New answers tagged

1

I can't say I've heard it called Matsumoto's theorem, but there are quite a few references having this information. A classic is Reflection Groups and Coxeter Groups by Humphreys, and a more recent one is Combinatorics of Coxeter groups by Bjorner and Brenti, which has quite a bit about $S_n$ (though most of the book works in a more general setting).


2

Suppose $n$ is an integer at least $2$. If $G_n$ is a subgroup of $S_{12}$, then it is a normal subgroup. In particular, it must be either $A_{12}$ or $S_{12}$. We cannot have $G_n=A_{12}$, because $n$ would have to be even and thus $G_n$ would contain a transposition. On the other hand, we have $G_n=S_{12}$ if and only if $n$ is a multiple of the exponent ...


2

$\;\left|S_4/N\right|=6\implies S_4/N=C_6\;\;or\;\;S_3\;$, because these two are the only groups of order six up to isomorphism. Yet the first option is possible iff $\;[S_4,S_4]=A_4\le N\;$ , so it actually is the second one.


1

The solution is correct, the number of permutations with a fixed length of the cycle containing 1 is $(n-1)!$. The length may be any number between $1$ and $n$, i.e. $n$ choices, so the total number of permutations is $n\times (n-1)!=n!$, as expected. Because this $(n-1)!$ is independent of $k$, we may say that all the lengths of the cycle including $1$ are ...


1

Try using a direct algorithmic proof instead. Sketch: Given any $\alpha \in S_n$, recall that we can write $\alpha$ as a product of transpositions. Hence, if we know how to express any $2$-cycle as a product of $(0,1)$ and $(0,1,\ldots, n-1)$, then we are done. Given any transposition $(a, b) \in S_n$ with $a < b$, notice that we can write it as a ...


1

We can write $$c_\lambda =\sum_{g\in P_\lambda ,g^\prime\in Q_\lambda}sgn(g^\prime )\,e_{gg^\prime}$$ It is enough to show that coefficient of $e_{id}$ (where $id$ is the identity) in the above sum is non zero; in fact, we will show that the coefficient is $1$. This will follow from the fact that $gg^\prime =id$, for $g\in P_\lambda , g^\prime\in ...


1

This is easy to do on a computer. For example in magma http://magma.maths.usyd.edu.au/calc/ The following code: for k in [3..6] do X:=Sym(k); for n in [k..2*k] do G:=Sym(n); for sub in Subgroups(G:OrderEqual:=Factorial(k)) do H:=sub`subgroup; if IsIsomorphic(H,X) then [k,n]; Order(Centraliser(G,H)); end if; end for; end for; end for; will give you all ...


0

Note : $n\geq 4$ as you observed correctly. If $H \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$, then all elements in $H$ has to have order $2$. So all the elements in $H$ are either $2$ cycles or product of $2$ Cycles. Now, note that if $H$ contains a $2$ -cycle, then it has to be disjoint. (Why?) Think of an example. What if $H$ ...


1

You've misunderstood the notation. Refer to Definition 1: $\text{Aut}(TREE)$ is being used by the authors to denote the set of isomorphism classes of automorphism groups of (presumably finite) trees. So the notation $\Psi \in \text{Aut}(TREE)$ means that $\Psi$ is the full automorphism group of a particular tree. In fact Theorem 2 is just a restatement of ...


1

There is no definition of an "abstract finite group"; the term has no precise mathematical meaning. Generally speaking, if a mathematician refers to an "abstract group", they probably just mean "group", and are including the word "abstract" for emphasis, to clarify that any group is allowed, rather than some particular special kind of group that they have ...


1

Yes. A little easier example may help. Consider the two finite groups: $\mathbb Z/4\mathbb Z$ - the integers modulo 4 $\{1,-1,i,-i\}$ - the set of 4th roots of 1 These "look" very different, right? The first involves addition and remainders; meanwhile, the second has us multiplying complex numbers. However, they have the same "structure." What do we ...


1

Your permutation group is imprimitive, preserving the partition $\{1,3\},\{2,4\}$. This means that every element in this group sends each part into another part. Indeed, it is enough to check that the generators preserve the partition: $(1\;3)$ sends $\{1,3\}$ and $\{2,4\}$ to themselves, whereas $(1\;2\;3\;4)$ switches them.


0

If we denote $a = (1234)$ and $b = (13)$, one can easily check that $a^4 = e$, $b^2 = e$ and $ab = ba^{-1}$ which are precisely relations that define dihedral group $D_4$. Thus, subgroup generated by $a$ and $b$ in $S_4$ is isomorphic to quotient of $D_4$, and thus it's order is less or equal than $8$. Since $|S_4| = 4!$, obviously $a$ and $b$ can't generate ...


7

The partition $\{\{1,3\},\{2,4\}\}$ is invariant under the action of the two proposed generators, but not under all of $S_4$, so they cannot generate all of $S_4$.


1

Take a permutation $\sigma$ on $X$. Now extend $\sigma$ to a permutation $f(\sigma)$ on $Y$ by setting $f(\sigma)(y) = y$ for all $y \in Y - X$. Then $f$ is an injective morphism and $f(S_X)$ is a subgroup of $S_Y$.


1

Yes, it's correct. Note that you only need to generate the first row. The first column needs to be strictly increasing and thus follows immediately. So your first row should be indeed 123, 124, 125, 134, 135, 145 (in what I consider to be a more orderly fashion).


0

Consider transpositions. Number 4 is a fixed point of $\tau$, while number 2 is a fixed point of $\sigma$. So you must send 4 into 2. This gives the first transposition, (24). Now this would send $\tau$ into $(143)(2)$, which is not correct. How can you change the first cycle into the desired $(134)$? There are 3 possibilities. You can change (14), or ...



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