**Question:** Calculate the group $ \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) $.

Here $KO$ denotes the real k-theory spectrum and $M\mathbb Z/l\mathbb Z $ denotes a Moore Spectrum associated to the cyclic group of order $l$.

Assume for the following that $2$ divides $l$ (otherwise there are no problems).

Using the short exact sequence for Moore spectra one can easily calculated all the coefficients but $ \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) $ . There one gets only a short exact sequence: $$\mathbb Z/2\mathbb Z \hookrightarrow \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) \twoheadrightarrow \mathbb Z/2\mathbb Z $$

Note that for $l=2$ we have: $ \pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(\mathbb RP^2) $

But I'm neither able to calculate the latter group.

What techniques are used to solve problems like this?

**Answer:** (following the answer by Tom Goodwillie)

Idee: Use $S$-Duality to calculate $\pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(C_l)$ where $C_l$ denotes a Moore space to the cyclic group of order $l$ (and $C_2\simeq\mathbb RP^2$).

The $S$-dual of $C_l$ is $\Sigma^{-3}C_l$ (use that $C_l$ can be realized as the cofibre of $S^1\to S^1$).

By $S$-duality we have: $KO_{3}(C_l) \cong KO^0(C_l)$.

The latter group can be calculated by comparing $C_l$ and $\mathbb RP^2$.

In the end we get that the coefficients are as follows:

Note that the extension splits for some $l$ which I find surprising.