Subgroup of Additive Group Modulo m is Ideal of Ring

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m \in \Z: m > 1$.

Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.


Then every subgroup of $\struct {\Z_m, +_m}$ is an ideal of the ring of integers modulo $m$ $\struct {\Z_m, +_m, \times_m}$.


Proof

Let $H$ be a subgroup of $\struct {\Z_m, +_m}$


Suppose:

$(1): \quad h + \ideal m \in H$, where $\ideal m$ is a principal ideal of $\struct {\Z_m, +_m, \times_m}$

and

$(2): \quad n \in \N_{>0}$.


Then by definition of multiplication on integers and Homomorphism of Powers as applied to integers:

\(\ds \paren {n + \ideal m} \times \paren {h + \ideal m}\) \(=\) \(\ds \map {q_m} n \times \map {q_m} h\) where $q_m$ is the quotient mapping
\(\ds \) \(=\) \(\ds \map {q_m} {n \times h}\)
\(\ds \) \(=\) \(\ds \map {q_m} {n \cdot h}\)
\(\ds \) \(=\) \(\ds n \cdot \map {q_m} h\)


But:

$n \cdot \map {q_m} h \in \gen {\map {q_m} h}$

where $\gen {\map {q_m} h}$ is the group generated by $\map {q_m} h$.

Hence by Epimorphism from Integers to Cyclic Group, $n \cdot \map {q_m} h \in H$.

The result follows.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Subgroup_of_Additive_Group_Modulo_m_is_Ideal_of_Ring&oldid=526620"