Relation between the $2^n$-Bockstein homomorphisms

Question

The $2^n$-Bockstein homomorphism $$\beta_{2^n}:H^*(-,\mathbb{Z}/{2^n})\to H^{*+1}(-,\mathbb{Z}/2)$$ is associated to the short exact sequence $$0\to\mathbb{Z}/2\to\mathbb{Z}/{2^{n+1}}\to\mathbb{Z}/{2^n}\to0$$ In particular, $\beta_2=Sq^1$ is the Steenrod square.

My question: what is the relation between $\beta_{2^n}$ and $\beta_{2^{n+k}}$ for $n\ge1$, $k>0$?

I guess $$\beta_{2^{n+k}}\alpha=\beta_{2^n}$$ where $\alpha:H^*(-,\mathbb{Z}/{2^n})\to H^*(-,\mathbb{Z}/{2^{n+k}})$ is induced from $$\mathbb{Z}/{2^n}\to\mathbb{Z}/{2^{n+k}}$$ which maps the generator of $\mathbb{Z}/{2^n}$ to $2^k$ times the generator of $\mathbb{Z}/{2^{n+k}}$.

Can you help me to prove or disprove it?

Thank you!

Answer 1

I think you are correct. Consider the following map of short exact sequences $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z}/2 @>>> \mathbb{Z}/2^{n+1} @>>> \mathbb{Z}/2^n @>>> 0 \\ @. @| @VVV @VVV @. \\ 0 @>>> \mathbb{Z}/2 @>>> \mathbb{Z}/2^{n+k+1} @>>> \mathbb{Z}/2^{n+k} @>>> 0. \end{CD} $$

The connecting homomorphism induced by the first row is $\beta_{2^n}$, the connecting homomorphism induced by the second row is $\beta_{2^{n+k}}$, and the third vertical homomorphism induces what you call $\alpha$.

By the naturality of connecting homomorphisms, we have $$\require{AMScd} \begin{CD} H^*(-, \mathbb{Z}/2^n) @>\beta_{2^n}>> H^*(-, \mathbb{Z}/2) \\ @V\alpha VV @| \\ H^*(-, \mathbb{Z}/2^{n+k}) @>>\beta_{2^{n+k}}> H^*(-, \mathbb{Z}/2), \end{CD}$$ which proves your assertion.