I am trying to see why this is true, this is done in Brown's Cohomology of groups:
(In terms of notation $\tau(y)$ is the divided polinomyal algebra).
Consider the study of $H_*(G,\mathbb{Z}_p)$, consider the split-exact universal coefficient sequence
$0\rightarrow H_2(G) \otimes \mathbb{Z}_p \rightarrow H_2(G,\mathbb{Z}_p)\rightarrow Tor(H_1G,\mathbb{Z}_p)\rightarrow 0$
We have that $H_2(G)\cong \bigwedge^2 (G)$, and its also true that $\bigwedge^2(G)\otimes \mathbb{Z}_p\cong \bigwedge^2_{\mathbb{Z}_p}(G_p)$, where $G_p=G\otimes \mathbb{Z}_p = G/pG$ . Also, $Tor(H_1G,\mathbb{Z}_p)=Tor(G,\mathbb{Z}_p)=p^G=\{g\in G : pg=0\}$, where this can be seen using the long-exact sequence for $Tor$.So our sequence will take the form $0\rightarrow \bigwedge^2(G_p)\rightarrow H_2(G,\mathbb{Z}_p)\rightarrow p^G \rightarrow 0$
Choose a splitting $p^G\rightarrow H_2(G,\mathbb{Z}_p)$ of this sequence, this will extend to a $\mathbb{Z}_p-$algebra homomorphism $\phi : \tau(p^G)\rightarrow H_*(G,\mathbb{Z}_p)$ comptabile with divided powers.Combining $\phi$ with the map $\psi : \bigwedge(G_p)\rightarrow H_*(G,\mathbb{Z}_p)$ seen before we obtain an algebra map $\alpha : \bigwedge(G_p)\otimes \tau(p^G)\rightarrow H_*(G,\mathbb{Z}_p)$ given by $\alpha(x\otimes y)=\psi(x)\phi(y)$. We can now state: The map $\alpha:\bigwedge(G_p)\otimes \tau(p^G)\rightarrow H_*(G,\mathbb{Z}_p)$ is an isomorphism.
I get all the steps and I am just having some trouble proving this last statement of this isomorphism, I know that the idea is to start with cyclic groups and then finitely generated subgroups and then use direct limits to get to groups, but I cant even see why this is true for cyclic groups, I am also not very familiar with graded algebras and such. I guess first I am having some difficulty interpreting what $\bigwedge G_p \otimes \tau(p^G)$ is going to be as a graded algebra. Because I already know that for cyclic groups $ H_i(G,\mathbb{Z}_p)\cong $
\begin{array}{ll} \mathbb{Z}_p & i=0 \\ G_p & i odd \\ Tor(G,\mathbb{Z}_p)=p^G & i even \\ \end{array}
But since I dont quite understand the other graded ring strucutre I cant really see if it is an isomorphism, it looks like it is going to be but I am having some trouble making this formal. Thanks in advance.