TITLE: Periodicity of the cohomology of $\mathbb RP^\infty$
QUESTION [5 upvotes]: I want to prove that there is an isomorphism
$$H^*(\Sigma^{2^j}\mathbb RP^b/\mathbb RP^a:\mathbb Z/2)\cong H^*(\mathbb RP^{b+2^j}/\mathbb RP^{a+2^j}:\mathbb Z/2)$$
compatible with the Steenrod operations for any natural numbers $a, b$, and $j$ with $a\leq b$ and $2^j\geq b-a$.
I know that I have to use the equality
$$\binom{n+2^j}{k}=\binom{n}{k}\mod 2$$
for $2^j>k$ but I don't know how to apply it in this case. Any hint?
My attempt:
Both spaces are path connected, so they both have the $0$-th cohomology isormorphic to $\mathbb Z/2$. Therefore, sending generator to generator, we can construct an isomorphism on the 0-th cohomology. By composing suspensions isomorphisms, we have isomorphisms on reduced cohomology
$$\sigma^{2^j}:\widetilde{H}^*(\mathbb RP^b/\mathbb RP^a:\mathbb Z/2)\to \widetilde{H}^{*+2^j}(\Sigma^{2^j}\mathbb RP^b/\mathbb RP^a:\mathbb Z/2) $$
Moreover, there are also isomorphisms
$$\widetilde{H}^*(\mathbb RP^b/\mathbb RP^a:\mathbb Z/2)\to \widetilde{H}^{*+2^j}(\mathbb RP^{b+2^j}/\mathbb RP^{a+2^j}:\mathbb Z/2)$$
Therefore, we can construct the isomorphism
$$\phi:H^*(\Sigma^{2j}\mathbb RP^b/\mathbb RP^a:\mathbb Z/2)\to H^*(\mathbb RP^{b+2^j}/\mathbb RP^{a+2^j}:\mathbb Z/2)$$
at least on a cohomology groups level.
We want to find which conditions on the parameters $a,b,j$ let $\phi$ be compatible with the Steenrod operations. We have that $\phi$ is compatible with $Sq$ if the diagram 

commutes. Since $\phi$ is an isomorphism at each degree, we just need to check this on the generators. For $x$ algebra generator of $H^*(\mathbb RP^\infty)$,  
$$Sq(x^n)=(Sq(x))^n=(x^2+x)^n=\sum_{k=0}^n\binom{n}{k}x^{n+k}=\sum_{k=0}^n\binom{n+2^j}{k}x^{n+k}$$
is it the right direction?

REPLY [2 votes]: You asked for hints, so here are some: if $x$ is the algebra generator of $H^*(\mathbb{R}P^\infty; \mathbb{F}_2)$, then $\text{Sq}(x)=x+x^2$, where Sq is the total Steenrod square, $\text{Sq} = \sum \text{Sq}^i$. The operator Sq is an algebra map.
Edit: further hints: the mod 2 cohomology groups of the real projective spaces are generated (as vector spaces) by the powers of $x$, and the equation $\text{Sq}(x^n) = (\text{Sq}(x))^n$ lets you compute the action of the Steenrod operations on $x^n$. If you expand the right side of this, you will see binomial coefficients...