Prove fundamental group is the direct product Suppose that $A$ is a retract of $X$ with retraction $r : X \rightarrow A$. Also suppose that $i_*(\pi(A,a))$ is a normal subgroup of $\pi(X,a)$. Prove that $\pi(X,a)$ is the direct product of the subgroups image $(i_*)$ and $\ker(r_*)$.
I think there is something wrong with the question. Shouldn't it be '$\pi(X,a)$ is isomorphic to the direct product of the subgroups image $(i_*)$ and $\ker(r_*)$'? 

If I'm reading correctly, your problem is with the wording, using "is" rather than "is isomorphic to" in the phrasing? If so, note that group isomorphism is an equivalence relation. Therefore, one can implicitly work with the isomorphism class of a group and use "is" in reference to the isomorphism class being the same when represented by either of the two groups rather than "is isomorphic to" for the actual sets themselves.
Basically, "is" is shorthand for "is isomorphic to" since we're usually only looking for isomorphic structures rather than strictly equal sets.
