# Tag Info

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Here is a combinatoric proof (which I am not sure if it is from permutation group). Count the number of 'reversed pair' in the list. What I mean by reversed pair is pair of numbers which the one bigger in value comes before the smaller one. For example, in $[2,1,3,4]$, $(2,1)$ is a reversed pair but $(1,3)$ is not. Now observe the change of the number of ...

6

Hint If $m$ is odd then $$m^2-1 \equiv 0 \pmod{8}$$

3

An algebraic closure must be a splitting field of all polynomials over itself (coefficients from the algebraic closure). So existence and uniqueness of a splitting field of all polynomials over a field $K$ does not trivially imply existence and uniqueness of an algebraic closure of $K$. However, it is not hard to prove that an algebraic extension of an ...

2

When you take the splitting field of all irreducible polynomials in a given field $K$ then all those polynomials split in the bigger field, but there is no guarantee that all polynomials in the bigger field split since there are many more new polynomials now. The difficulty with proving an algebraic closure exists is precisely this: it is easy to add the ...

2

Wlog, assume that $|G|\geq 3$. Let $f:G\rightarrow S_n$ be a permutation representation induced by $G$ acting on the left cosets of $H$. Since $G$ is simple and its kernel is contained in $H$ and $H$ is proper, its kernel must be trivial. Hence $f$ is injective. Now note that $A_n$ is normal in $S_n$, hence $f^{-1}(A_n)$ is normal in $G$. Suppose $f^{-1}(... 2 One way to dissolve the confusion here is to think of it this way. A point$(z_1,...,z_n)$is a function$z:\{1,...,n\}\to Z$. Hence, the point$(z_{\sigma(1)},...,z_{\sigma(n)})$, is the composition$z\circ \sigma$. Now we can think of the domain of$f$as the set of functions $$Z^n\simeq \{z:\{1,...,n\}\to Z\}\;,$$ and this way you can rewrite the action ... 2 (I edited it because my former answer was wrong) It is true that $$\left(\tau\left(\sigma f\right)\right)\left(z_1,\dots,z_n\right)=\left(\sigma f\right)\left(z_{\tau(1)},\dots,z_{\tau(n)}\right)$$ However, we should be careful; we multiply$\sigma f$by$\tau$. Since $$\left(\sigma f\right)\left(z_1,\dots,z_n\right)=f\left(z_{\sigma(1)},\dots,z_{\sigma(n)}\... 2 Your flips of [1,\ldots,n] are precisely transpositions of \{1,\ldots,n\}, so proving your statement about flips is immediately equivalent to proving the powerful statement about permutation groups that the identity is not the product of any odd number of transpositions. So it is inevitable to either rely on these powerful results or rederive them. I'll ... 2 Let \mathbf{2} be the category with two objects and one morphism between them. Then the arrow category of any category \mathbf{C} is isomorphic to the functor category \text{Funct}(\mathbf{2}, \mathbf{C}); consequently, limits and colimits can be computed from the general facts about limits of functors. In particular, if \mathbf{C} has all (co)... 2 Hint: every proper nonzero subgroup of \mathbb{Z} is isomorphic to \mathbb{Z} About your points. A quotient a\mathbb{Z}/b\mathbb{Z} can be simple, precisely when a divides b and b/a is prime. It is true that, assuming K is a subgroup of P and H is normal in P, then K/H is normal in P/H if and only if K is normal in P. This is ... 2 Actually consider the ideal I:=(b^2, a^2-ab)\in \mathbb{Z}_2[a,b]. Then$$ a^3=(a+b)(a^2-ab)+ab^2\in I $$Then the RHS ring is in fact \mathbb{Z}_2/(b^2, a^2-ab). The LHS ring (simplified version) is \mathbb{Z}_2[x,y,z]/(x^2, x+y+z, yz)\simeq \mathbb{Z}_2[x,y]/(x^2, y(x+y)). Also since the base ring is \mathbb{Z}_2 you have x^2-xy=x^2+xy, and the ... 2 \mathbb Z_2 is abelian. Thus conjugation can't change the first member of an element of \mathbb Z_2\oplus S_3. 1 Your first reformulation of the problem is not correct; it suffices to show that one of the quotients is not simple, not all of them. It is true that K/H is normal in P/H if and only if K is normal in P. But you don't need this fact because all groups concerned are abelian, so all subgroups in all quotients are normal. The last statement you wish ... 1 Let \ell=(x_1,x_2,\ldots, x_n) be a list of the numbers in [n]. For each k\in[n] denote by p_k the number of x_i\ (1\leq i<k) with x_i>x_k\,, and call \beta(\ell):=\sum_{k=1}^n p_k the badness of \ell. It is easy to convince oneself that a transposition ("flip") \tau:\>\ell\mapsto\ell' changes the badness by \pm1. It follows ... 1 The complete transposition graph X_n is defined to be the graph whose vertex set is the symmetric group S_n and two vertices are adjacent whenever they differ by a transposition. It is easy to see that this graph is bipartite, with bipartition the set of even permutations and the set of odd permutations. You want to prove that any walk in this graph ... 1 Here's a different answer that is, in some way, very discoverable and scalable. Let F be a field not of characteristic 2. Then in F^n (the product of n copies of F), every element with entries only from \{1,-1\} satisfies a^2=1, and would be a root of X^2-1. This is quite different from the integers mod 8 since that ring is indecomposable, and ... 1 Whether the operations are finitary or infinitary, it is not possible to determine whether an algebra is finite knowing only whether or not it is finitely generated. In general, it may not even be possible to determine whether an algebra is finite given any amount of information. For finitely generated groups, which have finitary operations, given a set of ... 1 If the image is not in A_n, then the image must have a non-trivial normal subgroup, so apply the correspondence thm, we get a normal subgroup in G, which is a contradiction. 1 Your "functions with finite support" approach is the standard approach to take to this, and is very likely what any author who considers the question trivial has in mind. The correct notion of "finite" to take when defining monomials is the usual one, as you have done (this is necessary for the polynomial ring you get to be the free commutative k-algebra ... 1 Here is an option which takes place between your two suggestions, without appealing to choice: Direct limit of finitely generated subrings. We define k[X] to be the direct limit of the system \{ k[Y]\mid Y\in[X]^{<\omega}\} with inclusions as the embeddings. Now to check that the operations are well-defined we only need to check they restrict ... 1 I just want to add a couple of observations to Jose Brox's answer. Firstly your question had the additional assumption that A has an anti-involution * fixing each of the x_i, but this doesn't make a difference. Indeed consider unital \mathbb C-algebra with presentation$$ F=\mathbb C\langle x_1,\ldots,x_n \mid x_i^2=x_i,\;x_1+\ldots+x_n=1\rangle.$...

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