In algebraic topology, one studies the homology and cohomology of spaces. However, when we study group homology/cohomology, we almost exclusively talk about cohomology. Why is this? Is there an aesthetic reason to prefer cohomology in this context? Does group cohomology just happen to be easier to compute?
I understand that the cup product gives cohomology more structure than homology, but this alone doesn't seem to justify the lopsided amount of attention that group cohomology gets.
Just some random thoughts on homology vs. cohomology.
I think in algebraic topology, cohomology often comes together with homology, as historically, cohomology is usually introduced by explicitly dualizing the homological singular complex $C_\bullet (X)$ (cellular complex, etc.): $$C^\bullet (X;A) = \operatorname{Hom}_\mathbb{Z} (C_\bullet (X), A).$$ In group cohomology, to calculate $$H^n (G, M) = R^n (-)^G (M) \quad\text{and}\quad H_n (G,M) = L_n (-)_G (M)$$ (do you find invariants $(-)^G$ to be more natural and interesting than coinvariants $(-)_G$?), one also usually works with some specific complex (the bar-resolution) and dualizes it to get cohomology, so we have the same "duality" (the universal coefficient theorem; when the action of $G$ is trivial): $$ 0\to \operatorname{Ext}_\mathbb{Z}^1 (H_{n-1} (G,\mathbb{Z}),M) \to H^n (G,M) \to \operatorname{Hom} (H_n (G,\mathbb{Z}), M)\to 0.$$ Thus far, it's not much different from topology.
In general, cohomological theorems and calculations have their homological counterparts, and many texts discuss both, e.g. Chapter 6 of Weibel's book and Brown's "Cohomology of groups".
Probably (?) one exception is interpretation of lower-dimensional (co)homology groups. For instance, $H^1 (G,M)$ often appears in explicit terms, as some $1$-cocycles modulo $1$-coboundaries (crossed homomorphisms modulo principal crossed homomorphisms). Something similar should exist for $H_1 (G,M)$, but the result is not quite as nice, in my opinion.
In some settings homology and cohomology appear together, for instance in Tate cohomology $\hat{H}^n (G,M)$, when $G$ is finite. Tate cohomology puts together homology and cohomology, it also has cup-products defined for all $n\in \mathbb{Z}$, etc.
As in topology, we also have cap-products between cohomology and homology $H^p (G,M) \otimes H_q (G,N) \to H_{q-p} (G, M\otimes N)$.
So I think that the situation with (co)homology of topological spaces is not that different, and it's not quite true that nobody cares about group homology.
If you read, say, Brown's book, it becomes quite clear that cohomology is better than homology. Firstly, as Don Alejo noted, cohomology comes equipped with the cup-product, but also in regard to various finiteness conditions for groups; Brown has the entire chapter (chapter 8) about this. One first defines finiteness conditions in terms of projective resolutions but then identifies it in cohomological terms; at the same time, there is no clear-cut homological interpretation. For instance, a group $G$ has cohomological dimension $cd(G)$ (over ${\mathbb Z}$) at most $n$ if there exists a finite resolution by projective ${\mathbb Z} G$-modules $$ 0\to P_n \to ... \to P_0 \to {\mathbb Z}\to 0. $$ (Why this is the "right" definition is not immediate, read Brown's book to see an answer.) Equivalently, $G$ has cohomological dimension over ${\mathbb Z}$ at most $n$ if $H^i(G, M)=0$ for all $i>n$ and all ${\mathbb Z} G$-modules $M$. There is the "dual" notion of homological dimension defined via injective resolutions and group-homology but it is not as nicely-behaved. For instance Stallings proved that $cd(G)=1$ if and only if $G$ is a nontrivial free group. On the other hand, other locally free groups also have homological dimension $1$, like the group $G={\mathbb Q}$. In higher dimensions, $n\ge 3$, cohomological dimension of a group turns out to be equal to its geometric dimension: A group $G$ is said to have geometric dimension $\le n$ if there exists an $(n-1)$-connected complex $X$ and a (properly discontinuous) free cellular action $G\times X\to X$. In contrast, the best one gets (for $n\ge 3$) with the homological dimension $hd(G)$ is that if $hd(G)=n$ then the geometric of $G$ is $\le n+1$ (not the equality).
