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4

You could try Lagrange's Theorem which tells you something about the order of a subgroup. Count the elements of $H$.


3

What you have written is not a presentation. In general it is not a good or efficient strategy to just write down powers of elements hoping that you will get a presentation. The best known presentation of $S_5$ is on the generators $a=(1,2)$, $b=(2,3)$, $c=(3,4)$, $d=(4,5)$: $$\langle a,b,c,d \mid a^2,b^2,c^2,d^2,(ab)^3,(bc)^3,(cd)^3,(ac)^2,(ad)^2,(cd)^2 ...


2

A way that doesn't use group actions (which are awesome, by the way, so try to use them when you can): We want to know the order of the centralizer of $g,\ |C(g)|$. Now $x \in C(g)$ means that $xgx^{-1} = g$. Since $x(1\ 2\ 3)x^{-1} = (x(1)\ x(2)\ x(3))$ we can immediately see two things: $x$ must map $\{1,2,3\} \to \{1,2,3\}$ and thus $\{4,5\} \to ...


2

Let $G$ acts on itself by conjugation. Consider the orbit of $g$, it's the conjugacy class of $g$ in $S_{5}$, so the set of $3$ cycles. Now, $H$ is the stabilizer of $g$. If $x \in H$, then $xg=gx$ so $xgx^{-1}=g$. And if $x \in Stab(g)$, then $xgx^{-1}=g$ so $xg=gx$ so $x \in H$. Now the orbit stabilizer theorem tells us that $|G|=|Orb(g)||Stab(g)|$, so ...


3

I doubt it. Presumably that's a typo, and it should read "geometric reasons". The usual proof identifies the coefficients of the product of two Schubert polynomials on the basis of Schubert polynomials as intersection numbers of transverse subvarieties, which are therefore non-negative integers. It has been generalised in various directions by increasingly ...


0

Some kind of lexicographic ordering: I will illustrate this for n=4 (partially!) That is first list all permutations starting with 1. There are 6 of them, and so those with 2 at 2nd position come first, and then those with 3 at 2nd position: We use recursion to list them, [1,2,3,4], [1,2,4,3], [1,3,2,4],[1,3,4,2] Now the 6 permutations with 2 at first: ...


2

I ran a small magma program and it turns out that the answer is $n=9$, when there are suddenly 14 extra solutions (on top of the usual two trivial ones). For each of these, the number of non-trivial conjugacy classes involved is between 7 and 10 (out of a possible 17) and the number of elements involved is either half or a third of all the elements. For ...


4

I think it is known that the number of subgroups of an elementary abelian $2$-group is larger than the number of subgroups of a symmetric group of roughly the same order. For reference, the corresponding OEIS sequences are: https://oeis.org/A006116 and https://oeis.org/A005432 In particular, you can see that, already, the elementary abelian $2$-group ...


4

No. Since any group with order $n$ is isomorphic to a subgroup of $S_n$ (Cayley's theorem), and any subgroup of a subgroup is a subgroup of the original group.


4

Apply Burnside's Lemma: Consider the action of $G$ over $[n]$. The number of orbits is $$|[n]/G|=\sum_{g\in G}\frac{[n]^g}{|G|}$$ where $[n]^g=\{k\in[n]:g(k)=k\}$. The hypothesis says that $|[n]^g|=1$ for all $g\neq e$. Moreover, $[n]^e=[n]$. Then $$|[n]/G|=\frac n{|G|}+\sum_{g\in G,g\neq e}\frac{1}{|G|}=\frac{|G|+n-1}{|G|}>1$$ This means two things: ...


1

Your claim is not true. For example, consider the subgroup $G < S_5$ generated by $(123)$ and $(12)(45)$, which is isomorphic to $S_3$. Here every element of $G$ fixes a point but there is no common fixed point for $G$.


1

The reason the left multiplication action works is because for fixed $g\in G$, the map $\psi_{g}: G\to G$ given by $x\mapsto gx$ is bijective (only as a set theoretic map!). Hence, the correspondence $G\to S_{|G|}$ given by $g\mapsto \psi_{g}$ is a group homomorphism (which then you easily check has a trivial kernel, so that completes Cayley's theorem). ...


1

The induced (by the conjugation action) morphism is one-to-one if and only if the center of the group is trivial.


1

Given any distinct $a,b,c$, we want to express $(abc)$ as a product of $3$-cycles of the given form. Well, let's play around a bit. Multiplying two arbitrary $3$-cycles of the given form together yields: $$ (12x)(12y) = (1x)(2y) $$ Hmm, that didn't get us anywhere. In hindsight, if we want $a$, $b$, and $c$, then it makes sense that we'll need at least three ...


0

Hint. You know (or you can prove) that the transpositions $(1i)$ generate $S_n$. As $A_n$ is the set of permutations written as an even number of transpositions, we just have to generate the products $(1i)(1j)$ of two transpositions. So let's do it... For $i,j \ge 2$ and $i \neq j$ you have $(1i)(1j)=(1ji)$. You also have for $i,j \ge 3$ and $i \neq j$ ...


0

I don't have enough reputation to comment, but this is for @user16924 $\phi(y)$ can be written as a product of p-cycles. We don't need to know how many of them we have! Let's decompose them in transpositions. A p-cycle is $p-1 \equiv 0 (2)$ transpositions, so $t(p-1) \equiv 0 (2)$. And we are done.


7

This is probably overkill, but here goes: In view of the orders of $S_7$ and $A_8$, it suffices to show that $A_8$ has no subgroup $H$ of index $4$. If there were such an $H$, then $A_8$ would act transitively on the set of four left cosets of $H$, and the kernel of that action would be a normal subgroup of $A_8$ of index at least $4$ and at most $4!=24$. ...


2

@dREaM was faster than me, I shall try to detail a bit his answer. Let $\sigma \in A_5$. Decompose it into a product of disjoint cycles, $\sigma = c_1 c_2 \dots c_n$ (this decomposition is not unique, but this does not bother us). Let the order of cycle $c_i$ (equal to its length, i.e. the number of elements that it does not fix) be $o_i$. Since these cycles ...


5

If it did then $A_5$ would have an element of order $6$, this element would need to have a $3$-cycle and at least two cycles of even length (at least one so the order is even, but at least two because we are in $A_5$), which is impossible in $A_5$.



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