TITLE: Examples of the varying strengths of topological invariants
QUESTION [8 upvotes]: In my first algebraic topology class, I remember being told that the simplest reason for homology was to distinguish spaces. For example, if is X=circle and a Y= wedge of a circle and a 2-sphere then X and Y have the same fundamental group, so the fundamental group isn't strong enough to distinguish them. We need to look at the other homotopy groups or homology to tell them apart. I'm looking for a variety of other examples of this nature. The examples I'm wondering about are 

Same homology groups
Same cohomology groups, but different cohomology rings
Same cohomology rings (but maybe different Steenrod operations?)

If I put more thought into it, I could come up with others questions like these. Any other examples/thoughts along these lines would be very welcome! (I have examples for the first one, but I'm wondering what others will say.)

REPLY [6 votes]: To change up the nature of the responses some, IMO a good theorem to think about is the Kan-Thurston theorem. It states that given any space $X$ you can find a $K(\pi, 1)$ space $Y$ and a map $f : Y \to X$ inducing isomorphisms $f_* : H_i Y \to H_i X$, $f^* : H^i X \to H^i Y$ for all coefficients (it can be souped-up to allow local coefficients) and all $i$.  The map $\pi_1 Y \to \pi_1 X$ is onto. 
So from the point of view of cohomology algebras with Steenrod operations, these spaces are the same.  One way to "spin" this would be to say the fundamental group is a far stronger invariant than anything (co)homological.