Tag Info

Hot answers tagged

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

$\mathbb{Z_6}[X]/(2x+4)\simeq\mathbb{Z_2}[X]/(2x+4)\times\mathbb{Z_3}[X]/(2x+4)\simeq\mathbb{Z_2}[X]\times\mathbb{Z_3}$



Only top voted, non community-wiki answers of a minimum length are eligible