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The cyclic group is not isomorphic to
Counterexample
The groups and both have four elements, but they are not isomorphic: the first has an element of order , whereas every nonidentity element of the second has order .
Facts & Assumptions
Given: The additive quotient groups and .
The residue classes modulo form the quotient group of the additive integers (For every , the congruence-class group is the quotient group ).
In a direct product, a pair with finite component orders has order the unique positive natural whose canonical integer image is (If and have finite orders and , then in ).
A group isomorphism is a bijective homomorphism, and a homomorphism preserves powers (Group isomorphisms, automorphisms and the set , A group homomorphism automatically satisfies and , and for every ; for monoid homomorphisms preservation of the identity must be assumed).
Refutation
The class in has order , while the nonzero class in has order .
By [L2], each nonidentity pair has component orders or . Their least common multiple is , so every nonidentity pair has order .
A group isomorphism preserves the least positive exponent at which a power is the identity, by [L3]; it therefore cannot send the order- element of step 1.1 to any element of the target.
Hence and are not isomorphic.
Depends on
- If $g$ and $h$ have finite orders $m$ and $n$, then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$
- For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$
- A group homomorphism automatically satisfies $f(e) = e'$ and $f(g^{-1}) = f(g)^{-1}$, and $f(g^{n}) = f(g)^{n}$ for every $n \in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed
- Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 84 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Judson, Abstract Algebra: direct products (standard reference, not scraped)