In geometry, $\cong$ means congruence of figures, which means the figures have the same shape and size. (In advanced geometry, it means one is the image of the other under a mapping known as an "isometry", which provides a formal definition of what "same shape and size" means) Two congruent triangles look exactly the same, but they are not the ...
Prove that $\mathbb Z_ {m}\times\mathbb Z_ {n} \cong \mathbb Z_ {mn}$ implies $\gcd (m,n)=1$. This is the converse of the Chinese remainder theorem in abstract algebra.
abstract algebra - Prove that $\mathbb Z_ {m}\times\mathbb Z_ {n} \cong ...
I went through several pages on the web, each of which asserts that $\operatorname {Aut} A_n \cong \operatorname {Aut} S_n ; (n\geq 4)$ or an equivalent statement without proof, and many of them seem to regard it as a trivial fact.
This approach uses the chinese remainder lemma and it illustrates the "unique factorization of ideals" into products of powers of maximal ideals in Dedekind domains: It follows $-1 \cong 10-1 \cong 9$ hence you get a well defined map
$\Bbb Z [i]/ (a+bi)\cong \Bbb Z/ (a^2+b^2)$ if $ (a,b)=1$. Gaussian ...
I've just started to learn about the tensor product and I want to show: $$ (\mathbb {Z}/m\mathbb {Z}) \otimes_\mathbb {Z} (\mathbb {Z} / n \mathbb {Z}) \cong \mathbb ...
The proof given for the theorem is the following: "The description of line bundles in terms of their cocycles provides us with an isomorphism $\operatorname {Pic} (X) \cong \check {H^1} (X,\mathcal {O}^*_X) $ ".