# Tag Info

11

The automorphism group of $\mathbb{R}$ is trivial. The automorphism group of $\mathbb{C}$ is extremely complicated (at least if you accept the axiom of choice) : see http://www.jstor.org/stable/2689301?seq=1#page_scan_tab_contents for instance. The automorphism group of $\mathbb{H}$ is $\mathbb{H}^*/\mathbb{R}^*$ : according to the Skolem-Noether theorem, ...

4

The triangle with angles $\frac \pi \beta$, $\frac \pi \beta$ and $\frac \pi \gamma$ is an isosceles triangle. Tracing the height to the uneven side, two triangles with angles $\frac \pi 2$, $\frac \pi \beta$ and $\frac \pi {2 \gamma}$ are obtained. And the assertion follows.

4

By the first isomorphism theorem, $Im(f)\cong G/Ker(f)$. Now $N<Ker(f)$ since $N$ is abelian, so $|G/Ker(f)|=|Im(f)|$ is relatively prime to $|Aut(N)|$. But $Im(f)<Aut(N)$, so $|Im(f)|$ divides $|Aut(N)|$. Therefore the only possibility is that $Im(f)=\{e\}$, so that $Ker(f)=G$, i.e. $N$ is in the center.

3

It's the subgroup $\langle xyx,y \rangle$ of $\langle x,y \mid x^2,y^\beta,(xy)^{2\gamma} \rangle$. The index of this subgroup $2$, so checking that the subgroup has the presentation $\langle z,y \mid z^\beta,y^\beta,(xy)^\gamma \rangle$ (with $z=xyx$) is routine.

3

Let $x\in G$, $x\ne0$; then $\langle x\rangle$ is a (non trivial) cyclic subgroup of $G$. Since $\langle x\rangle\cong G$, we have that $G$ is cyclic.

2

Generically you cannot show anything better than $g(K)\le g(H\rtimes K)\le g(H)+g(K)$ as both bounds are attained in examples.

2

Write down explicitly: $$\forall\,g\in G\;,\;\;gHg^{-1}:=\{\,ghg^{-1}\;:\;h\in H\,\}$$ Now, since (your definition) $\;H\lhd G\implies ghg^{-1}\in H\;,\;\;\forall\,h\in H$ , and thus indeed $\;gHg^{-1}\subset H\;$ since every element in the definition written above is in $\;H\;$ .

1

First, the statement $g H g^{-1} \in H$ (which you had originally written, and then edited) is meaningless: here $g H g^{-1}$ is the set $\{g h g^{-1} : h\in H\}$ of elements of the form $g h g^{-1}$ for $h\in H$, and it doesn't make sense to ask whether it belongs to $H$. (Well, in set theory, it does make sense, because everything in mathematics is a set, ...

1

$D_{12} \cong C_2 \times S_3$ is nonabelian of order $12$, has at least $2$ normal subgroups ($C_2 \times 1$ and $1 \times S_3$), and they are not all abelian.

1

To each lie algebra $\mathfrak{g}$ there is a unique simply connected Lie group $G$ having $\mathfrak{g}$ as its lie algebra, and furthermore any other Lie group $H$ having lie algebra $\mathfrak{g}$ is covered by the universal one $G$, in other words is a quotinet of $G$ by some discrete central subgroup $K$. (In fact, covering space theory goes on further ...

1

$H$ is a normal subgroup of $G \Leftrightarrow \forall g \in G : gHg^{-1}\subseteq H$. Notice that $gHg^{-1}$ is just the notation we use for the set $\{ ghg^{-1} | h \in H \}$. So this formulation is in fact equivalent to your first definition: $\forall g \in G \forall h \in H : ghg^{-1} \in H$.

1

Regarding $R,$ let $h:F \to G$ be an isomorphism between subfields $F,G$ of $R .$ If $\forall x\in F\; (x>0\implies \sqrt x\in F),$ then $h=id_F.$ Because, for $x,y\in F,$ we have $$x>y\implies 0\ne h(x)-h(y)=(h(\sqrt {x-y}))^2 \implies h(x)>h(y).$$ And $h|Q=id_Q.$ So $\{q\in Q:q<x\}=\{q\in Q:q<h(x)\}$ for all $x\in F.$ In particular if ...

1

No. Set $m=\dfrac ab$. An element $(x, \dfrac cd)$ has finite order in $(\mathbf Z\times\mathbf Q)/M$ if there exists $n\in\mathbf N, z\in \mathbf Z$ such that $$n\frac cd=\frac ab z\iff nbc=adz$$ Just take $n=\dfrac{ad}{\gcd(ad,bc)}$.

1

I suppose $M=\mathbb Z\times m\mathbb Z$ with $m\ne0$. Then $(\mathbb Z\times\mathbb Q)/M=(\mathbb Z\times\mathbb Q)/(\mathbb Z\times m\mathbb Z)\simeq\mathbb Q/m\mathbb Z$. Write $m=\frac ab$. Now pick an element $x\in\mathbb Q$, $x=\frac cd$. Then $(da)x=cbm\in m\mathbb Z$, so the order of $x$ is finite.

1

Let $<p/q + \mathbb Z>$ be ANY element of $H_2$ (using your notation). Then as you have shown, you can write $p/q = k/n$ for some $k$, which also means $<p/q + \mathbb Z> = <k/n + \mathbb Z>$. But all elements of this latter form are already in $H_n$, so you have shown that $H_2 \subset H_n$. And since they have the same finite cardinality, ...

1

There are (infinitely generated) noncyclic torsion-free groups $G$ such that $Aut(G)\cong {\mathbb Z}/2$, see introduction to J. T. Hallett, K.A. Hirsch, Torsion-free groups having finite automorphism groups. I. J. Algebra 2 (1965) 287–298. and references given there (various examples are due to de Groot, Hulanicki, Fuchs, Sasiada). The paper itself ...

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