An algebraic group is not necessarily a topological group, since for an algebraic group, the product map $G \times G \to G$ is continuous when $G \times G$ is given the Zariski topology, which is finer than the product topology.
So does this tell you that the additive group $\mathbb{G}_a$ over an infinite field $k$ is not a topological group? The underlying variety of $\mathbb{G}_a \times \mathbb{G}_a$ is $\mathbb{A}_k^1 \times_k \mathbb{A}_k^1 = \mathbb{A}_k^2$, whose Zariski topology is the cofinite topology, where every open set is also open in the product topology (since $\mathbb{A}_k^1$ also has the Zariski topology).
Let $G = \mathbb G_a$. By definition closed subset in $G \times G$ are finite unions of points and set on the form $g \times G$ and $G \times h$ for some $g, h \in G$. Clearly if $k$ is infinite then the "anti-diagonal" $\Delta' = \{(g,-g) \mid g \in G\}$ is not closed. But this is $+^{-1}(0)$, so $G$ is not a topological group.
(Of course, $G$ is a topological group if $k = \Bbb C$ and if you take the usual topology instead of the Zariski topology)