Take the 2-minute tour ×  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?  share|improve this question I don't know if this
is necessarily relevant, but Kijti Rodtes (a student of John
Greenlees) did enormous calculations of $ko_*(BSD_{16})$ and
related structures in his thesis, using
$\text{Ext}_{\mathcal{A}(1)}^*$ as a tool. –  Neil Strickland
Feb 23 '12 at 21:22 Have you looked at the memoir or book by
Bruner-Greenlees? they might remark on it. –  Sean Tilson Feb 24
'12 at 7:11 add comment  1 Answer  up vote 9 down vote accepted
Not a 'topological' explanation, but A(1) is the 8 dimensional
member of the family of semi-dihedral algeb