Is $\mathbb{Z}[G^n]$ isomorphic to $\bigoplus_n\mathbb{Z}[G]$?

Let $G$ be a group, $\mathbb{Z}[G]$ be it's group ring and $G^n$ the direct product of $n$ copies of $G$. Is the group ring $\mathbb{Z}[G^n]$ isomorphic to $\bigoplus_n \mathbb{Z}[G]$? If not, is it a free $G$-module anyway?

I started verifying that the homomorphism $\phi$ defined by:

$$\sum n_i (g^1_i,...,g^n_i) \mapsto (\sum n_i g^1_i, ..., \sum n_i g^n_i)$$

is a isomorphism. I think it's clear that it is a homomorphism:

$$\sum (n_i + m_i) (g^1_i,...,g^n_i) \mapsto (\sum (n_i + m_i) g^1_i, ..., \sum (n_i + m_i) g^n_i) = (\sum n_i g^1_i, ..., \sum n_i g^n_i) + (\sum m_i g^1_i, ..., \sum m_i g^n_i)$$

I also believe that surjectivity is clear, the reverse map is simply reversing the arrows from the the definition. How should I show injectivity? One thing confusing me here is notation: that is what makes me doubt also about the validity of my steps above.

• Tensors, not sums. – Pedro Tamaroff May 23 '16 at 1:36
• @PedroTamaroff wow, tensors? I was really off then. May I have a reference to that result, so I can read it through? – Henrique Augusto Souza May 23 '16 at 1:52
• There is an evident morphism you should prove is an isomorphism. – Pedro Tamaroff May 23 '16 at 2:09
• An answer to my second question shows that $\mathbb{Z}[G^n]$ is a free $\mathbb{Z}[G]$ module: math.stackexchange.com/questions/1502018/… – Henrique Augusto Souza May 23 '16 at 4:26

$\mathbb{Z}G^n$ is a free $\mathbb{Z}G$-module, but its rank is $|G|^{n-1}$. Here is a proof:

If $X$ is a $G$-set, $X$ breaks as a disjoint union of orbits $X \cong_G \amalg_i Gx_i$, and each orbit $Gx_i$ is $G$-isomorphic to $G/G_{x_i}$, where $G_{x_i}$ is the isotropy of the point $x_i$. That is, $X \cong_G \amalg_i G/G_{x_i}$. This implies that $\mathbb{Z}X$ is isomorphic to $\bigoplus_i\mathbb{Z}(G/G_{x_i})$ as $\mathbb{Z}G$-modules.

If we see $G^n$ as a $G$-set with the left action of $G$ by coordinate-wise multiplication, then

$$G^n \cong_G \coprod_{g_1,\ldots,g_{n-1} \in G} G (1,g_1,\ldots,g_{n-1}) \cong_G \coprod_{g_1,\ldots,g_{n-1} \in G} G$$

where the second isomorphism holds because the isotropy of the element $(1,g_1,\ldots,g_{n-1})$ is trivial, and I am writing $G$ instead of $G/\{1\}$. Therefore we conclude with an isomorphism of $\mathbb{Z}G$-modules

$$\mathbb{Z}G^n \cong \bigoplus_{g_1, \ldots, g_{n-1} \in G} \mathbb{Z}G$$

Now addressing your argument, the map you define is a homomorphism of $\mathbb{Z}G$-modules, it is well-defined and surjective. But it is not injective in general.

For instance, consider $\phi((1,g)-(1,h)) = (0,g-h)$ and $\phi((g,g)-(g,h)) = (0,g-h)$. You can see in this example the problem with your inverse.

Note: It is common (in group cohomology books, etc) to see a different basis of $\mathbb{Z}G^n$ as a $\mathbb{Z}G$. Instead of choosing the elements $(1,g_1,\ldots,g_{n-1})$ one takes the elements of the form $(1,g_1,g_1g_2,\ldots,g_1g_2\cdots g_{n-1})$ which are often denoted by $[g_1|g_2|\ldots|g_{n-1}]$. This is called the bar notation.