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1

For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$. For $n=5$, the eigenvalues are the roots of $(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - ...


0

Hint: Since $h$ is continuous, you simply have to show that it stays bounded as $y$ tends to infinity (uniformly in $x$ -- here $z=x+iy$). Consider the expansion $f(z) = \sum_{n\geq 1} a_n {\rm e}^{2\pi i n z}$, which starts at $n=1$ since $f$ is cuspidal. As $y\to\infty$, what happens for each terms in the series ? How fast do they decay ?


1

We have $$x^3 = y^2 = (xy)^2 = 1,$$ so $$yx = (yx)^{-1} = x^2y,\text{ }yx^2 = x^2yx = x^4y = xy.$$ a. Fix a nonzero $u\in V$, and let $$v = u+xu+x^2u = (1+x+x^2)u,\text{ }w=u+yu = (1+y)u.$$ Since $\{1,x,x^2\},\{1,y\}$ are subgroups of $G = S_3$, we have $xv = x^2v = v$ and $yw = w$. We have the operator identities $$y(1+x+x^2) = y+yx+yx^2 = y+x^2y+xy = ...


3

Let $G = S_n$ and let $H$ be the stabilizer of $1$ in $G$ $($i.e., the permutations $\sigma$ so that $\sigma(1) = 1)$; thus $H$ is isomorphic to $S_{n-1}$. We show that any $G$=representations with an $H$-fixed vector is either isomorphic to the trivial representation or to the representation on $\{(x_1, \dots, x_n)\text{ }|\text{ }\sum x_i = 0\}$. Let ...


1

This is one of the places where generalizing makes structure more visible. If you are willing to consider $S_n$ for all $n$ at once, i.e., study $$ \coprod_n BS_n, $$ then the cohomology algebra has two products making it a Hopf ring (a ring object in the category of coalgebras). Giusti, Salvatore, and Sinha studied this and include explicit rules for ...


1

Is this a counterexample? # sage def SGA(n): return SymmetricGroupAlgebra(QQ, n) def j(a, b): SGA = a.parent() res = SGA.zero() for m, c in a: for mm, cc in b: lap = m.left_action_product(mm) if max(lap.cycle_type()) <= 2: res += c * cc * SGA.basis()[lap] return res SGA4 = ...


1

$B$ is a $7-$ cycle hence $B^7=I\implies B^8=B$ Now from $B^4$ find $B^8$


2

A Sylow 2-subgroup contains at most 3 elements of order 2 because it is of order 4. If there were only 3 such subgroups, there would be at most 9 elements of order 2. Since there are 15, 3 is excluded. $N_G(H)$ is the normalizer of $H$ in $G$, meaning the set of all elements $g$ such that $gHg^{-1}=H$. $G$ acts on the set of 2 Sylow subgroups transitively ...


2

Write $G = S_n$. Recall that $\mathbb{C}^n = P \oplus \text{span}(1, 1, \dots, 1)$. Let $\phi: \mathbb{C}^n \to P$ be the projection and $e_1, \dots, e_n$ the standard basis for $\mathbb{C}^n$. $($Therefore, $\phi$ is given explicitly as $\phi(x_1, \dots, x_n) = (x_1 - \overline{x}, \dots, x_n - \overline{x})$ where $\overline{x} = (\sum x_i)/n$.$)$ Let $f_i ...


1

These questions can be answered with a little bit of character theory. If you can work out the character of $P$, then the character of $\wedge^2 P$ is not so bad. For instance look at the bottom of the second answer of this question. Symmetric and exterior power of representation Once you can compute the character of the relevant representations, finding ...


1

Hint: Try breaking $X \times X$ into $A:=\{(x,y) \in X \times X:x=y \}$ and $B=\{(x,y) \in X \times X:x \neq y \}$.


1

Here is a way to construct an homomorphism from $S_4$ to $S_3$, which has kernel $H$. Consider the three possible partitions of the set $\{1,2,3,4\}$ into two sets: $$A=\{1,2\}\cup\{3,4\}; \quad B=\{1,3\}\cup \{2,4\}; \quad C=\{1,4\} \cup \{2,3\}.$$ Now you can see what does an element $p\in S_4$ to those partitions. For instance the element $p := (1234)$ ...


0

I wrote this as comment, but it probably looks better as answer. In general, given a group $G$ and a subgroup $H\leq G$, for $x\in G$, we have $x^G\cap H=\bigcup_{i=1}^n x_i^H$, for some $x_1,...,x_n\in H$ and $n\leq |G:H|$. Notice that this intersection may be empty, in which case $x$ is called derangement (a derangement always exists). In your case, ...


1

In $S_n$, conjugacy classes are determined by cycle type. For example, the conjugacy class $C$ of $(123)$ in $S_4$ consists of all 3-cycles. The $S_4$-conjugacy class $C$ is contained in $A_4$, where it becomes the union of two $A_4$-conjugacy classes. These are $$ \{ (123), (421), (243), (341) \} $$ and $$ \{ (132), (412), (234), (314) \} . $$ Notice that ...


0

Consider the sign homomorphism $\varphi:S_n\rightarrow \{\pm 1\}$, which maps even permutations to 1 and odd permutations to -1. Notice that $\ker \varphi = A_n$, and since $\varphi$ is surjective, \begin{align*} \lvert \ker \varphi \lvert = \frac{\lvert S_n \lvert}{\lvert \{\pm 1\}\lvert}=\frac{n!}{2}. \end{align*}


2

There is an old result of Jordan that a transitive subgroup of $S_n$ that contains an element of prime order $p$ with $n/2 < p < n-2$ is equal to $A_n$ or $S_n$. This applies to groups satisfying your hypothesis provided that there exists such a prime, which is true for $n \ge 8$.


3

(i) is correct. And for the second one you just see from the co-domain to domain.. that is, \begin{align} (P_1\circ P_2)^{-1}&=\begin{pmatrix}1&2&3&4&5&6\\ 2&4&5&1&6&3\end{pmatrix}^{-1}\\ &=\begin{pmatrix} 2&4&5&1&6&3\\ 1&2&3&4&5&6\end{pmatrix}\\ &=\begin{pmatrix} ...



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