# Is $M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$ free?

I'm trying to solve a question which asks me to determine whether the quotient $$\mathbb{Z}$$-module $$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$ is free. I'm then supposed to find some element of $$M$$ of infinite order.

I'm really new to modules and finding them hard to wrap my head around, so would appreciate any help you could offer/hints you could give me.

EDIT: while a similar question has been asked here - Show that $M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$ as $\mathbb{Z}$-modules - the answers there use the fact $$M$$ is isomorphic to $$\mathbb{Z} \oplus \mathbb{Z}_4$$. I'd like to show that it isn't free more directly if at all possible. Would appreciate any help anyone could offer.

• $(1,1,1)$ is obviously a torsion element. Free modules are torsion-free. – rschwieb Feb 22 at 20:36
• Also, applying elementary row and column operations really is very short and direct, as done at the duplicate. – Dietrich Burde Feb 23 at 12:36
• – ancientmathematician Feb 24 at 13:37