TITLE: Coefficients of real k-theory with coefficients
QUESTION [7 upvotes]: 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.

REPLY [9 votes]: By $S$-duality the group $KO_3(\mathbb RP^2)$ is isomorphic to $KO^0(\mathbb RP^2)$. The latter group is not killed by $2$. Let $L$ be the nontrivial real line bundle on $\mathbb RP^2$. The bundle $L\oplus L$ on $\mathbb RP^2$ is nontrivial, even stably nontrivial, because its second Stiefel-Whiney class is not zero.