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The question is like this:

IF $G=S_5$ and $g=(1\quad 2\quad 3)$, determine the number of elements in $H=\{x\in G:xg=gx\}$.

To do the question, first it says $$x(4)=(x(1\quad 2\quad 3))(4)=(1\quad 2\quad 3)x(4)$$ so $g(4)=4\ or\ 5$. Similarly, $g(5)=4\ or\ 5$.

Hence $g(\{4,5\})=\{4,5\}$ and also $g(\{1,2,3\})=\{1,2,3\}$.

My understanding for this part is that this is because the permutation $(1\quad 2\quad 3)$ has nothing to do on 4 or 5. Is this correct ?

But then it says this means $x$ must be one of the following: $$id,\ (1\quad 2\quad 3),\ (1\quad 3\quad 2),\ (4\quad 5),\ (1\quad 2\quad 3)(4\quad 5),\ (1\quad 3\quad 2)(4\quad 5)$$ I'm wondering how these are found? How do we find those $x$ satisfying $xg=gx$?

For example, what if I change the question as $G=S_6$ and $g=(1\quad 2\quad 3\quad 4)$ or $g=(1\quad 2\quad 3)$?

And how can we determine if two permutations commute without actual calculation?

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  • $\begingroup$ A general case is considered in this question. $\endgroup$ – Alex Ravsky Nov 23 '17 at 3:42
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What your first observation amounts to is that $H$ is (isomorphic to) a subgroup of $S_3 \times S_2$, where the first factor acts on $\{1,2,3\}$ and the second factor acts on $\{4,5\}$. This only has $12$ elements, which is a drastic reduction from the $120$ elements of $S_5$ we would have to test by "brute force".

Moreover, it can be seen that any element (corresponding to an element) of the form $(e_{S_3},\sigma)$, for either of the two elements $\sigma \in S_2$ will be in $H$, since disjoint cycles commute. Hence:

$gx = xg \iff (g_1,\sigma)((1\ 2\ 3),e_{S_2}) = ((1\ 2\ 3),e_{S_2})(g_1,\sigma)$

and since the second coordinate is always $\sigma$:

$\iff g_1(1\ 2\ 3) = (1\ 2\ 3)g_1$

So it really boils down to which elements of $S_3$ commute with $(1\ 2\ 3)$. Clearly, any power of $(1\ 2\ 3)$ commutes with $(1\ 2\ 3)$, and since these form fully half of $S_3$, and $S_3$ is non-abelian, this must be all of them (since any subgroup of $S_3$ which contains more than half the elements of $S_3$ must be $S_3$ itself).

Thus $H = \langle (1\ 2\ 3),(4\ 5)\rangle \cong \langle(1\ 2\ 3)\rangle \times \langle(4\ 5)\rangle \cong \Bbb Z_3 \times \Bbb Z_2 \cong \Bbb Z_6$; in fact, $H$ is cyclic with generator $(1\ 2\ 3)(4\ 5)$, and the six explicit permutations you listed are the powers of this generator.

I hope this makes it clearer how to extend this reasoning to $S_n$ for the same 3-cycle. For a 4-cycle like $(1\ 2\ 3\ 4)$, you'd have to look at which elements of $S_4$ commute with that 4-cycle, which takes a little more work than with a 3-cycle.

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  • $\begingroup$ Thank you for your answer. But I didn't quite understand. What do you mean by powers of this generator? And how do you conclude permutations like (1 2) or (2 3) does not satisfy this without calculation? Or can I say the permutations commutative with another permutation and with no other additional elements must be of the same length (except identity)? $\endgroup$ – J.Y May 4 '16 at 16:34
  • $\begingroup$ Let's pretend we're just working with $S_3$ for a moment. What elements of $S_3$ commute with the 3-cycle $(1\ 2\ 3)$? Well, it's clear any power of $(1\ 2\ 3)$ commutes with $(1\ 2\ 3)$, and it's not hard to show that the set of all elements of $S_3$ that commute with $(1\ 2\ 3)$ form a subgroup of $S_3$, called the centralizer of $(1\ 2\ 3)$. Now the powers of $(1\ 2\ 3)$ are: $\{e, (1\ 2\ 3), (1\ 3\ 2)\}$, and this is a subgroup (it's called $A_3$) of $S_3$. If any of the remaining elements commuted with $(1\ 2\ 3)$, then the centralizer of this 3-cycle would be all of $S_3$, $\endgroup$ – David Wheeler May 4 '16 at 22:36
  • $\begingroup$ But if this is so, then $(1\ 2\ 3)$ would be central (contained in the center of $S_3$, where the center is the subgroup of every element that commutes with all of $S_3$). In particular, $(1\ 2\ 3)$ would commute with $(1\ 2)$. But $(1\ 2)$ and $(1\ 2\ 3)$ generate $S_3$, so if $(1\ 2\ 3)$ commutes with $(1\ 2)$ then $S_3$ would be abelian, and this is false. $\endgroup$ – David Wheeler May 4 '16 at 22:46
  • $\begingroup$ This is so clear. Thank you! $\endgroup$ – J.Y May 5 '16 at 19:53

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