New answers tagged

8 votes
Accepted

Is every ordered abelian group the additive group of an ordered ring?

No, and the classification of subgroups of $\mathbb{Q}$ can be used to find a counterexample. Consider the subgroup $A$ generated by $\frac{1}{p}$ as $p$ runs over all primes (or more generally ...
2 votes

Exponent of noncyclic finite abelian group

Hint One of the $p_i$ has to be repeated in the expression for $G$ as a product of cyclic groups of order $p_i^{\alpha _i}$ (by the fundamental theorem of finite abelian groups). Otherwise we get ...
  • 3,812
0 votes
Accepted

Does every embedding between finite free $\mathbb Z/p^2 \mathbb Z$-modules split?

This follows from the fact that $\mathbb{Z}/p^2\mathbb{Z}$ is injective as a module over itself. More generally, for any $n\neq 0$, $\mathbb{Z}/n\mathbb{Z}$ is injective over itself. To prove this, ...
-1 votes

Show that every commutative group of order 6 is cyclic

There are only two groups of order 6 less isomorphism (there are more groups, but the rest will be isomorphic to only two groups) $S_3$ and the $Z_6$. To reach this conclusion it is enough to employ ...
0 votes

Show that every commutative group of order 6 is cyclic

You can use Cauchy's theorem or its proof. By this theorem, there must be elements of order $2$ and elements of order $3$ in the group. Let $g_2$ and $g_3$ be elements of order $2$ and $3$, ...
0 votes

Show that every commutative group of order 6 is cyclic

The outline of a proof is like that: Assume there is no element of order 6. So all elements except e are of order 2 or 3. a. Show elements of order 3 exist in pairs. b. Show there is 4 element of ...
  • 844
2 votes
Accepted

Do indecomposable abelian groups of inacessible cardinal rank exist?

According to Theorem 2.1 of the following paper, there are indecomposable torsion-free abelian groups of any infinite cardinality. Shelah, Saharon, Infinite Abelian groups, Whitehead problem and some ...
3 votes

Determine the invariant factors of $\mathbb{Z^2}$

For a response that doesn't use exact sequences (although this is the simplest way to get the answer if you have learned them): First, note that there is no torsion, since the first coordinate of $n(x,...
3 votes
Accepted

Determine the invariant factors of $\mathbb{Z^2}$

We have a short exact sequence $$0 \to \mathbb{Z} \overset{y \mapsto (0, y)}{\longrightarrow} G \overset{(x, y) \mapsto x}{\longrightarrow} \mathbb{Z} \to 0.$$ Moreover, since $\mathbb{Z}$ is a free ...
1 vote
Accepted

What does this notation mean: $(\chi(D),p^j)$

I am sorry, but the comment of David A. Craven help me to understand that it is in fact the greatest common divisor but in algebraic number field. In fact, if we accept that this is a GCD, then the ...
2 votes
Accepted

How to finish the proof of: If $N\unlhd G=H\times K$, then $N$ is abelian or $N$ intersects $H \times \{e\}$ or $\{e\} \times K$ non trivially

Suppose $N$ is normal, and nonabelian. Since it is nonabelian, there exists an element $(h,k)\in N$ such that either $h$ is not central in $H$, or $k$ is not central in $K$ (if both $h\in Z(H)$ and $k\...
2 votes

Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

This is done in three steps: $H,K\unlhd G$ since each subgroup of an abelian group is normal. Let $g\in H\cap K$. Since $g\in H$, we have $\phi(g)=g$, but $g\in K$, so $0=\phi(g)=g$. Thus $H\cap K=\{...
  • 40.3k
1 vote

Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

Hint The short exact sequence $$1\to K\hookrightarrow G\overset{\phi}{\to }H\to1$$ is left split. The morphism $\psi:G\to K$ being $\psi(x)=x-\phi(x)$.
  • 3,812
3 votes
Accepted

Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

Let $x\in G$. Since $\phi:G\to H$, know that $\phi(x)\in H$ so we can think of $\phi$ as extracting an $H$-content of $x$. Then $\phi(\phi(x))=\phi(x)$ because $\phi$ is identity on $H$. This is the ...
1 vote

How to formally prove that an abelian group generated by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

Say we have $k$ (column) vectors in a space $F^d$, where $F$ is a field. They are linearly independent over $F$ if and only if the $d\times k$ matrix $M$ formed with these $k$ columns has rank $k$, ...
  • 48.6k
1 vote

How to formally prove that an abelian group generated by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

An approach using linear algebra only. Let $V$ be the $\Bbb{R}$-span of $A$. The familiar Gram-Schmidt process (but without the normalization step!!) gives us an orthogonal basis $u_1,u_2,\ldots,u_k$ ...
5 votes
Accepted

How to formally prove that an abelian group generated by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

The setup and desired condition are both invariant under change of coordinates (because $GL_n(\mathbb{R})$ acts by homeomorphisms) so up to a suitable change of coordinates we can assume that the $v_i$...
1 vote

How to formally prove that an abelian group generated by $k$ linearly independent vectors $v_1,\cdots, v_k$ in $\mathbb R^d (k\le d)$ is discrete?

If you can have an integer linear combination that comes arbitrary close to the origin you can always divide by the maximal coefficient, to get a bounded set of coefficients so that the linear ...
  • 24.4k
2 votes

Why does $P_{17}$ being abelian imply that $P_{17} \leq N_G(P_3)$?

Here's another way to see it, without using automorphism groups. Since there is a unique Sylow $17$-subgroup $P$ it is normal in $G$, and $G/P$ has order $15$. It is easy to see that groups of order $...
2 votes

Why does $P_{17}$ being abelian imply that $P_{17} \leq N_G(P_3)$?

I propose another solution. We know that $P_{17}\lhd G$. Let $f:G\to \operatorname{Aut}(P_{17})$ be a homomorphism defined by the rule $f(g)(x)=gxg^{-1}$. Since $ |\operatorname{Aut}(P_{17})|=16$ and $...
  • 5,883
2 votes

An "almost" metric via a generalized definition of absolute values

If you stare at the metric space axioms carefully you can see exactly what is being used about $[0, \infty)$ that allows us to state them: all we need is the operation $+$ and the order $\le$. So we ...
2 votes
Accepted

Abelian torsion-free group whose every quotient is reduced

Suppose $G$ is a nontrivial torsion-free abelian group. Then $G$ has a quotient $H$ which is torsion-free of rank $1$ (just take the image of any nonzero homomorphism $G\to\mathbb{Q}$), which we may ...
0 votes

Let $G$ be a finite abelian group with a transitive faithful action on set $X,|X|=n$. Prove $|G|=n.$

By action's transitivity and group's abelianess, the kernel of the action is given by $\bigcap_{y\in X}\operatorname{Stab}(y)=$ $\bigcap_{g\in G} g\operatorname{Stab}(x)g^{-1}=$ $\operatorname{Stab}(x)...
  • 216
0 votes
Accepted

What are the torsion coefficients of $\Bbb Z_{30}\oplus \Bbb Z_{18}\oplus\Bbb Z_{75}?$

Using only what you know, we have $$(30,18,75) \to (2,3,5,2,9,3,25) \to (3,30,450).$$.
  • 81.1k
3 votes

Prove that the group generated by sum of two subgroups is isomorphic with their cartesian product

Lemma: Let $G$ be a group and $H\le G, K\triangleleft G$ then $H\vee K=HK$ Proof: $HK\subset H\vee K$. $HK\le G$. $H\vee K$ is the smallest subgroup of $G$ containing both $H$ and $K$. $\square$ ...
  • 10.7k
2 votes

Prove that the group generated by sum of two subgroups is isomorphic with their cartesian product

One way to define the (internal) direct product $G=H\bowtie K$ of two subgroups $H,K$ of a group is that all of the following is satisfied: $G=HK$. $H\cap K=\{e\}.$ $H,K\unlhd G$. It is a standard ...
  • 40.3k

Top 50 recent answers are included