Choose essential curves $\alpha,\beta$ on $T^2$ with $i(\alpha,\beta)=1$. They fill the torus. Let $[f]$ be central in $\operatorname{Mod}(T^2)$. For every curve $\gamma$,
$$
fT_\gamma f^{-1}=T_{f(\gamma)}.
$$
We also use that equality of Dehn twists about essential curves implies equality of their unoriented isotopy classes. Since $[f]$ commutes with $T_\alpha$ and $T_\beta$, it preserves both $\alpha$ and $\beta$.

The structure graph of this filling pair has one intersection vertex in the embedded union. Its orientation-preserving symmetries induced by a torus homeomorphism are the identity and simultaneous reversal of both curves. By the <Alexander method>, the corresponding mapping classes are the identity and the elliptic involution
$$
\iota:T^2\longrightarrow T^2,
\qquad x\longmapsto-x.
$$
The involution commutes with every torus mapping class, as is also clear from the identification $\operatorname{Mod}(T^2)\cong SL_2(\mathbb Z)$ where it is $-I$. Therefore the <center of the mapping class group of the torus> is
$$
\boxed{Z(\operatorname{Mod}(T^2))=\{1,\iota\}\cong C_2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 156 / 1 / c / Solution

{title2=$Z(\operatorname{Mod}(T^2))=\{1,\iota\}$}

The center of $\operatorname{Mod}(T^2)$ has order two. A central class commutes with twists about two curves intersecting once, hence preserves both curve classes. The <Alexander method> reduces it to the identity or the elliptic involution $\iota:x\mapsto-x$, and both are central.

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 Incoming links: Center of the mapping class group of the torus