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 …

Dec 24, 2025 · The proof that every countable linear order embeds in $ (\mathbb {Q},<)$ can be transposed to proving $ (\mathbb {R},<)\cong (\mathbb {R}\setminus\mathbb {Q},<)$ by assuming …

Originally you asked for $\mathbb {Z}/ (m) \otimes \mathbb {Z}/ (n) \cong \mathbb {Z}/\text {gcd} (m,n)$, so any old isomorphism would do, but your proof above actually shows that $\mathbb {Z}/\text {gcd} …

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 Operators B...

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. Any help would be appreciated.

May 4, 2025 · Yes, this is correct. The isomorphism is not just of $\mathbb {Z}$ -modules (which would just be the same as additive groups), but of rings. However, as user Lullaby points out, your …

Sep 28, 2011 · The symbol $\cong$ can in principle be used to designate an isomorphism in any category (e.g., isometric, diffeomorphic, homeomorphic, linearly isomorphic, etc.).

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 isomorphism: $$ R/ReR …

Jan 1, 2025 · 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 …

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 $ {\mathbb Z}_ …