Tag Info

New answers tagged

2

I've been sleeping with Silverman and Tate's Rational Points on Elliptic Curves under my pillow for years now, with no effect. So here's the simple country coder's approach. (Spoiler: yes, 108 points.) To construct the field $\mathbb{F}_{125}$ we need a cubic irreducible over $\mathbb{F}_5$. As irreducibility for a cubic amounts merely to the absence of ...


0

As others have noted it is not hard to show that $\phi: G \to T$, $g \mapsto (g,g)$ is a homomorphism. However I'd like to explicitly point out that there is an easy way to show this is bijective, when asked if something is bijective it is sometimes easier to just show it has an inverse function: Define $\psi: T\to G$ by sending $(g,g)$ to $g$. Alternative ...


1

The real group algebra of the quaternion group of order $8$ is the direct sum of $4$ copies of the real field and one copy of the $4$-dimensional division algebra of real quaternions.


2

No. There is a famous counter-example, due to Malcev. A necessary condition that $M$ embeds in a group is that, for any elements $a,b,c,d,x,y,u,v\in M$, from $$ax = by, cx = dy, au = bv,$$ it follows that $$cu = dv.$$ Malcev calls this property "Z". It is easy to see that any monoid that embeds in a group satisfies property Z. Malcev's counter-example ...


4

Hints: (1) Consider the sign of a permutation. (2) What is the order of $S_2$?


4

Define $f: T\to G$ by $f(g,g)=g$. Then $f((g_1,g_1)(g_2,g_2))=f((g_1g_2,g_1g_2))=g_1g_2=f((g_1,g_1))f((g_2,g_2))$ So $f$ is a group homomorphism. To show oneone ; $f((g_1,g_1))=f((g_2,g_2))\Rightarrow g_1=g_2\Rightarrow (g_1,g_1)=(g_2,g_2)$ To show onto: $\forall g\in G$ we have $(g,g)\in T$ and $f((g,g))=g$ So $f$ is an isomorphism.


2

Outline: Define $\varphi: T\to G$ by $\varphi(g,g)=g$. Show that $\varphi$ is an isomorphism. Checking that $\varphi$ is an isomorphism will not be difficult. Note that $T$ is not a subgroup of $G$, it is a subgroup of $G\times G$. There is no need to assume that $G$ is Abelian. The result holds for all groups.


3

Hint: Let $\phi\colon G \to P$ be the mapping that sends each element in $G$ to the corresponding set that it belongs to in $P$. It's not too hard to see that $\phi$ is a surjective homomorphism. Thus, it follows by the First Isomorphism Theorem that: $$ G / \ker \phi \approx \phi(G) = P $$ which is what we want, since kernels are always normal subgroups.


2

You've determined that the part of $P$ containing the identity, call it $N$, is a subgroup. Next on the list of things to prove: (1) $N$ is normal, (2) parts of $P$ are cosets of $N$, (3) the group operation. (1): You want to show $gNg^{-1}=N$ for any given $g\in G$. Consider $[g]\cdot N\cdot [g^{-1}]$. (2) Say $g\in G,n\in N$. You want to show ...


5

Hint make a Cartesian product of $Z_3$ by itself enough times.


0

So thinking of groups as having something to do with numbers isn't really a great idea. Yes, the integers are an example of a group, but they're a very degenerate / misleading one in a number of ways. The definition of a group abstracts the idea of a group of symmetries: a group element is a symmetry (or "automorphism") of some other object. A symmetry is ...


3

In virtue of the Cauchy's theorem for groups, every group with order $23$ is cyclic, hence isomorphic to $C_{23}$. On the other hand, $C_{25}$ and $C_5\times C_5$ are not isomorphic, since in the latter there is no element with order $25$.


2

$G = G_\alpha G_\beta$ is equivalent to $G_\alpha$ acting transitively on $B$, which implies that any conjugate of $G_\alpha$ in $G$ is transitive on $B$. But the conjugates of $G_\alpha$ are the stabilizers $G_x$, so we get $G = G_x G_\beta$ for all $x \in A$. Then, conjugating this equation by elements of $G$ gives $G = G_x G_y$ for all $x \in A$, $Y \in ...


3

An orthogonal matrix $O^TO={\bf 1}$ must automatically have $\det O = \pm 1$. Thus the set $O(n)$ of orthogonal matrices consists of (at least) two disconnected components, one with determinant $+1$ and one with determinant $-1$. One may show that each of the two components is a $\frac{n(n-1)}{2}$ dimensional real manifold. If we look at the component ...


0

You have proven that $H$ is closed under inverses. To prove that $H$ is not closed under addition, consider $(x, y)+(-x, -y)$. (Considering this product actually gives you two reasons why your subset isn't a subgroup! Also, you should realise that this subset is precisely $\mathbb{R}^2\setminus(0, 0)$...)


0

What you’ve written so far seems good to me, (modulo the now-fixed mix-up). To go on: Try to find a group arrow $G → F$. Use the universal property of the free group on $X$ (of which $G$ is a factor group). Then use the first isomorphism theorem (or the universal property of factor groups, whatever you call it). Then don’t show that $f$ is monic, but that ...


1

With AllSubgroups, the correct form is shown below: gap> g:=GL(2,7); GL(2,7) gap> f:=AllSubgroups(g);; gap> Length(last); 1704 gap> h:=Filtered(f,t->Size(t)=42);; gap> Length(h); 96 gap> h[1]; Group([ [ [ Z(7)^2, Z(7)^3 ], [ 0*Z(7), Z(7) ] ], [ [ Z(7)^4, Z(7)^2 ], [ 0*Z(7), Z(7)^2 ] ], [ [ Z(7), Z(7)^2 ], [ Z(7)^5, Z(7)^3 ] ] ]) ...


2

The other answers have shown that such a map must be identically zero. If we only require the map to be a homomorphism on the positive integers, the question is slightly more interesting: the positive integers as a multiplicative monoid are generated by the primes, and FTArithmetic means that any set map from the primes to $\mathbb N_0$ extends uniquely to a ...


2

Interestingly, there are no such maps if we remove that zero from the picture. That is, there are no maps $\phi:\mathbb{N}\rightarrow\mathbb{N}$ such that $\phi(ab)=\phi(a)+\phi(b)$ for all $a,b\in\mathbb{N}$. This is because, by taking $a=1=b$, we again see that $\phi(a)=0$ for all $a$, a contradiction as $0\not\in\mathbb{N}$. This argument does not imply ...


9

I may be missing something- but it seems that the only map with this property is identically $0$. The proof is as follows: $\phi(0.0)=\phi(0)+\phi(0)$ which implies that $\phi(0)=0$. Now, for any $a \in \mathbb{N}$, $\phi(a.0)=\phi(a)+\phi(0)$ which means $\phi(0)=\phi(a)+\phi(0)$ which implies $\phi(a)=0$.


1

Number of choices for $b$ are $7$ and that for $a$ are $6$, therefore the total number such matrices is $42$. This shows $|G|=42$. Use the following to get the the matrices of different orders. $$ \begin{bmatrix} 1 & b\\ 0 & a \end{bmatrix}^n = \begin{bmatrix} 1 & b\left(\frac{a^n-1}{a-1}\right)\\ 0 & a^n \end{bmatrix} $$


2

Note that there is an obvious bijection $G\to \Bbb Z_7\setminus\{0\}\times\Bbb Z_7$ (can you find it?). It follows that $$ |G|=|\Bbb Z_7\setminus\{0\}|\times|\Bbb Z_7|=6\times 7=42=2\times 3\times 7 $$ Thus the only primes dividing $|G|$ are $2$, $3$, and $7$. For the second part of your question note that $$ A = \begin{pmatrix}1 & b\\ 0 & ...


2

Hint. For elements of order $2$ you need $$\pmatrix{1&b\cr0&a\cr}^2=\pmatrix{1&0\cr0&1\cr}\quad\hbox{but}\quad \pmatrix{1&b\cr0&a\cr}\ne\pmatrix{1&0\cr0&1\cr}\ .$$ That is, $$\pmatrix{1&b+ab\cr0&a^2\cr}=\pmatrix{1&0\cr0&1\cr}\quad\hbox{but}\quad ...


1

Let $R$ be a ring and $M$ an $R$-module. Suppose $u,v\in R^\times$ are units. Then for any $\pi\in R$, $$\frac{M}{u\pi\color{Green}{vM}}=\frac{M}{u\pi\color{Green}{M}}\cong \frac{\color{Blue}{u^{-1}M}}{\pi M}=\frac{\color{Blue}{M}}{\pi M} \implies \frac{M}{u\pi v M}\cong\frac{M}{\pi M}.$$ The isomorphism is given by multiplication-by-$u^{-1}$. Let $M=\Bbb ...


2

$H$ is a subgroup of $K$, so consider $K/H$, the set of cosets of $H$ in $K$. This set is finite because $[K:H] = [G:H]/[G:K]$. Moreover, if $[G:H] = [G:K]$ then this forces $[K:H] = 1$, so $K=H$.


3

Notice that you can get from $(0,0)$ to $(1,1)$ by following one red arrow and one blue arrow. Repeat this operation with the aid of the diagram until you get back to $(0,0)$, and count the number of steps. That will give you the order of $(1,1)$.


1

Let $x,y\in S$, and consider the sets $x^{-1}S$ and $y^{-1}S$, now remark that $1\in x^{-1}S\cap y^{-1}S$ as $\{aS\}_{a\in G}$ is a partition of $G$, then $x^{-1}S=y^{-1}S$, since $x^{-1}\in x^{-1}S$ we get $x^{-1}\in y^{-1}S$, there exists $s\in S$ such that $x^{-1}=y^{-1}s$, so $x^{-1}y=s\in S$. It follow that $S$ is a subgroup of $G$.


1

If you are not quite satisfied with $\Bbb Q$ under addition, you can also take the Prufer $p$-group for some prime $p$. The Prufer $p$-group is usually realized as a subset the unit circle of the complex plane under complex multiplication, as seen here. That a generating set for this group doesn't contain minimal generating set lends itself to be more ...


0

Non-artinian modules with infinitely many generators will give you a counter example then( not all of them but thats where you check first). like @MJD said $\langle \Bbb Q, +\rangle$ gives us counter example.


1

Hints: Assuming the group operation on $P_D$ is the symmetric difference, don't bother looking for a counter-example. Note that the identity element of $P_D$ is the empty set $\emptyset$ and each element is its own inverse. Apply the subgroup criterion: a subset $H$ of a group $G$ is a subgroup if, and only if, it is non-empty and $xy^{-1}\in H$, for all ...


1

There is a whole area, originally due to Conway and extended by Borcherds, in which positive integral lattices (quadratic forms) are embedded in indefinite or Lorentzian lattices. This is being continued, in particular, by Daniel Allcock at U. Texas, Austin. I can't say I understood everything, but help from Borcherds and Allcock led to a nice little paper; ...


5

Lagrange's theorem says the order of an element divides the order of the group, so $$\text{ord}(x)|pq\implies \text{ord}(x^q)|pq/q=p$$ so $x^q$ is in some group of order $p$. However, there is not necessarily only $1$. Take, for example, $S_3$, the symmetric group on three symbols, which has three conjugate groups of order $2$, which one would you pick ...


1

Hint: if $x^q$=e then $x^q\in G_p$ If $x^q\neq e$, then the order of $x$ is $p$....


1

Okay, so over here I found a nice solution: $$x^2ax=a^{-1}$$ $$xax=x^{-1}a^{-1}$$ $$xaxax=x^{-1}a^{-1}ax=e$$ $$xaxaxa=a=(xa)^3$$ This doesn't seem to rely directly on that hint in the problem, but it looks like it works. Also, you may want to mark my question as a duplicate since I discovered a very similar question in that link. Thanks


6

The theory of $L$- and $\zeta$-functions gives the following result (tributed to some combination of Hasse, Weil and Davenport). The number of points on an elliptic curve defined over $\Bbb{F}_q$ can be written in the form $$ |E(\Bbb{F}_q)|=q+1-\omega_1-\omega_2, $$ where the complex numbers $\omega_1,\omega_2$ are conjugates of each other, and satisfy ...


1

To expand on their hint a bit: $(xax)^3$ $= (xax)\cdot(xax)\cdot(xax)$ $= xax^2axxax$ $=xa(x^2ax)xax$ $= xaa^{-1}xax$ $=\ldots$? Of course, once you have this, you'll have a cube root for $a^{-1}$; you need to show why this implies that $a$ itself has a cube root. You should be able to use what you know about cube roots in $\mathbb{R}$ as a model for this; ...


2

Hint $\,\ a^{\large 3}= 1\,\Rightarrow\,a^{\large 3n}=1\,\overset{\times\, a}\Rightarrow\,a^{\large 1+3n} = a\, $ is a square if $\ 2\mid 1\!+\!3n,\,$ e.g. $\, n = \,\ldots\ $ QED Remark $ $ Here is the intuition. $\ a^{\large 3}= 1\ $ implies that exponents on $\,a\,$ can be considered mod $\,3,\,$ $$\color{#c00}{a^{\large 3}= 1}\ \Rightarrow\ a^{\large ...


2

$(xax)(xax)=xax^2ax=xaa^{-1}=x$, so $(xax)^3=(xax)(xax)(xax)=x^2ax=a^{-1}$


0

p-regular languages are commonly known as (regular) group languages in the literature since their syntactic monoid is a finite group. If a language is accepted by a permutation automaton, then its minimal DFA is also a permutation group, but this group is transitive (since every state is accessible from the initial state). Thus your subclass is actually ...


1

The "standard" direct summands of $\mathbb Z\oplus\mathbb Z$ are $(1,0)\mathbb Z$ and $(0,1)\mathbb Z$. (It's easy to check that $(1,0)\mathbb Z+(0,1)\mathbb Z=\mathbb Z\oplus\mathbb Z$ and $(1,0)\mathbb Z\cap (0,1)\mathbb Z=\{(0,0)\}$.) Note that if you make a $2\times 2$ matrix with the two vectors you get the identity matrix. Use the above example as a ...


4

Hint: Multiply by $a$ to find $$a^4 = a$$


14

Hint: Try $x$ as a power of $a$ (since these are the only things being assuredly in $G$).


0

If $G_i$ is a group for each $i\in I$ then you can define as product of these groups the set of functions $I\rightarrow\bigcup_{i\in I}G_i$ that send each $i\in I$ to an element of $G_i$. Equipping this set with composition $f.g$ that is prescribed by $i\mapsto f(i)g(i)$ for each $i\in I$ makes it a group.


2

If you mean direct product, answer is yes and you can find the detail in http://en.wikipedia.org/wiki/Direct_product_of_groups


5

As Mike pointed out, $\operatorname{Hom}(G,H)$ is not in general a group. However, it is a group if $H$ is abelian, and this observation can lead you to a rabbit hole. It has a category-theoretical intepretation. Suppose we have a category $\mathcal C$ satisfying some natural assumptions (existence of finite products). Then an object $G$ in $\mathcal C$ is ...


0

Both determinants are 7 and therefore the quotient is a group of order 7. Since such a group is unique, the quotients are isomorphic.


0

In general, the quotient is isomorphic to $\prod_{k=1}^{n}\frac{\mathbb Z}{d_k\mathbb Z}$ where the $d_k$ are the "elementary divisors", i.e. the diagonal coefficients in the Smith Normal Form. To expand on Mariano’s answer in your example : the Smith normal form for $((1,2),(4,1))$ is ${\sf diag}(1,7)$ and $H_1={\sf span}((7,0),(-3,1))$. Similarly, the ...


6

There is, in general, no natural group structure on $\text{Hom}(G_1, G_2)$. If $G_2$ is an abelian group there is one: if $\varphi, \psi \in \text{Hom}(G_1, G_2)$, define $(\varphi+\psi)(x)=\varphi(x)+\psi(x)$. In verifying that this is indeed an (abelian!) group, you will likely see what fails when $G_2$ is not abelian. $\text{Hom}(G_1, G_2)$ is of course ...


1

There are at least two distinct values of "distinct" that I can imagine you might mean, but unless I've misinterpreted your question, I think it's fairly straight-forward in either case. If you mean "distinct up to isomorphism", then take the groups $$C_{n}\times C_{2}^{\infty},\;\;n\in\mathbb{Z}^{+}, n > 1$$ where $C_n$ is cyclic of order $n$. No two ...



Top 50 recent answers are included