# a question on symmetric group $S_5$

Given that $x=(1 2)(3 4 5)\in S_5$ so its order $6$, and its a product of two cycle,I want to know whether $x$ commutes with all elements of $S_5$ and is it conjugate to $(4 5)(2 3 1)$? Thank you for the help.

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well, I was thinking how to show that $x$ is in the center of $S_5$ if it is commutes with every elements of $S_5$, But I could not get it. – La Belle Noiseuse Jul 9 '12 at 19:02
You don't have to show that $x$ is in the center of $S_5$ if it commutes with every element of $S_5$, because that is the definition of the center. – MJD Jul 9 '12 at 19:16
Any two permutations are conjugate if and only if they have the same cycle structure – Host-website-on-iPage Jul 10 '12 at 8:58

There is a rule for conjugating in $S_n$. If you have two permutations $\sigma$ and $\omega$, then to compute $\sigma\omega\sigma^{-1}$, you just permute the list of numbers in $\omega$ by $\sigma$. For instance $$(234)(12)(34)(234)^{-1}=(13)(42)$$ because $(234)$ sends $2$ to $3$, $3$ to $4$ and so forth. You should be able to work out that those two elements you have are conjugates. Once you know that you can answer the second question. If $x$ commutes with all elements of $S_5$, how big is its set of conjugates?

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according to you $[(1 2)(3 4 5)](4 5)(2 3 1)[(1 2)(3 4 5)]^{-1}= (5 3)(1 4 2)$ right? so $x$ is conjugate to $(4 5)(2 3 1)$ – La Belle Noiseuse Jul 9 '12 at 19:16
Uh? you just proved that $\,x(45)(231)x^{-1}\neq (45)(231)\,$, which does not prove $\,x\,$ is not conjugate to $\,(45)(345)\,$... – DonAntonio Jul 9 '12 at 19:26
Joe Johnsson,its size is 12? – La Belle Noiseuse Jul 9 '12 at 19:36
@Patience If an element of a group is in the center, how many elements can be in its conjugacy class? – Joe Johnson 126 Jul 9 '12 at 20:10
And $Z(S_n)=\{e\}$for $n\ge 3$ got it than you all – La Belle Noiseuse Jul 9 '12 at 22:54

If you want to prove $\,x=(12)(345)\,$ conjugate to $\,(45)(231)\,$ choose an element $\,\sigma\in S_5\,$ s.t. $$\sigma(1)=4\,,\,\sigma(2)=5\,,\,\sigma(3)=2\,,\,\sigma(4)=3\,,\,\sigma(5)=1\Longrightarrow \sigma=(14325)$$and then $$\sigma(12)(345)\sigma^{-1}=(14325)(12)(345)(15234)=(123)(45)$$

From here that your second question's answer is yes. About the first one: choose some easy permutation, say $\,(13)\,$ , and check whether $\,x(13)x^{-1}=(13)\,$ ...

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There's also a rule for recognizing conjugates in $S_n$: two permutations are conjugate if and only if they have the same cycle structure. So for example $(12)(345)$ is conjugate to $(45)(231)$ and to $(14)(235)$, but not to $(12)(34)$ or $(1234)$.

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thank you for the information,I am still not getting how to show whether $x$ commutes with every elements of $S_5$, well in that case $xy=yx\forall y\in S_5$ so $xyx^{-1}=y$ meaning every elements of $S_5$ is conjugates with $x$ but that is not true because, $S_5$ has elements with different cycle structure. am I right?? – La Belle Noiseuse Jul 9 '12 at 19:25
I think Joe Johnson's suggestion is pretty good for calculating the center of $S_5$. – MJD Jul 9 '12 at 19:27
@Patience: Dear Patience, Your above comment demonstrates some very basic confusions about the concepts you are working with. If $x$ is in the centre of a group $G$, then any conjugate of $x$, which by definition is an element of the form $y x y^{-1}$ for some $y \in G$, is actually equal to $x$. Thus $x$ has no other conjugates in $G$ besides itself. This is more or less the opposite of your suggestion that "every element ... is conjugates with $x$", a comment which is in turn based on a misinterpretation of your (correct, but not so useful) formula $x y x^{-1} = y$. Regards, – Matt E Jul 9 '12 at 23:17
I understand dear sir, thank you very much, I have revised these concepts, :) – La Belle Noiseuse Jul 9 '12 at 23:20
@Patience: Dear Patience, Great! Best wishes, – Matt E Jul 9 '12 at 23:21

Another hopefully-intuitive way of understanding how you might find a counterexample for whether $x$ commutes with all other elements of $S_5$ is by thinking of the action of $S_5$ and what your permutation $x$ represents. You know that if $x$ commutes with all elements, then $x$ conjugates each element to itself: $xcx^{-1} = c$ for all elements $c$, by right-multiplying the definition of commutation $xc=cx$ by $x^{-1}$ on both sides. (Similarly, left-multiplying both sides of the equation by $c^{-1}$ gives you $c^{-1}xc = x$, which shows why the answer to your second question is an answer to the first). The mental image of a 15-puzzle or a Rubik's cube might help here: imagine two 'rings', one with the numbers $1$ and $2$ on it and one with the numbers $3, 4, 5$. Then applying $x$ simultaneously 'rotates' both of these rings clockwise (taking $1$ to $2$ and vice versa, and rolling $3\rightarrow 4\rightarrow 5\rightarrow 3$). Applying $x^{-1}$ is equivalent to rotating these rings counterclockwise, so in the absence of any interim terms the two rings get back to their original positions: $xx^{-1} = e$; your challenge is to find some operation $c$ so that inserting $c$ in between the two applications of $x$ will shuffle things up a bit. Trying a $c$ that explicitly jumbles the two 'rings' seems like the best way of throwing a gear in the works, so take for instance $c=(13)$ and run the calculations...

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