# Tag Info

1

As has been noted in the comments, there is no such formula. The order of $ab$ depends on the common elements and their positions in the two cycles.

1

Upon taking a power of a permutation, a $k$-cycle can only arise from an $m$-cycle when $k|m$. Since $m$ is limited to $7$ in $S_7$, the $4$-cycle must have arisen from a $4$-cycle. That leaves $3$ other elements, which must be in cycles whose length divides $3$, since otherwise the cycles would survive the cubing. The only possibilities are one $3$-cycle or ...

6

Take two different transpositions, for example $\alpha=(1\ 2)$ and $\beta=(1\ 3)$.

6

In order to show that $A_n ≤ [S_n,S_n]$, you just have to write every $3$-cycle as a commutator in $S_n$, because the $3$-cycles generate $A_n$, if $n \geq 3$ (if $n \leq 2$, then this is trivial). This is true because $(a \; b \; c) = (a\; c\; b)^2=(a\; c)(c\; b)(a\; c)(c\; b)$.

1

Enumerating the elements of $G$ we get $$\{e,(34),(45),(35),(345),(543)\}\cup \{(12),(12)(34),(12)(45),(12)(35),(12)(345),(12)(543)\}$$ Verify that the subgroup generated by $(34)$ and $(45)$ is normal, and the subgroup generated by $(12)$ is normal. Obviously they intersect trivially, and the suggestively written union above shows that they generate the ...

1

One can define various structures such as graphs of valency $k$ (for some fixed $k \ge 3$), bipartite graphs, strongly regular graphs, $k$-chromatic graphs (for fixed $k \ge 2$), or $k$-connected graphs (fixed $k \ge 1$). A structure $\mathcal{C}$ is said to be universal if every finite group is the automorphism group of some graph in $\mathcal{C}$. Each ...

2

In general, suppose that $F$ is a field and $G$ is a group that can be expressed as a direct product $G=H\times K$. Let $\rho$ and $\sigma$ be representations of $H$ and $K$ over $F$, respectively. Then a corresponding representation of $G$ over $F$ may be constructed from $\rho$ and $\sigma$ by using tensor products. Suppose that $\rho$ and $\sigma$ ...

0

Answering my own question here... You can't have an infinite planar graph with the properties I describe unless they're chosen carefully. You can fill the plane with an infinite number of complexes such as a square face with a vertex at one corner and two edges adjacent to that vertex. Put a bunch side by side and you see the vertices and edges get shared ...

0

For $n>4$: a subgroup of index $2$ is normal. Show any (non-trivial) normal subgroup $H$ (non-trivially) intersects $A_n$. Then $H \cap A_n$ is normal in $A_n$ and by simplicity of $A_n$ the claim follows. For $n=4$ you can look at the subgroups directly.

0

It does not give proofs, but given the tone of the original question a good graphical tool for the OP would be Group Explorer. It does a lot of the donkey work, and can show you various visualisations, including the multiplication tables in helpful ways. http://groupexplorer.sourceforge.net/ For $D_6$, that $Z_2$ is the center is (to me a least) kinda ...

2

We now treat the case of an alphabet of $k$ letters rather than a binary alphabet. We use two classes of letters, the pattern being represented by $WWY_0$ and an additional sequence of letters from $Y_1$ to $Y_{k-2}.$ In this way we obtain $k$ letters total. Now there are several cases, the easiest is if $W$ does not ocur at all. These are ...

2

In the present case (binary necklace, forbidden pattern $110$) we have a simple observation (which does not generalize). This is if we divide the necklace into adjacent segments consisting of repetitions of one and the same symbol we cannot have a run of two or more ones since these would form the pattern $110$ with the zero following the run of ones. ...

2

For even $n\gt2$ the alternating group $A_n$ is transitive but contains no $n$-cycle.

4

If $p$ is prime, then every transitive subgroup $G\leq S_p$ contains a $p$-cycle. This is because the order of $G$ is divisible by $p$ by the orbit-stabilizer theorem, hence $G$ contains an element of order $p$ by Cauchy's theorem, and this element must be a $p$-cycle. To see that this is not the case for general $n$, consider the subgroup $$\{1,(12)(34),(... 0 Here is another answer using the standard action of S_4 on the set \{1,2,3,4\}. We use the following lemma (proof below) Lemma: If (C_2)^3 acts on a set X with four elements, then there is at least one nontrivial element of (C_2)^3 which acts trivially. If (C_2)^3 were a subgroup of S_4, then (C_2)^3 would act on \{1,2,3,4\} by ... 1 You are almost there! Let S be a subgroup of S_4 of order 8. The orbit of S under the conjugation of S_4 has size 24/|Stab(S)|. Now note that the stabilizer of S for this action is N(S), the normalizer of S in S_4 (i.e., the biggest subgroup of S_4 in which S is normal). In particular N(S) has size 8 or 24 since it contains S, ... 1 If A is symmetric and negative definite (i.e., all its eigenvalues are negative), then -A is symmetric and positive definite (i.e., all its eigenvalues are positive). From the symmetry of -A, we conclude that it is diagonalizable, that its eigenvalues are real, and that its eigenvectors are orthogonal. Hence, -A has an eigendecomposition -A = Q \... 3 Easily we see that -A is symmetric with positive eigenvalues and by spectral theorem it's diagonalizable in orthonormal basis: there is an orthogonal matrix P and \lambda_1,\ldots,\lambda_n>0 such that$$-A=P DP^T$$where D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). Let \Delta=\operatorname{diag}(\sqrt\lambda_1,\ldots,\sqrt\lambda_n) then ... 0 (abc) maps a to b, b to c, and c to a. Now, all we need to do is map c to a (so b \mapsto c \mapsto a), a to d (c \mapsto a \mapsto d), and d to c — i.e., (cad). Thus, (ab)(cd) = (cad)(abc). 2 Try (ab)(cd) = (ab)(bc)(bc)(cd)= (abc)(bcd). 1 \;\gamma^{-1}\; will be the permutation mapping each cycle of \;\alpha\; to a corresponding cycle of \;\beta\; of the same type, thus for example:$$\gamma^{-1}:=\begin{cases}1\to2\\2\to3\\5\to4\\.......\\4\to1\\6\to5\\.......\\3\to6\end{cases}\implies\gamma=(123654)$$and we can now check that$$\gamma^{-1}\alpha\gamma=(123654)(125)(46)(3)(145632)=(...

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 ...

Top 50 recent answers are included