Tag Info

Hot answers tagged

9

Let $G = S_5$, the symmetric group of 5 elements. Let $x_1 = (1,2), x_2 = (3, 4), x_3 = (1, 5), x_4=(2, 4), x_5 = (3, 5).$


6

The group theoretic interpretation is as the numbers game for the Coxeter group of type $\widetilde{A}_4$, with generators $s_1,s_2,s_3,s_4,s_5$, where the relations are $$s_1^2=s_2^2=s_3^2=s_4^2=s_5^2=1\mathrm{,}$$ $$(s_is_j)^2=1$$ if $i\not\equiv j\pm 1\pmod{5}$, and $$(s_is_{i+1})^3=1$$ for all $i$, where $i+1$ is taken modulo $5$. This is an infinite ...


5

No. Consider elementary matrices. Let $e_{ij}$ denote the matrix with $(i,j)$ entry $\delta_{ij}$. Let $e_{ij}(\lambda)=1+\lambda e_{ij},i\neq j$. Assume $n>2$. We have the so called Steinberg relations $$\begin{align}&(1)& e_{ij}(\nu)e_{ij}(\mu)&=e_{ij}(\nu+\mu)\\ &(2)& [e_{ij}(\nu),e_{jk}(\mu)]&=e_{ik}(\nu\mu)&\text{ if ...


3

By Cayley's theorem, a group $G$ of order $30$ is isomorphic to a subgroup of $S_{30}$ is such a way that no non-identity element of $G$ has any fixed point. There is an element $t$ of order $2$ in $G$ which is represented by a product of $15$ $2$-cycles, so as an odd permutation. The elements of $G$ which are represented as even permutations of $S_{30}$ ...


3

There is a version of the tensor product for nonabelian groups, but this notion is much more specialized. See http://www-irma.u-strasbg.fr/~loday/PAPERS/87BrownLoday%28vanKampen%29.pdf, section 2. In the construction at some point you do a mod out, which you cannot do in general if you do take the free group instead of the free abelian group. (You see a free ...


3

I will assume the characteristic of a non-unital ring is simply the exponent of the additive group, i.e. the smallest $n$ such that $na=0$ for every $a \in A$. Then the argument that the characteristic is prime for integral domains follows the usual way: So assume that $n=kl, \;\;\; 0<k , l <n$ is a composite number. Choose $a \in A$ such that $ra ...


3

Prime = irreducible in, e.g., GCD domains. (In fact, an integral domain is a UFD if and only if it is a GCD domain satisfying the ascending chain condition on principal ideals.) Now let's give an example of GCD domain which is not a UFD: $\mathbb Z+X\mathbb Q[X]$. (For instance, this ring doesn't have ACC on principal ideals: ...


2

Actually, you have an error in understanding the category $\mathcal{G}$ obtained from $G$. It is not the full subcategory of $\mathbf{Grp}$ whose only object is $G$: i.e. it is not the category with one object whose arrows are endomorphisms of $G$. Instead, it is the abstract category with one object whose arrows are elements of $G$. The product of arrows ...


2

If $G$ is a group of order 30, then the number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 30, so it must be 1 or 6. And the number of Sylow 5-subgroups is congruent to 1 modulo 3 and divides 30, so it must be 1 or 10. If there is a unique subgroup of order 3 or 5, it is normal and we are done. If not, there are 6 subgroups of order 5 and 10 ...


2

HINT: Consider the set $S=\{1\le m\le n\vert \gcd(m,n)=1\}$. Note that if $\gcd(m,n)=1,\,\,\exists\text{ integers }p,q$ such that. $mq+pn=1$ (as gcd can be expressed as a linear combination) Reduce both sides modulo n. Can you see where to go from here?


2

$\newcommand{\tensor}{\otimes}\newcommand{\from}{\colon}$Let $M \tensor N$ be an $A$ module, and $f \from M \times N \to M \tensor N$ be a bilinear map that satisfy the universal property of the tensor product. The elementary tensor $m \tensor n$ is defined to be $f(m,n)$. We will prove using just the universal property and not the construction that: $M ...


1

There is a proof that the universal property determines the tensor product of modules up to isomorphisms. The initial construction is only used to prove the existence of the tensor product. It is not a bad thing to visualize by tensors but sometimes it is better to strictly use the universal property. $\require{AMScd}$ \begin{CD} A\times B ...


1

$\begin{eqnarray}{\bf Hint}\quad\ x^n &\in&\! \langle x^{n-1},\ldots,x,1\rangle\, =: M \\ \Rightarrow\ x^{n+1} &\in& \langle x^n,\,\ldots,x^2,x\rangle\, \subseteq\, M\ \ {\rm by}\ \ x^n\in M\\ &\vdots&\\ \Rightarrow\ x^{n+k} &\in& M \end{eqnarray}$ Remark $\, $ Instead of using the division algorithm, we interpret the ...


1

Hints. $\operatorname{ht}\mathfrak{p}+\dim A/\mathfrak{p}\leq \dim A$ for any prime ideal $\mathfrak p$ (why?). $\dim A/\mathfrak a=\sup_{\mathfrak{p}\in V(\mathfrak{a})}\dim A/\mathfrak p$.


1

$|G|=|Ker(\varphi)|\cdot|Im(\varphi)|$ and $|Im(\varphi)|\cdot |H:Im(\varphi)|=|H|$. Hence their common divisor is at least $|Im(\varphi)|$.


1

By the Sylow theorems, the number of Sylow $p$-subgroups of the group is congruent to $1$ mod $p$ and divides the order of the group. Suppose $n=kp+1$ and $n\mid q$. Then since $q$ is prime we in fact have $kp+1=q$ or $k=0$. Assume $k\neq 0$. Then $q-1=kp$, meaning that $p\mid q-1$. Since we assume that this is not the case, we must have that $k=0$, hence ...


1

This proof works. As @bof noted in comments, there is a more direct approach, not using the more advanced result about the normal subgroups of $S_n$ for $n>4$. @hof's answer works for all $n$.


1

A element in an integral domain is irreductible if it is not a unit nor a product of non-units. An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab \in pA$ then $a$ or $b$ is in $p A$ which is ...



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