$\newcommand\Sq{\mathrm{Sq}}$I am trying to compute $ [\mathbb{HZ}/4,\mathbb{HZ}/4 ]$ the mod 4 Steenrod algebra. For some reason, I need to work it out till dimension 6 or so. My approach is to use the cofiber sequence
$\mathbb{HZ}/4 \to \mathbb{HZ}/2 \xrightarrow{\Sq^{1}}\Sigma \mathbb{HZ}/2 $
twice.
I did the computation till first five degrees. A description would be as follows, Let $g$ be the generator of $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{0}$, for convenience let us denote $g' = 2g$. So whenever there is an element with $g'$ would mean that it has $2$ torsion.
Degree | elements
$ 1: \beta g$
$ 2: \Sq^{2} g' $ (this would mean that $2\Sq^{2} g' =0$)
$ 3 : \Sq^{3}g$, (but satisfies $2\Sq^{3}g =0$ ), $\Sq^{2} \beta g'$
$ 4 : \Sq^{3} \beta g$( $2 \Sq^{3} \beta g = 0$), $\Sq^{4}g'$
$ 5 : \Sq^{5}g$( $2 \Sq^{5} g = 0$), $\Sq^{4} \beta g' = 0$
I do not know if they are right, but if somebody has done it please verify. I suspect the all the groups $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{n}$ are two torsion except $n=0,1$. Is this some sort of known result? Also another thing that I am worrying about is the equivalent Cartan formula in mod 4 Steenrod algebra? Is there a way to detect the Cartan formula?