# The first cohomology of group

I would like to ask if G is a group of order $p^4 (p\neq 2)$ as form $C_{p^3}\rtimes C_p$ (a semidirect product of cyclic group of order $p^3$ by a group of order $p$). Then can we obtain the first co-homology $H^1(C_p,C_{p^3} )$? Is there any upper bound on the order of $H^1(C_p, C_{p^3})$?

yours,

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(If you want cohomology, you should put the index as an upper index lest the wrath of the Gods Of Notation fall upon thee) – Mariano Suárez-Alvarez Sep 4 '12 at 20:25
Do you want the action on $C_p$ to be trivial? Do you want the cohomology of what group? You mention $G$ in the first sentence byt in the end you ask about the cohomology of $C_{p^3}$... – Mariano Suárez-Alvarez Sep 4 '12 at 21:17
Are you sure you don't mean $H^1(C_p,C_{p^3})$, which would correspond to the conjugacy classes of complements in the semidirect product? If so, then the answer depends on the action of $C_p$ on $C_{p^3}$. – Derek Holt Sep 4 '12 at 22:23
Your edit did not clarify what role the group G plays in your question. – Mariano Suárez-Alvarez Sep 5 '12 at 6:28
You have still not explained what is the connection between the group $G$ and your question about a cohomology group. – Derek Holt Sep 5 '12 at 11:14

There must be a mistake in your notation. Cohomology groups $H^r$(G, A) are defined only for a group G and a G-module A, i.e. an abelian group A on which G acts. If A is not abelian, one can still define $H^1$(G, A) in certain cases, but it is just a set (see Serre's "Galois Cohomology", chapter 3).
If your $H^1$ notation means that the semi-direct product acts trivially on $C_p$ , then the inflation-restriction exact sequence (= beginning of the Hochschild-Serre spectral sequence) should do the job.