# Tag Info

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 $(a \; b \; c) = (a\; c\; b)^2=(a\; c)(c\; b)(a\; c)(c\; b)$.

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 $$\{1,(12)(34),(... 3 Yes, your reasoning and answer are correct. 2 We now treat the case of an alphabet of k letters rather than a binary alphabet. We use two classes of letters, the pattern being represented by WWY_0 and an additional sequence of letters from Y_1 to Y_{k-2}. In this way we obtain k letters total. Now there are several cases, the easiest is if W does not ocur at all. These are ... 2 In the present case (binary necklace, forbidden pattern 110) we have a simple observation (which does not generalize). This is if we divide the necklace into adjacent segments consisting of repetitions of one and the same symbol we cannot have a run of two or more ones since these would form the pattern 110 with the zero following the run of ones. ... 2 For even n\gt2 the alternating group A_n is transitive but contains no n-cycle. 2 In general, suppose that F is a field and G is a group that can be expressed as a direct product G=H\times K. Let \rho and \sigma be representations of H and K over F, respectively. Then a corresponding representation of G over F may be constructed from \rho and \sigma by using tensor products. Suppose that \rho and \sigma ... 1 Yes, your solution is correct. The number of ways to choose the three fixed points is {8 \choose 3}, the number of ways to choose the transposition from the remaining 5 points is {5 \choose 2}, and the number of ways to form a 3-cycle from the remaining 3 points is (3-1)!=2. Thus, the number in question is {8 \choose 3} \cdot {5 \choose 2} \... 1 In |S_{20}|, the highest power of 7 which divides 20! is 7^2. So it is clear that the Sylow-7 subgroup of S_{20} is of order 7^2. Group of order 7^2 is either cyclic or isomorphic to Z_7\times Z_7. If it is cyclic, then S_{20} will have an element of order 49, and it should be product of disjoint cycles. Check, whether this is ... 1 Enumerating the elements of G we get$$\{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 ...

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

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

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 ...

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