# Tag Info

3

Correct. Alternatively stated: $H$ is normal so it is a union of conjugacy classes; all those ccls have order dividing $G$, and hence have size either $1$ or bigger than $p$ (by definition of $p$ as the smallest prime dividing $|G|$). Therefore it is a union of size-1 ccls.

2

Your guess is correct. Note that the $gcd(m,n)$ can be written as an integer linear combination of $m,n$. Let $m$ be relatively prime to $n$. (i.e $gcd(m,n)=1$). Then there are $x,y \in \mathbb{Z}$ such that: $xm+yn=1 \Rightarrow xm = 1 \text{mod} \, n$ So $x\, \text{mod} \, n$ is the inverse of $m$ in $\mathbb{Z}_n$. In particular $m$ is invertible, ...

2

Here is the pertinent question: What about $\sqrt{6}$?

1

Add in terms of the form $d\sqrt 6$ and you get a field, because $\sqrt 3\cdot\sqrt 6 = 3\sqrt 2$ etc.

1

An abelian group $A$ is a vector space over $\mathbb{Q}$ if and only if it torsion free and is divisible. That is, for each $n$ in $\mathbb{N}$ and each $b$ in $A$ (i) $nb= 0 \Rightarrow b=0$ (ii) there exists $a$ in $A$ such that $b=na$ (see below for the definition of $nx$) Proof: Suppose that $A$ is a vector space over $\mathbb{Q}$. Then trivially for ...

1

Here is a hint: show that $n$ divides $|X \setminus X^G|$. To show this, let $G$ act on $X$ and show that every orbit has length $1$ or a multiple of $n$. The following spoiler contains another hint:

1

Let $X$ be a $G$ set, $X$ is the disjoint of the orbits of $G$. Let $x\in X$ the cardinal of orbit $Ob(x)$ of $x$ is in bijection with $|G|/|G_x|=[G:G_x]$ where $G_x$ is the stabilizer of $x$, we deduce that $n$ divides the cardinal of $Orb(x)$ if $x$ is not fixed by $G$. $|X|=|X^G|+\sum_{Orb(x), x \neq X^G}|Orb(x)|$, since $|Orb(x)|=0 mod$ $n$ if $x\neq ... 1 This is typically not true for$n>2$. For instance, suppose$R=k$is a field,$M=k^2$, and the$M_i$are a bunch of distinct$1$-dimensional subspaces. Then the$M_i$satisfy your hypotheses, but$M/\bigcap M_i=M$is$2$-dimensional while$\bigoplus M/M_i$is$n$-dimensional. 1 You want to construct a bilinear map $$B \colon \mathrm{Hom}(V_1,W_1) \times \mathrm{Hom}(V_2,W_2) \rightarrow \mathrm{Hom}(V_1 \otimes W_1, V_2 \otimes W_2)$$ and then use the universal property to obtain$\psi$. In order to define$B(f,g)$, you will want to use the universal property again. Instead of defining$B(f,g) \in \mathrm{Hom}(V_1 \otimes W_1, ...

1

Let $n_{5}$ denote the number of Sylow $5$-subgroups. By Sylow Theorems, $n_{5} \equiv 1 \mod 5$ and $n_{5}|4$ and both conditions imply that $n_{5} = 1$. Hence the number of non-identity elements of order $5$ is $n_{5}(5-1) = 4$.

1

Let $P \in Syl_p(G)$. Note that $P \cong C_p$ and that $N_G(P)/C_G(P)$ injects homomorphically into $Aut(P) \cong C_{p-1}$. Since $p$ is the smallest prime dividing the order of $G$, we must have $N_G(P)=C_G(P)$. It now follows that $P \subseteq Z(N_G(P))$.

1

It's not trivial, but the answer is no. See this MathOverflow question. In fact, a more general result of B.H. Neumann implies that a group can't be the union of finitely many cosets of infinite index subgroups. More precisely, it states that if a group is the union of $n$ cosets of subgroups, then at least one of the subgroups must have index at most $n$.

1

The collection of all groups is very large. Saying it's "uncountable" is an understatement: for any infinite size (cardinality) of set in the mathematical universe, there is a group that large. No two groups of different sizes can even be isomorphic, much less identical. This collection is too big to form a set, given the systems of set theory that math is ...

1

I hastily wrote out a solution by hand today, as follows.

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