Let  $\sigma\in S_n$ . Describe the centralizer of $\sigma$.  Thought: If I conjugate $\sigma$, then $\tau^{-1}\sigma\tau=\sigma.$ This means that $\tau$ is a power of $\sigma,$ and $&lt;\sigma&gt;\subseteq C_{S_n}(\sigma).$ What should I do after? I guess I just want to know what are the elements that commute with $\sigma$.
122751
It is known that any two disjoint cycles in $S_n$ commutes. Therefore, any $\pi\in S_n$ which is disjoint with $\sigma$ is in the centralizer of $\sigma$: $C_{S_n}(\sigma)$. Also  $$ \sigma^i\pi\in C_{S_n}(\sigma), 0\leq i\leq|\sigma|. \tag{*} $$ Here are my questions:  Are these all the elements in $C_{S_n}(\sigma)$? If the answer is no, is there an method to find all the elements in $C_{S_n}(\sigma)$ other than check the elements one by one?    A quick search returns several relevant questions:        The method ($*$) is what I learned so far from these questions and their answers. An immediate example in which ($*$) does not work that I can come up with is $\sigma=(12)(34)(56)$ in $S_6$. No disjoint $\pi$ can be found in this case and  $$\langle(12),(34),(56)\rangle\subset C_{S_6}(\sigma).$$
55173
There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that  \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post  I don't see that...  My approach to this is:  Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing  $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.  But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...  I would be really thankful, if someone could give me quite a basic proof that even I understand...
74874
If $N$ is a normal subgroup of $G$ and $H$ a characteristic subgroup of $N$ (i.e., $\sigma(H)=H$ for all $\sigma \in \operatorname{Aut}(N)$). Show that $H$ is normal in $G$.     Does one needs to show that $\sigma(gHg^{-1})=gHg^{-1}$ for every g in G. Would that imply H is normal? is there a alternative way?
62887
I am studying Abstract Algebra on my own so I don't know if there are gaps in my knowledge. I didn't find any example to solve the following equation in my reference textbook. Find ${\tau\in S_7, \text{such that } \sigma'=\tau\sigma\tau^{-1} \text{ where } \sigma=(125)(3674), \sigma'=(314)(2765)}$ I tried the following. Right multiplication by ${\tau}$ and left multiplication by ${\sigma'^{-1}}$ gives ${\tau=\sigma'^{-1}\tau\sigma}$ ${\implies \tau=(341)(2765)^3\tau(125)(3674)}$ I tried the following approach. $\tau(1)=(341)(2765)^3\tau(\sigma(1))=(341)(2765^3)\tau(2)$ $\tau(2)=(341)(2765)^3\tau(\sigma(2))=(341)(2765)^3\tau(5)$ $\tau(5)=(341)(2765)^3\tau(\sigma(5))=(341)(2765)^3\tau(1)$ If I back substitute the values, I get $\tau(2)=(341)^2(2765)^6\tau(1) \therefore\tau(1)=(2765)\tau(1)$ Is this approach correct? Am I close? How do I proceed forward?   One observation I made was if $\tau(1)\in\{2,5,6,7\}, $ then the equation cannot be satisfied. Therefore, $\tau(1)\in\{3,4,1\}$. Is this correct?
2141
Find the center of the symmetry group $S_n$.   Attempt:  By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.  So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes.  $rp(i) = rj$    Can someone please help me, I am stuck?  Thank you.
117839
Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well.  Let $|G| = p^2$ for some prime $p$.  Consider $x \in G$. So, $|x| = p, $ or $ p^2$. If it is the latter, then, we are done.  So, if $|x|= p$, then, let, $H = [h|  h=x^i, i \in Z] \Rightarrow |H|= p$.  As $p$ is the smallest prime dividing $|G|$, $H$ is a normal subgroup.  Now, consider some $y \in G$, but not in $H \Rightarrow y^p= e$. Then, its easy to show that $G = \langle x,y \rangle$.  Now, consider a homomorphism $f:G \rightarrow \operatorname{Aut}(H), g \to c_g $, wherein $c_g$ represents conjugacy by $g$. So, for example, $c_g(x) = gx g^{-1}$.  Then, it is clear that, $H \subset \operatorname{Ker}(f) $, as it is cyclic and therefore abelian. So, we need only look at $G/H$.  Basically, I am asking for a hint as to why $c_y(x) = yxy^{-1} = x = c_e(x)$, where $x$ and $y$ are the generators of $H$ and $G/H$, respectively?  Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that $\operatorname{Im}(f)$ is trivial and therefore that $G$ is abelian.  Thank You!
12378
$\mathcal{C}_1$, $\mathcal{C}_2$ are collections of subsets of $X$,then Im having hard time seeing why the  following is true. Can someone explain them to me?  $\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $  $\sigma(\sigma( \mathcal{C} )) = \sigma( \mathcal{C} ) $  where $\sigma( \mathcal{F} ) $ is the sigma field generated by the collection $\mathcal{F}$ of subsets of $X$  thanks
120083
Which of the following are true?   $\sigma \circ \sigma(j)=j~\forall j, 1 \leq j \leq 5$. $\sigma^{-1}(j)=\sigma(j)~\forall j, 1 \leq j \leq 5$. The set $ \{k: \sigma(k) \neq k \}$ has even number of elements. The set $\{k:\sigma(k) =k \}$ has an odd number of elements .    Can someone tell me to solve it in $1$ or $2$ min....trick..or some concept behind it?
19105
Given a collection of sets $\mathcal{C}$ and $E$ an element in the $\sigma$-algebra generated by $\mathcal{C}$, how do I show that $\exists$ a countable subcollection $\mathcal{C_0} \subset \mathcal{C}$ such that $E$ is an element of the $\sigma$-algebra, $\mathcal{A}$ generated by $\mathcal{C_0}$?    The hint says to let $H$ be the union of all $\sigma$-algebras generated by countable subsets of $\mathcal{C}$....although I don't know why.
109739
Question 2.3.20 in Qing Liu's Algebraic Geometry and Arithmetic Curves:  Let $A$ be a ring, $G$ a finite group of automorphisms of $A$. Let  $$    p: \mathrm{Spec}\ A \to \mathrm{Spec}\ A^{G} $$  be the morphism of affine schemes induced by the obvious inclusion.     Show that, for $x_{1}$, $x_{2} \in \mathrm{Spec}\ A$,    $$ p(x_{1}) = p(x_{2}) \iff \exists \ \sigma \in G : \sigma(x_{1}) = x_{2}.$$   The '$\impliedby$' direction is easy, but I'm having trouble with the other direction (although I have no doubt that it is also simple).   My attempts so far have included assuming that there is no $\sigma \in G$ such that $\sigma(x_{1}) = x_{2}$, so that for each $\sigma$ there is $g_{\sigma} \in x_{1}$ with $\sigma(g_{\sigma}) \notin x_{2}$, and then trying to construct an element of $p(x_{1})\backslash p(x_{2})$ out of these $g_{\sigma}$, for example by taking $$   g = \prod_{\sigma \in G}g_{\sigma} $$ and then  $$   h = \prod_{\sigma}\sigma(g) $$ but this appears to fail as I can't ensure that $\sigma(g_{\tau}) \notin p(x_{2})$ for $\sigma \neq \tau$.  Any hints would be much appreciated!
118342
Let $\sigma = (a_1\ a_2) \ \ and \ \  \tau = (a_1\ b_2\ \ldots\ b_p)$. (We have $a_2 = b_i$ for some i.).We know that $S_p$ is generated by $\{ (a_1\ a_2) \ \ and \ \  (a_1\ a_2\ \ldots\ a_p) \}$. So  I want to show that $\tau^k(a_1) = a_2$ for some k. Please help me.  Thank you
27938
Let $G$ be a group, and $H$ a subgroup.  $H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$.   Prove $N(H)$ is a normal subgroup of G, or give counterexample.  Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.  Thans in advance for any assistance!
80822
Let $S=\{\{x\}\mid x \in \mathbb{R}\}$. Is $\sigma(S)$ included in the Borel $\sigma$-algebra on $\mathbb{R}$? Is $\sigma(S)$ equivalent to the Borel $\sigma$-algbrea on $\mathbb{R}$?    I think the answer to the first question is yes. Because the Borel sets include singletons, the Borel $\sigma$-algebra must contain the smallest sigma-algebra of the set of all singletons.   Here's my attempt at the second question: my answer would be no. Countable unions (or intersections) of countable sets are countable, so that any $X \in \sigma(S)$ is either countable or a compliment of a countable set. Therefore, the interval $(0,1) \not \in \sigma (S)$ but it is included in the Borel $\sigma$-algebra. My issue here is that I don't know how to formalize this statement, if it is even correct.
119060
For which $n≥2$ does there exist $\sigma\in A_n$ with $|\sigma| &gt; n$? I have a proof that I'd like to be verified. My proof that there exists $\sigma\in A_n$ with $|\sigma| &gt; n$ if and only if $n\geq 8$: For $n&lt;8$, one can manually verify for each $n$ that there is no $\sigma\in A_n$ with order greater than $n$. Suppose $n\geq 8$. Case 1: $n$ is even and $3\not\mid n$. Then the permutation $(1,2,3)(4,\dots,n)$ is even and has order $\operatorname{lcm}(3,n-3)=3n-9\geq n+16-9\geq n + 7$. Case 2: $n$ is even and $3\mid n$. Then $n\geq 12$, so $n-2\geq 8$ is even and is not divisible by 3. Applying Case 1, we see that $(1,2,3)(4,\dots,n-2)\in A_n$ has order at least $n+5$. Case 3: $n$ is odd. Then apply one of the previous two cases with $n-1$ to get an even permutation with order at least $n+4$. $\blacksquare$ Is my proof correct? It doesn't seem very elegant. Is there some slicker proof?
112043
Suppose I fix $n$ and let $\sigma_k$ represent the $k$th permutation of $S_n$ with respect to  some ordering (whatever ordering might serve my purpose). Is there an elementary formula for $\sigma_k(i)$ which requires only $i, k,$ and $n$?  Is one known for small $n$, perhaps even as small as 4?
10037
Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See   For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.  In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$  My question is How can we differentiate between the two orbits? Does it have geometric meaning?    I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$...
67510
So I am struggling to understand the definition of a coset. If I have the following symmetric group $S_3=\{1, \sigma, \sigma\tau, \sigma\tau^2, \tau, \tau^2\}$, where $\sigma$=$\left(\begin{array}{ccc}1 &amp; 2 &amp; 3 \\ 1 &amp; 3 &amp; 2\end{array}\right)$ and $\tau$=$\left(\begin{array}{ccc}1 &amp; 2 &amp; 3 \\ 2 &amp; 3 &amp; 1 \end{array}\right)$.  I am given the subgroups $H=${$1,\sigma$} and $K=${$1,\tau,\tau^2$} How would I find all the left cosets of these subgroups? Thanks