Essential simple curves on a torus are primitive

Give $T$ the one-vertex square cellulation with product generators $a=S^1\times \{1\}$ and $b=\{1\}\times S^1$. The cellular boundary of the square is the commutator $ab{a}^{-1}{b}^{-1}$, whose signed incidence on each 1-cell is zero; the 1-cell boundary is also zero. Thus $H_1(T;\mathbb{Z})=\mathbb{Z}[a]\oplus \mathbb{Z}[b]$, and the two product projections read the two coordinates. For a PL loop $c:S^1\to S^1$, lift its finitely many arcs successively to the covering line $\mathbb{R}\to S^1$; the endpoint displacement is an integer, its degree. A degree-zero lift closes and contracts linearly, so $c$ is null-homotopic; conversely a null-homotopy lifts over a finite subdivision and keeps the integer displacement zero. Applying this to both coordinate projections shows that if $C$ is not null-homotopic, at least one homology coordinate of $[C]$ is nonzero. If $C$ separated $T$, the closure of either complementary component would be an oriented PL surface with boundary $C$; its finite oriented triangulation gives a $2$-chain with boundary $C$, so $[C]=0$. Hence $C$ is nonseparating, and cutting along it leaves a connected compact surface with two boundary copies.

Choose a vertex $p$ of a triangulation of $C$, and let $p_{+}$ and $p_{-}$ be its two copies on the boundary of the cut surface. A shortest edge path joining these corresponding copies has no repeated vertices; push its interior off any intermediate boundary edges to obtain a properly embedded PL arc. After gluing the boundary copies back, its endpoints become the same point $p$ and the arc crosses $C$ transversely exactly once. Orient the resulting simple closed PL curve $D$ arbitrarily; its signed intersection with $C$ is $+1$ or $-1$.

For oriented transverse PL loops on an oriented torus, the signed crossing count is invariant under replacing either loop by a homologous cycle: across the boundary of each oriented triangle, the intersections with a closed transverse loop enter and leave with opposite signs. It is additive in each cycle. The product basis loops $a=S^1\times \{1\}$ and $b=\{1\}\times S^1$ meet once, while parallel copies of each are disjoint, so for classes $(p,q)$ and $(r,s)$ the pairing is $ps-qr$ up to the global orientation convention. Since $C\cdot D=\pm 1$, this determinant is $\pm 1$.

If an integer $k$ with $\mathrm{\mid }k\mathrm{\mid }>1$ divided both $p$ and $q$, it would divide $ps-qr$, contradicting its value $\pm 1$. Thus $\gcd (\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid })=1$, which says exactly that $[C]$ is primitive. Its nonzero status was established in step 1.1.