# Tag Info

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 $4!=24$. ...

6

Note that $$(1\,2\,3\,4\,5)(1\,2\,3\,5\,4)=(1\,3)(2\,4)$$

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 $A_5$.

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: ...

4

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

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.

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 $a=(1,2)$, $b=(2,3)$, $c=(3,4)$, $d=(4,5)$: $$\langle a,b,c,d \mid a^2,b^2,c^2,d^2,(ab)^3,(bc)^3,(cd)^3,(ac)^2,(ad)^2,(cd)^2 ... 3 The most direct way to solve this is to consider the complement of the given set - that is, the bijections that do have a fixed points.$$Y=\{f:S\rightarrow S\mid f\text{ is bijective and }f(x)=x\text{ for some } x\in S\}.$$Notice that if we define, for each s\in S the set$$Y_s=\{f:S\rightarrow S\mid f\text{ if bijective and }f(s)=s\}$$then we may ... 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 ... 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 ... 2 You can choose a permutation a (others will work, too) such that: a(1) = 5, a(2) = 6, a(3) = 1 and a(4) = 3. It doesn't matter what we choose for a(5),a(6), as long as we don't pick from the set \{1,3,5,6\}, since those values are already "taken". a(5) = 2, and a(6) = 4 will do. Thus a = (1\ 5\ 2\ 6\ 4\ 3) is one possibility. 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 ... 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 ... 2 A function satisfying the axioms you listed is called a length function. For S_n, the word length would be an example. The symmetric group S_n is generated by elements s_1, \ldots, s_{n-1}, where s_i is the transposition (i,i+1). Define the word length of an element w of S_n to be the smallest \ell for which there is a decomposition$$ ...

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 key fact to note is that two elements of the symmetric group are conjugate if and only if they have the same number of $k$-cycles in their cycle decomposition for all $k$. In particular, every conjugate of a $3$-cycle is a $3$-cycle, which solves your second exercise. You have shown in a previous question that $$a(i_1i_2\cdots ... 1 The reason the left multiplication action works is because for fixed g\in G, the map \psi_{g}: G\to G given by x\mapsto gx is bijective (only as a set theoretic map!). Hence, the correspondence G\to S_{|G|} given by g\mapsto \psi_{g} is a group homomorphism (which then you easily check has a trivial kernel, so that completes Cayley's theorem). ... 1 The induced (by the conjugation action) morphism is one-to-one if and only if the center of the group is trivial. 1 Given any distinct a,b,c, we want to express (abc) as a product of 3-cycles of the given form. Well, let's play around a bit. Multiplying two arbitrary 3-cycles of the given form together yields:$$ (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 ...

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