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## Hot answers tagged abstract-algebra

6

Take $G = H \times H \times H \times \cdots$ for $H$ any nontrivial group.

2

Let $G = \mathbb Z ^ \mathbb N$ (with pointwise addition as the product). Then let $f:G \times G \longrightarrow G$ be $$f(g,h)(n) = \begin{cases} g(k), &n = 2k \\ h(k), &n = 2k+1 \end{cases}$$ You can verify $f$ is an isomorphism.

2

For $n\in\mathbb Z$ and group $G$ denote $r_n(G)=\{g\in G:g^n=1\}$. Obviously, if $G$ abelian group, then $r_n(G)\leq G$. Group of invertible elements of the ring $R$ denote $R^*$. We can say, that we are looking for $|r_{m-1}(\mathbb{Z}_m^*)|$. Denote $n=m-1$. By the chinese remainder theorem, $$\mathbb{Z}_m^* \simeq ... 2 Given any matrix M\in \mathbb R^{m\times n} define a function T:\mathbb R^n\to \mathbb R^m as T(v)=Mv for all v\in \mathbb R^n. It is your exercise to show that T is well-defined (hence a function) and a linear transformation. 2 Yes, the elements of$$R := \Bbb Z [x] / \langle (x - 1) (x - 2) \rangle$$"look like" a + bx, a, b \in \Bbb Z. Put more precisely, each element of R has a unique representative in \Bbb Z of degree \leq 1, like you say exactly because of the division algorithm. This identification alone, however, does not determine the ring structure. The additive ... 2 They are not isomorphic. In (\mathbb Q\setminus \{0\},\cdot) you have an element that is its own inverse (-1). This does not happen in (\mathbb Z,+) 2 No. (\Bbb Z, +) is generated by \{1,-1\} and (\Bbb Q^\times, \cdot) is not finitely generated. 2 The Grothendieck group \mathcal{G}(M) of a commutative monoid M is the unique commutative group satisfying the following universal property: there is a monoid morphism i\colon M \to \mathcal{G}(M) such that for every monoid morphism f \colon M \to G, where G is a commutative group, there is a unique group morphism \mathcal{G}(f) \colon ... 2 Suppose that your regular polygon has a vertice on the x-axis. Then the first vertice counted counter clock wise has coordinates (\cos(\frac{2\pi}{n}),\sin(\frac{2\pi}{n})). Hence if you know the construction of the polygon, by projecting the first vertice on the x axis, which can be done with a ruler and a compass, you can get \cos(\frac{2\pi}{n}). ... 1 The proposition can be reformulated as if \mathfrak{a}+\mathfrak{b}=(1), then \mathfrak{a}\cap\mathfrak{b}=\mathfrak{a}\mathfrak{b} The word “provided” is used in the sense of “when it is given that”. In your case \mathfrak{a}+\mathfrak{b}=(2)+(2)=(2)\ne(1). 1 If a=b, then a^2+ab+b^2=3a^2=a^2=0 and a=0=b, we are done. Suppose that a\neq b. Observe that 0=(a-b)(a^2+ab+b^2)=a^3-b^3. Thus, a^3=b^3. We claim that a=0 and b=0. If a\neq 0, then (a^{-1}b)^3=1 and the multiplicative order of a^{-1}b in the multiplicative group F-\{0\} is 1 because 3\not\mid |F-\{0\}|=2^n-1. Hence, ... 1 aL=0 and bN=0 implies (ab)x=0 for all x\in M: 1 Let G=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}, where p_1<p_2<\cdots<p_n are distinct primes and a_i\geq 1 for all i. Let N_p denote the number of Sylow p-subgroups and let N denote the total number of Sylow subgroups, i.e. the sum of all N_{p_i}. By the fact that N_p\equiv 1\pmod{p} and N_p||G| we must have that$$N_{p_i}\leq ...

1

I guess (from the comment discussion mostly) your question is not really about free objects but rather: given an adjunction $F \dashv U$, how can I explicitly write the bijection $\hom(FA, B) \to \hom(A, UB)$ so that the special case $\mathsf{Set} \leftrightarrows \mathsf{Grp}$ gives me the restriction $\varphi \mapsto \varphi\restriction A$? Given $F ... 1 From a computational point of view it amounts to saying that, each time you meet$x^2$, you can replace it with$3x-2$,$x^3$will be replaced with$\,3x^2-2x=7x-6$, &c. 1 I like the rest above, so I just wanted to suggest an argument for b). Suppose there is an$\overline{x} \in G/P$with order$p$, then$(x+P)^p=P$. But this means$x^p \in P$, so$(x^p)^{p^k}=e$for some$k$. But this means$x\in P$and thus$\overline{x}$is actually the identity in$G/P$. But the identity cannot have order$p$, so this is not possible. ... 1 Showing a subgroup isn't too bad. Note$P$is nonempty as$e\in P$. Also,$a_1,a_2\in P$with$p^k,p^m$such that$a^{p^k}=e=a^{p^m}$and$(a_1a_2^{-1})^{p^{m+k}}=e$since$G$is abelian. If some element$\bar{x}$was of order$p$in$G/P$, then$x\in P$so$\bar{x}=e\in G/P$. If$|P|\neq p^n$, then$G/P$has an element of order$p^k$for some$k$by ... 1 You found an$a\in\mathbb Z$such that$p\mid a^2+1$. If$p$is prime in$\mathbb Z[i]$then, from$p\mid a^2+1$you get$p\mid a+i$or$p\mid a-i$, and both cases lead to a contradiction. This shows that$p$isn't prime in$\mathbb Z[i]$. Then$p=(m+ni)(m-ni)$, so$p=m^2+n^2$. Now just take the ideal generated by$m+ni$. 1 $$\dfrac{|x+3|+x}{x+2} >1$$ Assume$x\le -3$$$\dfrac{|x+3|+x}{x+2} >1\iff\dfrac{-(x+3)+x}{x+2} >1\iff -(x+3)+x<x+2\iff-5<x$$ So the first interval is indeed$(-5,-3]$Now let$-3\le x < -2$$$\dfrac{|x+3|+x}{x+2} >1\iff\dfrac{+(x+3)+x}{x+2} >1\iff +(x+3)+x<x+2\iff x<-1$$ So the second interval is$[-3,-2)$Now$x>-2$... 1 Let$G$be the trivial group, for the only finite example. 1 Put in very simple terms, as polynomials two polynomials are equal if and only if all their coefficients are the same. So in$\Bbb Z_3[x]$, we have:$x^8 + 1 \neq x^3 + 1$, because the coefficient of$x^8$in the first is$1$, but the coefficient of$x^8$is$0$in the second (it has no$x^8$term). However, in$\Bbb Z_3$, we do have an$a \in \Bbb Z_3$... 1$\mathbb{Z_6}[X]/(2x+4)\simeq\mathbb{Z_2}[X]/(2x+4)\times\mathbb{Z_3}[X]/(2x+4)\simeq\mathbb{Z_2}[X]\times\mathbb{Z_3}\$

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