The group $H^1(K(\mathbb Z/4,1);\mathbb F_2)$ is one-dimensional, so it suffices to consider its nonzero class $x$. This class is represented by the quotient homomorphism $\mathbb Z/4\to\mathbb Z/2$, which lifts to the identity homomorphism with coefficients in $\mathbb Z/4$. Its <Bockstein homomorphism> for
$$
0\longrightarrow\mathbb Z/2\longrightarrow\mathbb Z/4
\longrightarrow\mathbb Z/2\longrightarrow0
$$
therefore vanishes. Since the first <Steenrod square> is this Bockstein and $\operatorname{Sq}^1x=x^2$ for every degree-one class,
$$
x^2=0.
$$
The zero degree-one class plainly has square zero as well.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 4 / iii / Solution

{c}
{title2=$\operatorname{Sq}^i$}
{wiki}

The Steenrod squares are natural operations
$$
\operatorname{Sq}^i:H^n(X;\mathbb F_2)\to H^{n+i}(X;\mathbb F_2).
$$
For a degree-one class $x$, $\operatorname{Sq}^1x=x^2$, and $\operatorname{Sq}^1$ agrees with the mod-two Bockstein.


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 Incoming links: Steenrod square