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Can you prove that every element of $S_n$ is equal to a product of transpositions? If so, you just need to show that each of those generating sets contains all transpositions. Edit: To make this more direct, this task becomes straight forward once you understand how conjugation works in $S_n$.

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Not sure how to "convince" you, but I suspect you are worried because some of the 2-cycles do the job and others do not. The trick is in picking the right 4-cycle to go with your 2-cycle or vice versa. (1,2) works with (1,2,3,4) because 1 and 2 are adjacent in (1,2,3,4) whereas 1,3 is not. But (1,3) will work with (1,3,2,4). For more info: ...

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Just to clarify a few basic points. The full symmetric group $S$ on a countably infinite set $X$ is indeed well-defined up to group isomorphism. (The same applies to sets $X$ with any given cardinality.) $S_*$ is indeed a normal subgroup of $S$. It is easy to see that $g^{-1}hg \in S_*$ for all $g \in S$ and $h \in S_*$. In fact it can be shown that $S$ ...

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Check out this paper by Alperin/Covington/Macpherson and references there in (seems to be free from citeseer). They analyze the automorophism group of your mystery quotient, which should tell you a fair bit about the structure.

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Let $(x_1x_2x_3)$ and $(y_1y_2y_3)$ represent the two $3$-cycles. Assume that $x_4,x_5$, respectively $y_4,y_5$ are the remaining two symbols in the five-element set $S_5$ operates on. There is a permutation $\tau$ sending $x_i$ to $\tau x_i=y_i$ for $i=1,...,5$. This will be either a permutation with positive sign or with negative sign. But if it's of ...

3

Yes, the theorem works, but that would give the mapping $\gamma=(1\mapsto 3,\ 2\mapsto 5,\ 3\mapsto 2,\ 4\mapsto 4,\ 5\mapsto 1)$. (Can you write its cycle structure?) Edit: I found out that $\sigma^2$ is rather $(35241)$. Actually, there can be other solutions as well, if you rotate the written line of the cycle, namely if you write $\sigma^2$ as e.g. $(1\ ... 2 Besides to @Betty's points, there is another way for seeing why does this happen. We know that$S_4$can have the following presentation: $$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Let's satisfy$a=(1,2),~~b=(1,2,3,4)$in above relations. Indeed$a$and$b$can do that, but what will happen if we set$a=(1,3),~~b=(1,2,3,4)$? By this assumption, we see ... 2 Note that if$\ker(f) \neq \mathbb{Z}$, then$f(1) \neq e$($e$referring to the identity of$S_3$). Why? With the above established, we have two possibilities: either the order of$f(1)$is$2$, or the order of$f(1)$is$3$. In the first case, we have$f(2k) = f(2)^k = e^k = e$, so that$\ker(f) = 2\mathbb{Z}$. Similarly, in the second case, we have ... 2 Hint Let$X \subseteq S_n$be the set of all permutations having$1$and$2$in separate cycles and$Y \subseteq S_n$the set of all permutations having$1$and$2$in the same cycle. Show that the mapping$f : X \to Y$which "glues together" the two cycles containing$1$and$2$(cycles written down such that they start with$1$and$2$, resp.) is a ... 2 The length of the first row of a Plancherel-random Young diagram (with$n$boxes) has the same distribution as the longest increasing subsequence of a random permutation$\pi$in the symmetric group$S_n$(hint: use Robinson-Schensted correspondence). By Markov's inequality, the probability that$\pi$contains a decreasing sequence of length at least ... 2 We will use Pieri's formula which is a special case of the Littlewood-Richardson-rule. It states: Let$a+b = n$be natural numbers and$\alpha \vdash a$a partition of$a$. Then the induced representation$\mathrm{Ind}_{S_a\times S_b}^{S_n} V_\alpha \otimes V_{(b)}$(where$V_\alpha$is the irreducible representation corresponding to$\alpha$and ... 1 These permutations are called involutions. The counting function for involutions on$n$elements is documented at OEIS here. You'll be able to find explicit formulas, recurrence relations, asymptotics, and generating functions there, along with some references. OEIS (Online Encyclopedia of Integer Sequences) is a pretty nice resource in general for ... 1 A subgroup$G$of$E(n)$is called discrete if for each point$x\in \mathbb{R}^n$the family$\{g.x\mid g\in G\}$is locally finite. Since this family contains each point with multiplicity equal to the order of the stabiliser$G_x$of$x$, the family is locally finite if and only if the orbit$Gx$is discrete and the stabilizer$G_x\$ is finite. This shows ...

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Every dihedral group is generated by two elements of order 2. And every group generated by two elements of order 2 is dihedral. Furthermore the order of the generated dihedral group is the order of the product of the two elements. So if x,y each have order 2, then is dihedral with 2*n elements where n is the order of x*y. This is why dihedral groups ...

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