# $\text{Hom}(\mathbb{F}_p G, M)$ and $H^1(G,M)$

I'm trying to read (part of) "The Presentation Rank of a Direct Product of Finite Groups" / Cossey, Gruenberg, Kovacs (Journal Of Alegebra 28, 597-603 (1974)).

Here are some basic assertions I need help with:

(by context $p$ is a prime number and $G$ is a finite group)

For each $p$ dividing $|G|$, and each irreducible $\mathbb{F}_p G$-module $M$, let $E=\text{Hom}_G(M,M)$. Then: 1. $\text{Hom}_G(\mathbb{F}_p G,M) \cong E^{r_M}$, 2. $H^1(G,M) \cong E^{s_M}$, for certain non-negative integers $r_M$, $s_M$.

I am familiar with all terms except for "irreducible module". I know what a simple module is, and what an irreducible representation is. My guess is that irreducible module is just an old term for a simple module, but I'm not sure. So, I need help with this terminology issue and with the proof of both assertions.

Also note that this is my first encounter with group cohomlogy. I read the definition here: http://groupprops.subwiki.org/wiki/First_cohomology_group. I used to know a bit about cohomology in a topological context, but it's not fresh in my mind.

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I think you're right with the word 'irreducible', see mathworld.wolfram.com/IrreducibleModule.html. – InvisiblePanda Nov 19 '12 at 11:29
How is $\mathbb{F}_p G$ defined? – Adeel Nov 19 '12 at 12:13

First of all, if you are not familiar with group cohomology, $H^1(G,M)$ means $\operatorname{Ext}^1_{\mathbb{F}_pG}(\mathbb{F}_p,M)$ and irreducible means simple.

By Schur's lemma $E$ is a division ring over $\mathbb{F}_p$, in fact a finite field. Now note that for any $\mathbb{F}_p$-module $N$, both $\hom_{\mathbb{F}_p G}(N,M)$ and $\operatorname{Ext}_{\mathbb{F}_pG} (N,M)$ are $E$-modules by functoriality. Since any $E$-module is free, both sorts of groups are isomorphic (as abelian groups, or $E$-modules, or $\mathbb{F}_p$-modules) to a direct sum of $E$s.

Edit: $\hom_{\mathbb{F}_pG}(N,-)$ and $\operatorname{Ext}^1_{\mathbb{F}_pG}(N,-)$ are functors from the category of $\mathbb{F}_pG$-modules to the category of $\mathbb{F}_p$ vector spaces. How you describe functoriality of Ext depends on the way you choose to construct it, but it is a derived functor so it is a functor which ever way you do it :)

For example, you may construct Ext in such a way that elements of $\operatorname{Ext}^n_{\mathbb{F}_pG}(N,M)$ are equivalences classes of module maps $\phi: P_n \to M$, where $P_n$ is a certain projective resolution of $N$. Then if $f:M\to M$ is a module map, the class of $\phi$ acted on by $f$ is represented by $f \circ \phi$.

Alternatively, you may think of $\operatorname{Ext}^n_{\mathbb{F}_pG}(N,M)$ as consisting of equivalence classes of long exact sequences $$0 \to M \to L_n \to \cdots \to L_1 \to N \to 0$$ of $\mathbb{F}_pG$-modules beginning with $M$ and ending with $N$. The action of $f$ on the equivalence class of such a sequence is represented by a sequence as follows $$\begin{array}{cccccc} 0 & \to & M & \to L_n & \to L_{n-1} & \to \cdots \\ & & \downarrow f &\downarrow & & \\ 0 & \to & M & \to X & \to L_{n-1} & \to \cdots \end{array}$$ where $X$ is the pushout of $M \to L_n$ and $f:M \to M$.

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I'm not the person who originally answered this question. – m_t_ Nov 19 '12 at 12:37
Thank you. One claim is unclear to me: "$\text{Hom}_{\mathbb{F}_pG}(N,M)$ and $\text{Ext}_{\mathbb{F}_pG}(N,M)$ are $E$-modules by functoriality". Can you please say what the functors are? (from which category to which category). I can see how $\text{Hom}_{\mathbb{F}_pG}(N,M)$ is an $E$ module in a natural way, but not sure about $\text{Ext}_{\mathbb{F}_pG}^1(N,M)$. – user3533 Nov 19 '12 at 17:20
(i guess they are functors from $\mathbb{F}_pG$-modules to $E$-modules, but not sure why) – user3533 Nov 19 '12 at 17:24
are these the relevant functors, or are these $\text{Hom}_{\mathbb{F}_pG}(\mathbb{F}_pG,-)$ and $\text{Ext}_{\mathbb{F}_pG}^1(\mathbb{F}_p,-)$ the right ones? – user3533 Nov 19 '12 at 17:52
Like I said, by functoriality. If $f:M \to M$ is a module map then functoriality of $\operatorname{Ext}^n(N,- )$ gives a map $\operatorname{Ext}^n(N,f) : \operatorname{Ext}^n(N,M) \to \operatorname{Ext}^n(N,M)$ (not common notation, but my point is you are applying the functor $\operatorname{Ext}^n(N,- )$ to $f$). – m_t_ Nov 19 '12 at 17:58