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6

The kernel of a surjective homomorphism from $S_5$ to $S_4$ would have order $|S_5|/|S_4|=5.$ This is impossible because: $S_5$ has $1+4!=25$ elements of order $1$ or $5$; the image of each of those $25$ elements must have order $1$ or $5$ in $S_4$; but $S_4$ has no elements of order $5,$ so those $25$ elements must all belong to the kernel of the ...


6

The possible candidates for such an $H$ are the subgroups of $S_5$ that are cyclic of order 5. All elements of $S_5$ of order 5 are given by $5$-cycles. However, the subgroup generated by a 5-cycle is not normal, so no $H$ can exist, as desired.


5

Hint, as requested: the order of an element $\sigma$ in $\mathfrak{S}_n$ is the least common multiple of the cycle lengths of $\sigma$.


4

There are indeed $\frac156!=144$ different $5$-cycles in $S_6$ but note that a single $5$-Sylow accounts for 4 of them. Indeed $c$, $c^2$, $c^3$, $c^4$ are the 4 different non-trivial elements in $\langle c\rangle$. Thus the total number of $5$-Sylows is $\frac14144=36$.


3

Easily we see that $-A$ is symmetric with positive eigenvalues and by spectral theorem it's diagonalizable in orthonormal basis: there is an orthogonal matrix $P$ and $\lambda_1,\ldots,\lambda_n>0$ such that $$-A=P DP^T$$ where $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Let $\Delta=\operatorname{diag}(\sqrt\lambda_1,\ldots,\sqrt\lambda_n)$ then ...


3

You have $f:H \to H/K$ defined by $f(h)=hK$ is a homomorphism (the natural projection). Since $H/K$ is abelian, the image of $f$ is abelian and thus $H' \subset ker(f)$. But $ker(f)=K$ (because $f(h)=hK=K$ iff $h \in K$). Now consider a 3-cycle $(a_1a_2a_3)$. Since $n \geq 5$, there are some numbers $a_4$ and $a_5$ disjoint from $a_1,a_2,a_3$. Note that if ...


3

It sounds like you're describing the quotient space $\mathbb{R}^n/S_n$. Some people call this the $n^{th}$ "symmetric power" (of $\mathbb{R}$), although be a little careful with that terminology because it can be used to refer to two other related but different constructions. This quotient is not a manifold, but can be thought of as an orbifold. For ...


2

Suppose $\sigma: H\to\mathrm{GL}(W)$ is a complex irreducible representation of $H$. Call the trace $\chi$. Then character theory says that $\sum_{h\in H}\chi(h)$ equals $|H|$ if $\sigma$ is trivial and $0$ otherwise. Additionally, the sum $\sum_{h\in H}\sigma(h)$ is an interwtining operator, and Schur's lemma says the intertwining operators of a complex ...


2

With $q$ the total number of elements on the cycle that contains $1,2,\ldots,k$ we get the formula $$\sum_{q=k}^n \frac{q!}{q} {n-k\choose q-k} (n-q)!$$ which says that there are $(q-1)!$ cycles on the $q$ elements, we must choose the remaining $q-k$ elements for the cycle from the elements other than the first $k$ and combine this with any ...


2

Here’s what Ian Stewart has to say on p. 268 of his book Galois Theory*: The transitive subgroups [of $S_n$], up to conjugacy, have been classified for low values of $n$ by Conway, Hulpke, and MacKay (1998). … There is only one such subgroup when $n=2$, two when $n=3$, and five when $n=4,5$. The magnitude of the task becomes apparent when $n=6$: in this ...


2

Let $\gamma$ be the function that takes a number in the cycle of $\beta$ to the number in $\alpha$ in the corresponding position. That is, send $2 \rightarrow 1$, and $3 \rightarrow 2$, and $4 \rightarrow 5$ et cetera. Intuitively you can think of this as as you temporarily relabelling the numbers in $\beta$ (applying $\gamma$), applying the permutation ...


2

Note that $S_6$ is the symmetric group of $6$ symbols which consists of $6!$ element. We can use the theorem that two permutations are conjugate iff they have the same cycle type. As you have already simplified the permutations into the product of disjoint cycles, it is easy to see that the two permutations have the same cycle type. Fortunately, using the ...


1

If $A$ is symmetric and negative definite (i.e., all its eigenvalues are negative), then $-A$ is symmetric and positive definite (i.e., all its eigenvalues are positive). From the symmetry of $-A$, we conclude that it is diagonalizable, that its eigenvalues are real, and that its eigenvectors are orthogonal. Hence, $-A$ has an eigendecomposition $-A = Q ...


1

$\;\gamma^{-1}\;$ will be the permutation mapping each cycle of $\;\alpha\;$ to a corresponding cycle of $\;\beta\;$ of the same type, thus for example: $$\gamma^{-1}:=\begin{cases}1\to2\\2\to3\\5\to4\\.......\\4\to1\\6\to5\\.......\\3\to6\end{cases}\implies\gamma=(123654)$$ and we can now check that ...


1

One solution is to rewrite $\gamma\alpha\gamma^{-1}=\beta$ as $\gamma\alpha=\beta\gamma$. Then $\gamma\alpha(1)=\beta\gamma(1)$, and from $\alpha(1)=2$ we get $\gamma(2)=\beta\gamma(1)$. Similarly, $\gamma(5)=\beta\gamma(2)$ and $\gamma(1)=\beta\gamma(5)$. Combining these, we get $\gamma(2)=\beta\gamma(1)=\beta\beta\beta\gamma(2)$, so $\gamma(2)$ has ...


1

Let's go over how these representations are defined: We have a certain partition of the set $\{1,2,\ldots,n\}$; this partition is into an ordered collection of unordered sets. Such a partition might look like $[\{1,3\}, \{2,4\}]$; note this is exactly the same partition as $[\{3,1\}, \{2,4\}]$, but not the same as $[\{2,4\}, \{1,3\}]$. The problem with ...


1

$\sigma\cdot(2,3)=\sigma(2),\sigma(3))$ holds for all $\sigma\in S_3$. So for example if $\sigma=(1\,2\,3)$, then $\sigma(2)=3$ and $\sigma(3)=1$, so $\sigma\cdot(2,3)=(3,1)$.


1

Let $G=S_4,\ H=S_3$ where $H$ is regarded as pemutations of $\{1,2,3\}$ and so is a subset (and a subgroup) of $G$ regarded as permutations of $\{1,2,3,4\}.$ Then none of $(14),(24),(34)$ are in $H,$ so we can form the three possibly distinct right cosets $H(14),\ H(24),\ H(34)$ and none of these are $H.$ If we can show any two of these are in fact ...


1

We have $\sigma(1) = 2$ and $\sigma(2) = 3$, so the first factor $(x_1-x_2)$ becomes $(x_{\sigma(1)}-x_{\sigma(2)}) = (x_2 - x_3)$. Similarly for the rest. Does that clear things up?


1

If $\;n_p\;$ denotes the number of Syloy $\;p\,-$ subgroups, then we know that $$n_p=[S_p:N_G(P)]=\frac{p!}{|N_G(P)|}$$ and we know that $\;n_p=(p-2)!\;$


1

Found it! http://sheaves.github.io/Subgroup-Lattice/ This shows how to build the lattice in Sage, and at the very end of the third post there is a complete generator. (You can use Sage freely on websites like sagemath.org, or installing it on your computer.)


1

Every permutation of $X$ that leaves $Y$ invariant as a set is really composed of two distinct, independent and non-overlapping permutations: one is a permutation of $Y$, and the other is a permutation of $X \setminus Y$. Indeed, if a permutation of $X$ maps all the elements of $Y$ back into $Y$, then the restriction of the original permutation (on $X$) to ...


1

It's important that on the first page the paper states that, without loss of generality, all of their irreducible representations are unitary, so that $\rho(g)^\dagger=\rho(g)^{-1}$ for all $g\in G$. By the way, the beginning of section $3$ is in error. The sum $\sum_{h\in H}\rho(h)$ is not a projection, because it is not normalized correctly. Instead, ...


1

$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ is just the definition of the Frobenius norm of the matrix $\hat{f}(\rho)$.


1

What you call "exponentiation" means conjugacy: $\phi$ is a homomorphism $G\rightarrow Aut(V)$ for some vector space $V$; an (invertible) matrix is an automorphism of $V$ (once you choose a basis), and $\phi^A$ is the representation $x\mapsto A\phi(x)A^{-1}$ (or replace $A$ with $A^{-1}$ depending on the author's conventions). In other words, it's just the ...



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