# Tag Info

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


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 ...

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