Define the Steenrod algebra $A^\ast$ to be the algebra of all stable mod $p$ cohomology operations. Without actually computing $A^\ast$, is it possible to see that $A^\ast$ acts faithfully on $H^\ast(BC_p; \mathbb F_p)$?

My question is closely related to this one.

Here is an attempt at an argument. (**EDIT:** As Tyler's comment shows, this argument doesn't work! I'll leave it up, though, as an example of the kind of thing I might hope could be true) Let $\phi: H \to \Sigma^r H$ be a nonzero stable cohomology operation (where $H = H\mathbb F_p$ is the Eilenberg-MacLane spectrum). Then

$$\phi \wedge 1 : H \wedge H \to \Sigma^r H \wedge H$$

is a nonzero $H$-module map, and so is nonzero on homotopy. We have $H = \varinjlim_n \Sigma^{-n} K(\mathbb F_p, n)$ . It follows that

$$\phi \wedge 1: H \wedge \Sigma^{-n} K(\mathbb F_p, n) \to \Sigma^r H \wedge \Sigma^{-n} K(\mathbb F_p, n)$$

is nonzero on homotopy for some $n$. I *think* it's the case that $H_\ast(K(\mathbb F_p, n))$ is generated under Pontryagin product by $H_\ast(\Sigma K(\mathbb F_p, n-1))$ -- but I'm not sure if this is true, much less whether there is a non-computational reason for it. This ought to allow us to induct downwards to show that

$$\phi \wedge 1: H \wedge K(\mathbb F_p, 1)^N \to \Sigma^r H \wedge K(\mathbb F_p, 1)^N$$

is nonzero on homotopy for some $N$, which is *almost* the desired conclusion.