Let $\mathcal{A}(1)$ denote the subalgebra of the $\mathrm{mod}\ 2$ Steenrod algebra generated by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$.
The cohomology with $\mathbf{F}\_2$ coefficients of the semidihedral group
$$SD\_{16} = \langle g, h \mid g^8 = h^2 = 1, hgh = g^3\rangle$$
of order $16$ is isomorphic to $\mathrm{Ext}^\ast\_{\mathcal{A}(1)}(\mathbf{F}\_2, \mathbf{F}\_2)$. Is there a topological explanation for this isomorphism?



The semidihedral group of order 16 and ko
Not a 'topological' explanation, but A(1) is the 8 dimensional member of the family of semi-dihedral algebras, whose members of dimension $2^n$ for $n > 3$ are the mod 2 group rings of the semidihedral group of that order. Their cohomology ring is insensitive to the differences; that difference is reflected in the order of the higher Bockstein from H^3 to H^4.


Crawley-Boevey completely analyzed the structure of the modules over these algebras in


Crawley-Boevey, W. W. Functorial filtrations. III. Semidihedral algebras. J. London Math. Soc. (2) 40 (1989), no. 1, 31–39. 


where he applies his results from Functorial Filtrations I and II to determine all the representations of "semidihedral algebras" $A\_m = k\langle a,b | a^3 = b^2 = 0, a^2 = (ba)^mb\rangle$ (CAVEAT: when $k$ is a field with more than 2 elements). 


There is one of these of $k$-dimension $4n-1$ for each $n \geq 2$.
The 7 dimensional one is $A(1)$ modulo its socle ($a=Sq^2$ and $b=Sq^1$), and the $2^n-1$ dimensional one is
the group ring of the semidihedral group modulo its socle for larger $n$. 


Quotient by the socle is sensible here, since $A(1)$ and group rings are Frobenius, so that free modules are injective, hence split off. Modules with no free summand are all pulled back from the quotient by the socle.


There is a more recent (and I believe simpler) account of the representation theory as well, but I am embarrassed to admit that I cannot remember the author.


Since Crawley-Boevey's work actually classifies the modules over $A(1) \otimes GF\_4$, I have always thought it would be entertaining to do the Galois descent to $A(1)$.


