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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),(...


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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