6
votes
Accepted
Proving an algebraic problem categorically
That argument is correct.
Note that (more or less by definition) the $\mathrm{Ab}$-functor is left-adjoint to the inclusion functor from abelian groups to groups. From that perspective, the property ...
- 247
5
votes
Accepted
What is wrong with my argument that every group of order $pq$ is abelian?
The subgroups $H$ and $K$ may not be normal, so you cannot take the quotients and conclude that they contain the commutator subgroup.
For instance, when $G=S_3$ it has subgroups of order $2$ and $3$, ...
- 326k
5
votes
Accepted
Let $0\to\ker f\to A \xrightarrow fB\to C \to 0$ be an exact sequence of abelian groups. Can we say that $C\cong\operatorname{coker}f$?
The answer is yes. Let $g$ denote the surjection from $B$ onto $C,$ then $$C\cong B/\ker g=B/\operatorname{im}f.$$
- 29.6k
4
votes
Accepted
Let $p$ and $q$ be twin primes, both greater than or equal to $5$ , is every group of order $p^2q^2$ abelian?
Your argument has some errors. In particular you claim to have proved that every group of order $p^2q^2$ is isomorphic to $\mathbb Z_{p^2q^2}$, but there are other 3 non-ismorphic abelian groups of ...
3
votes
find the order of the element $a^{-1}b?$
The order of $a^2$ and $b^2$ are respectively $4$ and $5,$ which are coprime. The order of $(ab)^2$ is therefore $20,$ which is even. This proves that the order of $ab$ is exactly $40.$
- 29.6k
3
votes
Is this Monoid Finitely Generated?
The answer appears to be yes.
$\textbf{Lemma:}$ Let $A$ be a non-empty subset of $\mathbb{N}^n$ equipped with the partial order $(a_1, \ldots ,a_n)\leq(b_1, \ldots, b_n)$ iff $a_i\leq b_i$ for all $i=...
- 3,151
3
votes
Accepted
Let $A$ be a 2-torsion finite abelian group . Let $f: A\to (\Bbb{Q}/\Bbb{Z})^n$ be an arbitrary group homomorphism. $image(f)\le 2^n$?
The $2$-torsion part of $\mathbb{Q}/\mathbb{Z}$ is $\left\langle \frac{1}{2}+\mathbb{Z}\right\rangle\cong\mathbb{Z}/2\mathbb{Z}$.
Indeed, if $\frac{a}{b}+\mathbb{Z}$ has order dividing $2$, with $\gcd(...
- 388k
2
votes
Let $0\to\ker f\to A \xrightarrow fB\to C \to 0$ be an exact sequence of abelian groups. Can we say that $C\cong\operatorname{coker}f$?
Let $g\colon B\to C$ be the given map. We know that $g$ is surjective, hence $B/\ker g\cong C$. But $\ker g=\operatorname{im} f$.
- 18.3k
2
votes
Accepted
Let $A,B, C,A',B'$ be an abelian groups. Let $f:A\to B$ and $g:B\to C$ be group homomorphisms.
Preliminary fact.
Consider a commutative square of abelian groups as below, where the horizontal homomorphisms are isomorphisms: $\require{AMScd}$
$$\begin{CD}
M@>\sim>>M'\\
@V\alpha VV @V\...
- 766
2
votes
Let $p$ and $q$ be twin primes, both greater than or equal to $5$ , is every group of order $p^2q^2$ abelian?
With a more general spin: yes, for twin primes $p$ and $p+2$, where $p \geq 5$, all groups of order $p^2(p+2)^2$ are abelian indeed! In general, one can show that every group of order $n$ is abelian, ...
- 47.7k
2
votes
Accepted
If $G$ is an abelian group of order 187, then every nontrivial subgroup of $G$ has order 11 or 17.
The order of a subgroup divides the order of a group. Since $187=11\cdot 17$, the only possible subgroups have order $1$, $11$, $17$, or $187$. (For an abelian group, all four possibilities will occur....
- 18.3k
2
votes
Accepted
Canonical isomorphism $Hom_{\text{Ab}} (\bigoplus_i A_i, B) \overset{\cong}{\rightarrow} \prod_i Hom_{\text{Ab}}(A_i, B)$
There are canonical "inclusions" $\sigma_j : A_j \to \bigoplus A_i$ (given by $\sigma_j(x) = (x_i)_{i \in I}$ where $x_j = x$ and $x_i = 0$ for $i \ne j$). The system $(\bigoplus A_i, \...
- 72.9k
2
votes
Accepted
A question on commutative diagram of abelian groups
This is false. As someone mentioned in the comments (I noticed a lot of people are writing answers as comments lately, am I missing something?), you can use as both rows $0 \to \mathbb{Z}/2\mathbb{Z} \...
1
vote
Question about proof of uniqueness of decomposition of finite abelian $p$-groups
Identifying $G$ with $A_1\times\dots\times A_k$, an element $x=(x_1,\dots,x_k)$ satisfies $x^p=1$ iff $x_i$ satisfies $x_i^p=1$ in $A_i$ for each $i$. A cyclic group of order $p^n$ has $p$ elements $...
- 326k
1
vote
Accepted
If G is a finite abelian group, prove that $o(G) = o(\hat{G})$ and G is isomorphic to $\hat{G}$.
You wrote
"Since G is finite abelian, we may decompose G to cyclic groups, call them $A_i$, proved in an earlier problem, that $\hat{A}_i$ is also cyclic, and $A_i\cong \hat{A}_i$. As a result, $...
- 56
1
vote
Accepted
What is the definition of a homotopy of homotopies in the category of chain complexes?
I will give as mentioned above a description of a model of the $\infty$-category of chain complexes, by turning for any ring $R$ the category $\mathrm{Ch}_R$ of chain complexes of $R$-modules into a ...
- 1,105
1
vote
Accepted
What conditions do we need to calculate: $\mathrm{Ext}(\Bbb Q,M)\cong\left(\prod_{p\text{ prime}}\hat{M}_p\right)/M$?
I would not say Weibel made an error so much as that his notation is a little imprecise. When he writes $\left(\prod_{p\text{ prime}}\hat{B}_p\right)/B$, he really means the quotient of $\prod_{p\...
- 326k
1
vote
Accepted
The notation $A\otimes_{\Bbb Z}G$ in Robinson's "A Course in the Theory of Groups (Second Edition)".
The tensor product is defined on p. 235 in the chapter on representations (8.4), despite being used earlier.
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