# Tag Info

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### 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 ...
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### 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 $3$, ...
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### 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 $$C\cong B/\ker g=B/\operatorname{im}f.$$
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### 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 ...

### 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 $40.$

### 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 $\ker g=\operatorname{im} f$.
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### 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\...

### 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, ...
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### 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....
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1 vote

1 vote
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### 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 vote
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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 $\prod_{p\... 1 vote Accepted ### The notation$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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