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5

A simpler example is provided by ${\rm GL}(n,\mathbb{C})$ for any $n >1.$ We may choose a transversal to the conjugacy classes which consists completely of upper triangular matrices ( as every matrix in ${\rm GL}(n,\mathbb{C})$ is conjugate to an upper triangular matrix). This transversal generates a proper subgroup of $G,$ as the subgroup it generates ...


3

No. For example, there exist infinite groups with exactly two conjugacy classes. There might be easier counterexamples though.


3

Check out the new book (amazon-link) Tom Leinster, Basic Category Theory, Cambridge Studies in Advanced Mathematics, Vol. 143, 2014


2

FPE's answer and Adam Hughes' comment are great. Ignoring the fact that $\mathbb{C}$ is algebraically closed (and hence all polynomials over $\mathbb{C}$ reduce into linear factors), you can look at it in this perspective: Consider a polynomial $f(x) \in F[x]$. If $\deg(f) = 2$ or $3$, then $f$ is irreducible $\iff$ it has no roots in $F$ (Why?). From ...


2

We shall first consider how many different ordered bases $\Bbb F_q^n$ has. Recall that $|GL_n(\Bbb F_q)|=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$. Each element of $GL_n(\Bbb F_q)$ represents a linear map that carries the standard (ordered) basis $\{e_1, e_2, \ldots, e_n\}$ to another ordered basis. We can establish a bijection between the ordered bases of $\Bbb ...


1

It has to be true for every factorisation $f(x) = g(x)h(x)$, not only for the one you give. In particular, you can always write $f(x) = 1\cdot f(x)$, and $1$ is a unit, so your argument can't work. And remember that $\mathbb C$ is algebraically closed, so that any polynomial of degree at least $2$ is reducible, since it has roots in $\mathbb C$.


1

Suppose $a\ne0$; consider the canonical projection $\pi\colon R\to R/Ra$ and the monomorphism $\mu_a\colon R\to R$ defined by $r\mapsto ar$. If $R/Ra$ is injective, there exists $g\colon R\to R/Ra$ such that $g\circ \mu_a=\pi$. Now $$ \pi(1)=g\circ \mu_a(1)=g(a)=ag(1)=0 $$ because $a$ annihilates $R/Ra$, which is absurd unless $Ra=R$, that is, $a$ is a ...


1

You'll never find such an isomorphism, because $x+x=0$ for all $x\in\mathbb Z[i]/2\mathbb Z[i]$, but not for all $x\in\mathbb Z/4\mathbb Z$. The theorem you state is only true if $\gcd(a,b)=1$. See this answer from the same question.


1

Introduction to Category Theory by Harold Simmons is a nice and gentle way to get into category theory with plenty of exercises (and full solutions!). I'm an undergrad as well, and I worked through this book before moving on to Categories for the Working Mathematician because it is more leisurely. More to the point of your question, Intro to Category Theory ...


1

There is a canonical linear map $V^* \otimes W^* \to (V \otimes W)^*$. It is an isomorphism when $W=K$ (both sides identify with $V^*$ and the linear map becomes the identity). The class of $W$s for which it is an isomorphism is closed under finite direct sums - this is because both sides are additive (contravariant) functors in $W$. It follows that it is an ...


1

Restricting $*$ to the subset $H$ means forgetting (or not caring) what $a*b$ is when $a\notin H$ or $b\notin H$. In other words, the restricted operator is the function $*':H\times H\to G$ defined by $h_1*'h_2=h_1*h_2$ for all $h_1,h_2\in H$.


1

Consider $\mathbb N_0\times \mathbb N_0$ with componentwise multiplication. Then $\{\,(n,1)\mid n\in\mathbb N_0\,\}$ and $\{\,(n,0)\mid n\in\mathbb N_0\,\}$ are disjoint submonoids (with neutral elements $(1,1)$ and $(1,0)$, respectively). In fact, simply consider $\{0,1\}$ with multiplication and the submonoids $\{0\}$ and $\{1\}$.


1

"May a monoid have a different neutral element than its submonoid?" By the very definition of a submonoid, no. But of course there are sub-semigroups which happen to be monoids w.r.t. a different neutral element, for example $(\{0\},*,0)$ inside $(\{0,1\},*,1)$.



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