I came across this sign when reading some papers. I looked up Wikipedia. It says "The symbol "$\\cong$" is often used to indicate isomorphic algebraic structures or congruent geometric …

Oct 11, 2014 · 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 ...

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.

Feb 10, 2025 · Let $R$ be a ring with unity, and let $e$ be an idempotent element of $R$ such that $e^2 = e$. If $e$ is a central idempotent of $R$, then we obtain the following ring …

Aug 22, 2023 · Q1: Yes, this is the definition of the determinant of a one-dimensional vector space. Q2: Yes, the dual of the trivial line bundle is the trivial line bundle (for instance, use that …

If $Aut (G)\cong \mathbb {Z}_n$ then $Aut (G)$ is cyclic, which implies that $G$ is abelian. But if $G$ is abelian then the inversion map $x\mapsto x^ {-1}$ is an automorphism of order $2$.

The most obvious way to show this is by using the Chinese Remainder Theorem to see that $ {\mathbb Z}_ {12} \cong {\mathbb Z}_4 \times {\mathbb Z}_3$ (as rings) and therefore also $ …

Dec 5, 2025 · In the fifth example in page $19$ the author explains why $\mathrm {SU} \left (2 \right) \cong \mathbb {S}^ {3}$ as real Lie groups. There is a step I don't seem to understand …

Sep 28, 2024 · Claim: $\operatorname {Hom}_ {G} (V,W) \cong \operatorname {Hom}_ {G} (\mathbf {1},V^ {*} \otimes W)$ I'm looking for hints as to how to approach the proof of this claim.

In mathematical notation, what are the usage differences between the various approximately-equal signs "≈", "≃", and "≅"? The Unicode standard lists all of them inside the Mathematical …