Let $G$ be a group. If $M$ is any right $G$-module, then we can consider $M$ as left $G$-module also under the action $g.m:= mg^{-1}$, where $m \in M$ and $g \in G$. Consider $\mathbb{Z}[G^{n+1}]$ as right $\mathbb{Z}[G]$-module under the action $( g_0,g_1, \ldots, g_n).g=( g_0g,g_1g, \ldots, g_ng)$. Then it is true that $\mathbb{Z}[G^{n+1}] \otimes_{\mathbb{Z}[G]} \mathbb{Z} \cong \mathbb{Z}[G^n]$ as $\mathbb{Z}$-module. One can prove this using properties of tensor product but I want to see it explicitly.

My attempt is as follows

Define a map $ \phi: \mathbb{Z}[G^{n+1}]\otimes_{\mathbb{Z}[G]} \mathbb{Z} \to \mathbb{Z}[G^{n}]$ as $$\phi(( g_0,g_1, \ldots, g_n)\otimes 1)= ( g_0 g_n^{-1},g_1 g_n^{-1}, \ldots, g_{n-1}g_n^{-1}).$$

The inverse of the above map is given by $( g_0,g_1, \ldots, g_{n-1}) \mapsto ( g_0,g_1, \ldots, g_{n-1},1) \otimes 1.$

I want to know whether the above maps are correct or not? And are there any other maps to see the above isomorphisms, as I am not getting the desired result using these maps?

I should explain the problem too:

Let $a=\big((g_1,g_2) \otimes 1 \big)\otimes g \in (\mathbb{Z}[G^2] \otimes_{\mathbb{Z}[G]} \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Z}[G]$.

Now If I use the above isomorphism, then $a=a_1=(g_1g_2^{-1}g,g) \in \mathbb{Z}[G^2]$, where as if I use the associativity of tensor products then $a=a_2=(g_1g,g_2g) \in \mathbb{Z}[G^2]$, and clearly $a_1 \neq a_2,$ unless $g_2=e$.

I want to understand why this is happening?

  • $\begingroup$ Just to clarify things, is the $\mathbb Z[G]$-module structure of $\mathbb Z$ defined by $g\cdot n=0,\,\forall n\in \mathbb Z,\,g\in G$? $\endgroup$
    – awllower
    Feb 18, 2020 at 6:51
  • 1
    $\begingroup$ awllower: Yes, $\mathbb{Z}$ is trivial $G$-module. $\endgroup$
    – eyp
    Feb 18, 2020 at 6:52
  • $\begingroup$ If $\mathbb Z$ is a trivial $G$-module don't we have $$ (g_0,\dots,g_n)\otimes 1 = (1,g_1g_0^{-1}\dots,g_n g_0^{-1})\cdot g_0 \otimes 1 = (1,g_1g_0^{-1},\dots,g_n g_0^{-1})\otimes 0=0?$$ $\endgroup$
    – user347489
    Feb 18, 2020 at 8:00
  • $\begingroup$ $(g_0,\ldots,g_n) \otimes 1 = (g_0 g_n^{-1}, g_1 g_n^{-1}, \ldots, g_{n-1}g_n^{-1}) .g_n \otimes 1= (g_0 g_n^{-1}, g_1 g_n^{-1}, \ldots, g_{n-1}g_n^{-1}) \otimes g_n .1=(g_0 g_n^{-1}, g_1 g_n^{-1}, \ldots, g_{n-1} g_n^{-1}) \otimes 1$ $\endgroup$
    – eyp
    Feb 18, 2020 at 8:04
  • 2
    $\begingroup$ user347489: $\mathbb{Z}$ is trivial $G$ modules so $g.n=n, \forall g \in G$ and $n \in \mathbb{Z}$. $\endgroup$
    – eyp
    Feb 18, 2020 at 8:08

1 Answer 1


The isomorphism is fine; in fact, you can use any of the $n+1$ positions, giving $n+1$ different isomorphisms $$\alpha_i\colon \mathbb Z[G^{n+1}]\otimes_{\mathbb Z[G]}\mathbb Z\to\mathbb Z[G^n], \quad (g_0,\ldots,g_n) \mapsto (g_0g_i^{-1},\ldots,g_ng_i^{-1}). $$

Your problem is that you have messed up the bimodule structure. You are considering $\mathbb Z\otimes_{\mathbb Z}\mathbb Z[G]\cong\mathbb Z[G]$ as a $\mathbb Z[G]$-bimodule, but whereas it is the regular module with respect to the right action, it is actually the trivial module with respect to the left action. In other words, we have the actions $g\cdot y\cdot k=yk$ for $g,y,k\in G$.

So, in your problem at the end of the post, we can use $\alpha_0$ to obtain the isomorphism $$ \mathbb Z[G^2]\otimes_{\mathbb Z[G]}\mathbb Z[G] \to \mathbb Z[G^2], \quad (g,h)\otimes k \mapsto (hg^{-1},k). $$ The $\mathbb Z[G^2]$-$\mathbb Z[G]$-bimodule action on $\mathbb Z[G^2]$ is then $$ (g,h)\cdot (x,y)\cdot k := (hxg^{-1},yk). $$

Alternatively we can use $\alpha_1$ to get the map $(g,h)\otimes k \mapsto (gh^{-1},k)$, and the bimodule action is then $$ (g,h)\cdot (x,y)\cdot k := (hxg^{-1},yk). $$

  • $\begingroup$ Andrew Hubery: Thank you. I got your point. $\endgroup$
    – eyp
    Feb 19, 2020 at 7:31
  • $\begingroup$ Andrew Hubery: If we take the map $\mathbb{Z}[G^2] \otimes_{\mathbb{Z}[G]} \mathbb{Z}[G] \to \mathbb{Z}[G^2]$ as $(g,h) \otimes k \mapsto (gh^{-1}k,k)$, then it is a $\mathbb{Z}[G]$-module isomorphism under the given action( as in the question) of $G$ on $\mathbb{Z}[G^2]$. $\endgroup$
    – eyp
    Feb 19, 2020 at 7:34
  • $\begingroup$ Andrew Hubery: May I request you to please look at the question mathoverflow.net/questions/352118/…. I have spent many days to sort it out but still no progress. I think there too I am doing something wrong with tensor products. $\endgroup$
    – eyp
    Feb 19, 2020 at 7:39

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