Intereting Posts

Does (Riemann) integrability of a function on an interval imply its integrability on every subinterval?
Can you raise a number to an irrational exponent?
Proving that a sequence is Cauchy (sequence including factorials).
Can we found mathematics without evaluation or membership?
Question about trigonometry/trigonometry question?
Example of a simple module which does not occur in the regular module?
What is the accepted syntax for a negative number with an exponent?
Does the Levi-Civita connection determine the metric?
Are there any Combinatoric proofs of Bertrand's postulate?
Proof of Inequality using AM-GM
Can you make money on coin tosses when the odds are against you?
Can infinitely many primes lie over a prime?
Convergence/Sequences/Box Topology
Identify quotient ring $\mathbb{R}/(x^2-k), k>0$
Question about Geometric-Harmonic Mean.

What is an example of a $\mathbb{Z}$-module which has exactly three proper submodules?

- Can a cubic equation have three complex roots?
- System of linear equations having a real solution has also a rational solution.
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- Show that $G$ is cyclic
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- If $M$ is an artinian module and $f : M\to M$ is an injective homomorphism, then $f$ is surjective.
- Group isomorphism concerning free group generated by $3$ elements.
- Euler's remarkable prime-producing polynomial and quadratic UFDs
- Index of a sublattice in a lattice and a homomorphism between them

$\mathbb{Z}/6\mathbb{Z}$ has exactly three proper submodules, namely $\{0,2,4\}$ (which is isomorphic to $\mathbb{Z}/3\mathbb{Z}$, $\{0,3\}$ (which is isomorphic to $\mathbb{Z}/2\mathbb{Z}$) and $\{0\}$.

To see that this are all $\mathbb{Z}$-submodules, i.e. subgroups of $\mathbb{Z}/6\mathbb{Z}$, notice that because $\mathbb{Z}/6\mathbb{Z}$ is cyclic every subgroup is also cyclic. Then $0$ generates $\{0\}$, $2$ and $4$ generate $\{0,2,4\}$, $3$ generates $\{0,3\}$, and $1$ and $5$ generate $\mathbb{Z}/6\mathbb{Z}$ itself.

PS: Another examples is $\mathbb{Z}/8\mathbb{Z}$; the nice thing about this example is that we can generalize it to $\mathbb{Z}/2^n\mathbb{Z}$, which has exactly $n$ proper subgroups.

A $\mathbf Z$-module is but an abelian group. For any prime $p$, $\mathbf Z/p^3\mathbf Z$ is an example of an abelian groups with three proper subgroups, which are linearly ordered. These subgroups are

$$0\subsetneq p^2\mathbf Z/p^3\mathbf Z\simeq \mathbf Z/p\mathbf Z\subsetneq p\mathbf Z/p^3\mathbf Z\simeq\mathbf Z/p^2\mathbf Z\subsetneq \mathbf Z/p^3\mathbf Z.$$

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