New answers tagged

0

If H was not normal then for some h$\in$H and some g$\in$(G-H), g$^{-1}$hg must end in G-H and have order 3 which will imply that, g$^{-1}$h$^{3}$g = e i.e h$^{3}$g=g i.e. h$^{3}$=e contradiction to the Lagrange Theorem that h being an element of H of order '4' can't have order 3.


3

There are no nilpotent elements of order $3$ in $\text{Mat}_{2\times2}(\mathbb{R})$, while there are many of them in $\text{Mat}_{3\times3}(\mathbb{R})$. Hence, the two can't be isomorphic as rings.


0

Before I answer your question, let's calculate the cosets. To start with, let's pick some element not in $S$. $(1,0)$ will do for our purposes. We have $(1,0) + S = \{(1,0),(3,0),(1,3),(3,3)\}$. Note that $(3,0) \in S$, that is: $(3,0) + S = (1,0) + S$. Perhaps this might dissuade you from the notion that adding $(n,0)$ to $S$ will recover all the cosets. ...


0

Let $K$ act on $G$ by conjugation. Then the set of distinct conjugates of $x$ by elements of $K$ is just the orbit of $x$ under this action. It is a basic fact about group actions that: $$ |\textrm{orbit of } x| = \frac{|K|}{|\textrm{stabiliser of } x|}.$$ Where the stabiliser of $x$ is the set of elements of $K$ that fix $x$. Fixing $x$ under conjugation, ...


1

I think you have a typo. The isomorphism is $$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_{k+1}^{a_k}$$ Then, $\frac 12 = p_1^{-1}$ and so the corresponding polynomial is $-1x^0$. The point is that $(\mathbb Q_{>0}, .) \cong (\Pi_p \mathbb Z, +)$. That is exactly what $(\mathbb Z[x], +)$ is.


1

Take the element $\;w:=\frac16+\Bbb Z\;$ , as proposed in the comments. The subgroup generated by it is $$\langle w\rangle=\left\langle \frac16+\Bbb Z\right\rangle=\left\{\,\overline 0,\,w,\,2w,\,3w,\,4w,\,5w\,\right\}\;,\;\;\;\overline0:=0+\Bbb Z=\Bbb Z$$ and there you have a proper subgroup which doesn't fit in what you wrote.


0

In general, this type of problem can be very difficult. To see some of the difficulties, check out Wilkinson's Polyonmial, specifically, the conditioning and stability parts. In the case you describe, if you have real coefficients, then starting with $4$ distinct real roots, the roots will move continuously as the coefficients vary and remain real until ...


0

You want to construct a homomorphism $G \to GL(V)$: normally the easiest way of doing this is to choose a presentation for $G$ as $G = \langle X | R \rangle$, and choose matrices in $GL(V)$ to map the generators $x_i \in X$ to such that the relators $R$ are all satisfied. A simple example is $V = \mathbb R^2$ and $G = \langle x | x^5\rangle = C_5$, then we ...


0

Direct usage of L'Hospital rule would give you a defined term and on substitution you will have 4 as answer.


3

$5397$ isomorphism classes of groups. About 5 hours, under 1 GB in (development version of) GAP.


1

If $m \ne 0:$ $$M_m = \left\{(x,y) \in \mathbb{Z} \times \mathbb{Q} : \dfrac ym \in \mathbb{Z}\right\}$$ \begin{array}{lcc} f:\mathbb{Z} \times \mathbb{Q} & \to & \mathbb{Q}/\langle m\rangle \\ \phantom f:(x,y) & \mapsto & y + \langle m\rangle \end{array} \begin{align} (x,y) \in \ker f & \iff f(x,y) = ...


1

Assume that the angles of your triangles are all of the form $\pi/n$, $n\in {\mathbb N} - \{1\}$, of course is allowed to be different for different angles and triangles. Given a triangle $T$ with angles of this form, consider first the reflection group $\Gamma=\Gamma_T< Isom(H^2)$ generated by reflections in the edges of $T$. Then $T$ is a fudnamental ...


2

Since $S$ is isomorphic to a group of permutations on $26 - 9 = 17$ letters, its order is $17!$. Therefore it cannot contain an element of order $37$ or $38$, since neither of these numbers divide $17!$ (which is not divisible by any prime number greater than $17$). To show that it contains elements of order $36$ and $39$: factor these numbers into coprime ...


3

The condition is equivalent to being a torsion $\mathbb{Z}$-module. If $X$ is a torsion module, then any finitely generated submodule is a finitely generated torsion $\mathbb{Z}$-module : by the structure theorem for finitely generated modules over PID, it's finite. If $X$ is not a torsion module, then any non-torsion element generates an infinite ...


1

You've got the right idea using Lagrange's theorem, but your execution is not quite right, as pointed out by John Hughes. Instead, you might consider proving the more general fact: suppose $n$ is coprime to the order of $G$. Then the map $x \mapsto x^{n}$ from $G$ to itself is bijection. You'll want to use Bezout's lemma, as well as the fact that $G$ is ...


1

Hint: show that two elements of $S_{6}$ are conjugate if and only if they have the same cycle structure.


1

The answer to the first question is no. Let $H$ be any group of Fitting length $n > 1$ and $G = H \times H$. Then, taking $N = H \times \{1\}$, $f(N)=f(G)=n$, but $G/N \cong N$ has Fitting length $n$, so it is not nilpotent.


3

It might be clearer when using different symbols for the tow different actions talked about (and also different from the multiplication symbol in $G$). Even though the very axioms of group action are specifically so that this distinction feels unnecessary, it may be more helpful for this reasoning: Assume we have a left group action $G\times A\to A$, where ...


1

No. This is not true in general. If $G$ is a $\pi$-group, then any subgroup or quotient is a $\pi$-group as well. This follows easily from Lagrange's Theorem. But if you require a normal subgroup $P$ to be a $\pi$-group, then $|G:P|$ could miss some primes. For example, take $G=S_3$ and $P=A_3$. Then $P$ is a 3-group, but the quotient is certainly not.


1

A very group theoretic way to prove your claim is outlined in the following: Prove that for $n$ and $m$ co-prime, that is $\gcd(n,m)=1$, you have an isomorphism $\mathbb Z_{nm} \cong \mathbb Z_{n}\times \mathbb Z_{m}$ (where $\times$ denote the direct product of groups). Prove that the generators of $\mathbb Z_n \times \mathbb Z_m$ are exactly ordered ...


1

Here is a proof that does not use the axiom of choice. Let $e_n\in A$ denote the sequence whose $n$th term is $1$ and all other terms are $0$; we take it as known that any homomorphism $A\to\mathbb{Z}$ which vanishes on each $e_n$ is $0$ everywhere (the standard proofs of this certainly do not use choice). Suppose $A$ were projective. Let $F$ be the free ...


3

$D_{12} \cong C_2 \times S_3$ 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

$H$ 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

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, ...


0

Hint: Look at the orbits of $\left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ and $\left( \begin{matrix} 0 \\ -1 \end{matrix} \right)$. Which lines are missing? How can you conclude?


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.


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

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, ...


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

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

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 $n=\dfrac{ad}{\gcd(ad,bc)}$.


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.


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

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 ...


0

Prove that if $gcd(r_{1}, m) = gcd(r_{2},n) = 1$, and $gcd(m,n)=1$, then $gcd(r_{1}n+r_{2}m, nm) = 1$. Show that the converse also holds. Finish with a counting argument.


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, ...


5

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.


1

The nonzero vector $(x_1,x_2,\ldots,x_n)$ of $X$ generates an irreducible submodule isomorphic to $W_1$ if and only if there are $FG$-isomorphisms $\phi_i:W_1 \to W_i$ with $\phi_i(x_1) = x_i$ for all $i$. Otherwise it generates a reducible submodule isomorphic to a number of copies of $W_1$. The number will be equal to the maximum number of the $x_i$ that ...


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.


1

The statements in the book are not quite precise or correct. In the first statement, by "non-Euclidean spaces" the author means differential manifolds that are not open subsets of $\mathbb{R}^n$, not non-Euclidean geometry of curved Riemannian manifolds. The chain rule is that the differential of a composite function is the composition of the ...


1

Another argument: we can let $G$ act on $H$ by conjugation. Then the orbit-stabilizer theorem gives us $$p = |H| = \text{# of one-point orbits} + \sum (\text{sizes of nontrivial orbits})$$ Now $h \in H$ is in a one-point orbit iff $ghg^{-1} = h$ for every $g \in G$, iff $gh = hg$ for every $g \in G$, if and only if $h \in H \cap Z(G)$. Therefore, $$p = |H| = ...


0

I think your mistake is thinking that $\cdot$ is group multiplication, when it is in fact conjugation. For instance, $g\cdot e=geg^{-1}=e$ so the Orbit is $\{e\}$. $g\cdot e=e$ implies $geg^{-1}=e$ which holds for all $g\in G$, so the stabilizer is $G$.


2

Suppose $g$ does not commute with $x$. Then you get a non-trivial aciton of $K=\langle g\rangle$ on $H$ by conjugation. This means a non-trivial morphism $K\to Aut(H)$. Now try to see that $Aut(H)\simeq \mathbb{Z}/(p-1)\mathbb{Z}$, and try to show that any morphism $K\to \mathbb{Z}/(p-1)\mathbb{Z}$ must be trivial (using the fact that $p$ is the smallest ...


0

The following proof, interestingly enough, essentially comes from Section 1.1 of EBBINGHAUS-FLUM-THOMAS Mathematical Logic: First, fix a left identity $e$ (which is not unique, in principle). Let's show that for each $x\in S$ there exists $y\in S$ with $xy=e$. More precisely, any left inverse of $x$ is a right inverse of $x$. Given $x$, choose $y$ for ...


0

The group multiplication is a binary operation on the set of elements of the group, associating to each pair $\langle u,v\rangle$ of elements of $G$ a new element $uv$. When you have $b=c$, you are fixing, let's say, the right argument of this binary operation, and what you are left with is an ordinary function of one argument: the left-multiplication by ...


0

You've got to remember that you are assuming that $G$ is abelian from the start. Therefore the elements of $A$ commute with each other.


4

Of course the other direction holds. The other direction has nothing to do with groups or with anything really; it's just what equality means! Saying $a=b$ means that $a$ and $b$ are exactly the same thing; if $a$ and $b$ are the same thing then $ac$ and $bc$ are the same.


1

Yes the other way holds. This is essentially just the fact that multiplication is well-defined. The reason that they do not list the cancellation law as an if and only if is that the other way doesn't actually have to do with cancellation.



Top 50 recent answers are included