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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 ...

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

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

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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 ...

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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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