Show that the subalgebra of \(\mathcal{A}_2\) generated by \(Sq^i\) for \(i \leq 2\) has dimension 8 as a vector space over \(\mathbb{Z}_2\), with multiplicative structure encoded in the following diagram, where diagonal lines indicate left-multiplication by \(Sq^1\) and horizontal lines indicate left-multiplication by \(Sq^2\).
Solution
Write \[ a=Sq^1, \qquad b=Sq^2. \] The mod-2 Adem relations give \[ a^2=0, \qquad aba=b^2. \tag{1} \] From these two relations we obtain \[ b^2a=0, \qquad ab^2=0, \qquad b^4=0, \tag{2} \] and \[ abab=b^3=baba. \tag{3} \] Consequently every word in \(a,b\) reduces to a linear combination of \[ \boxed{ 1,\ a,\ b,\ ab,\ ba,\ b^2,\ bab,\ b^3.} \tag{4} \]
These eight elements are linearly independent in the Steenrod algebra. To see this, use the admissible basis. We have \[ ab=Sq^3, \qquad ba=Sq^2Sq^1, \qquad b^2=Sq^3Sq^1, \] while another Adem relation gives \[ bab=Sq^2Sq^3=Sq^5+Sq^4Sq^1, \] and hence \[ b^3=(bab)a=Sq^5Sq^1. \] These are nonzero combinations of distinct admissible basis elements, and the grading separates all cases except degree three, where \(Sq^3\) and \(Sq^2Sq^1\) are distinct admissible monomials. Thus (4) is a basis and the generated subalgebra has dimension \(8\).
It remains to record left multiplication. From (1)–(3): \[ \begin{array}{c|cccccccc} z&1&a&b&ab&ba&b^2&bab&b^3\\ \hline az&a&0&ab&0&b^2&0&b^3&0\\ bz&b&ba&b^2&bab&0&b^3&0&0. \end{array} \] Thus left multiplication by \(Sq^1=a\) and \(Sq^2=b\) gives exactly the diagonal and horizontal arrows in Hatcher’s diagram. Therefore \[ \boxed{\dim_{\mathbb Z_2}A(1)=8,} \] with the stated multiplicative structure.