7) , (1,2,3)(4,5,6,7)(8,9) \} $

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    Elements that are generated by these two elements also satisfy – 2017-02-06
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    What about (1,2,3)? or (8,9)? (or the identity)? There are many others – 2017-02-06
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    so there are three others then $\{ (1,2,3) , (4,5,6,7),(8,9) \} $ – 2017-02-06
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    There are more than three others. (1,3,2)? (4,6)(5,7)? – 2017-02-06
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    @spaceisdarkgreen those are $\sigma^3$ and $\sigma^4$. – 2017-02-06
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    Note that the elements $\theta$ the OP asks for form a group. It is called the centralizer (of $\sigma$). – 2017-02-06
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    @MarcBogaerts Not disagreeing with you. – 2017-02-06
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    yup so the set should also have the permutations generated by the cycles mentioned in the set. I am a bit sloppy now sorry for that. – 2017-02-06
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    @spaceisdarkgreen Sorry, it's $\sigma^8$ and $\sigma^6$. – 2017-02-06
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    @ChirantanChowdhury Indeed, because the product of two elements centralizing $\sigma$ also centralize $\sigma$. – 2017-02-06