Tag Info

New answers tagged symmetric-groups

0

$(abc)$ maps $a$ to $b$, $b$ to $c$, and $c$ to $a$. Now, all we need to do is map $c$ to $a$ (so $b \mapsto c \mapsto a$), $a$ to $d$ ($c \mapsto a \mapsto d$), and $d$ to $c$ — i.e., $(cad)$. Thus, $(ab)(cd) = (cad)(abc)$.

1

Try $(ab)(cd) = (ab)(bc)(bc)(cd)= (abc)(bcd)$.

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 ...

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 ...

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 ...

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

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 ...

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 $\frac14144=36$.

0

I know this post is old, but there's another elegant way to prove this - a subgroup of order 6 has index 2. We prove the following statement: Any subgroup of index 2 of a finite group must contain all elements of odd order. Let $G$ be finite and $H\subseteq G$ a subgroup of index 2. Any subgroup of index 2 is normal, so $G/H$ is a group and we write $\bar ... 0 Let$r\in[1\>..\>n-1]$be the length of the cycle containing the number$n$. Begin the listing of this cycle with$n$. There are$(n-1)(n-2)\cdots(n-r+1)$ways to choose the remaining entries of this cycle. Now there will be$n-r\geq1$numbers left over, one of them the largest. Begin the listing of the second cycle with this largest left-over number, ... 0 From the possible answers, it appears that the question intended to ask how many permutations have exactly two cycles in their cycle decomposition, including the$1$-cycles. If the smaller cycle has$k$elements, the greater has$n-k$, and$j$particular elements can form$(j-1)!$different$j$-cycles. Thus the desired count is $$... 0 Here is another proof, that actually proves a bit more: Let G,H be two non-abelian groups of order 6. Then G \cong H. First off, we seek to show each group has an element of order 3, and an element of order 2. Since both are non-abelian, we don't have any elements in either of order 6, for such an element would then generate the entire group, ... 0 There are four things you need when showing an isomorphism: \bullet Define a function \phi: A\rightarrow B. \bullet Prove that \phi is injective. \bullet Prove that \phi is surjective. \bullet Show that the homomorphism property \phi(xy)=\phi(x)\phi(y) holds. Generators do map to generators through isomorphism, but that is result of ... 0 D_3 can be embedded in S_3 by sending r to (123) and s to (23) and then extend homomorphically. You can see this intuitively by noticing that any element of D_3 is completely determined by its action on the three points of the triangle, and every element of D_3 then induces a bijection from this set of points to itself, label the points 1, ... 0 With a group this small you can almost get away with writing down an isomorphism explicitly as a table and verify by inspection that it is a homomorphism, and is injective and surjective. The only trouble is that there is 36 different products to check for the homomorphism condition f(ab)=f(a)f(b), and that's at the upper limit of where one would like to ... 0 A standard way to prove that these two sets are isomorphic is to prove that they satisfy the same defining relations. For this particular example, one can show without too much difficulty (i.e. just write out the full multiplication table) show that any group of order 6 such that there exist elements a and b where: order of a is 3, order of b is 2, ... 0 For H \subseteq G, if V is an irreducible representation of H and W is an irreducible representation of G, then Frobenius reciprocity says that$$ \mathrm{mult}_W(\mathrm{Ind}_H^G V) = \mathrm{mult}_V(\mathrm{Res}^G_HW). $$So we only need to determine the multiplicity of \mathrm{sgn}_{\mathfrak{S}_2} in the restrictions of the irreducible ... 0 Your mistake happened when you declared o(a)=9. Thats not true. It should read "|a| divides 9" ( sorry if you don't mind, I would like to use |a| notion instead of o(a) for order of a\in S_n ) This means |a|\in \{1,3,9\}. Let |a|=1. That means a=\varepsilon and hence (123)=a^3=\varepsilon^3=\varepsilon which is contradiction. Let ... 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 \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). 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 ... 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? 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 ... 0 Let H be normal subgroup of S_5 such that S_5/H is isomorphic to S_4. Then H=<\sigma> for some 5 cycle \sigma. Note that \tau(a_1\ a_2\ a_3\ ....\ a_k)\tau^{-1} = (\tau(a_1)\ \tau(a_2) ... \ \tau(a_k)). It is very useful find \tau \in S_5 s.t \tau H \tau^{-1} is not H 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. 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)|}$$and we know that \;n_p=(p-2)!\; 2 Here’s what Ian Stewart has to say on p. 268 of his book Galois Theory*: The transitive subgroups [of S_n], up to conjugacy, have been classified for low values of n by Conway, Hulpke, and MacKay (1998). … There is only one such subgroup when n=2, two when n=3, and five when n=4,5. The magnitude of the task becomes apparent when n=6: in this ... 0 Take \sigma\in S_n, we need to have \sigma^3(j)=j for all j\geq n. Take 4. We need to have \sigma^3(4)=4. Suppose \sigma(4)=4 then we are done. Suppose not, then we see that (4~l~m) belongs to decomposition of \sigma. Continuing this, way, neglecting elements which are mapped to itself, we are forced to have ... 1 Found it! http://sheaves.github.io/Subgroup-Lattice/ This shows how to build the lattice in Sage, and at the very end of the third post there is a complete generator. (You can use Sage freely on websites like sagemath.org, or installing it on your computer.) 0 The "standard" proofs of the simplicity of the finite classical groups like {\rm PSL}(n,q) make heavy use of permutation representations and the general theory of permutation groups. The final step in the proofs uses Iwasawa’s Theorem, which is criterion for a primitive permutation group to be simple. Of course the definition of these groups is in terms ... 1 Every permutation of X that leaves Y invariant as a set is really composed of two distinct, independent and non-overlapping permutations: one is a permutation of Y, and the other is a permutation of X \setminus Y. Indeed, if a permutation of X maps all the elements of Y back into Y, then the restriction of the original permutation (on X) to ... 5 Hint, as requested: the order of an element \sigma in \mathfrak{S}_n is the least common multiple of the cycle lengths of \sigma. 2 With q the total number of elements on the cycle that contains 1,2,\ldots,k we get the formula$$\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 ... 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 ... 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$\hat{f}(\rho)$. 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 ... 0 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 ... 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 ...

0

What you want to read is Serre's book on Linear Representation of Finite Groups. Section 8.2 of chapter 8 covers the case of a semi-direct product by an abelian group. For the more general case, you want to use the theory of induced characters in a more subtle way. This is not as trivial as it sounds though ; make sure you have the mathematical background to ...

Top 50 recent answers are included