In this MO answer, Geoff Robinson writes:(Punctuation error corrected)

Consider a group homomorphism $\phi:H\to A$, where $A$ is an Abelian group. Let $R$ be the group ring $GF(2)[A]$. Consider $\phi$ as a rank-$1$ representation of $H$ over $R$.

I assume Geoff means by $GF(2)$ the field $\mathbb{Z}/(2)$, which I prefer to write $\mathbb{F}_2$.

Now, if $\phi$ is a rank-representation of $H$ over $R$, then $A$ must be an $R$-module. But $R=\mathbb{F}_2A$ has characteristic $2$, whereas $A$ need not have exponent dividing $2$. So how can $A$ be an $R$-module?


1 Answer 1


You need to understand what the group algebra $R = \mathbb{F}_{2}A$ is. As vector space over $\mathbb{F}_{2}$, it has $\mathbb{F}_{2}$ basis $\{a: a \in A\}.$ It becomes an associative algebra over the field $\mathbb{F}_{2}$ by linearly extending the multiplication of $A$. That is, $\sum_{a \in A} \lambda_{a}a. \sum_{b \in A} \mu_{b}b = \sum_{c \in A} \left( \sum_{a \in A} \lambda_{a} \mu_{a^{-1}c} \right) c.$ In this case, this is a commutative algebra, since $A$ is Abelian. In particular, $R$ is a commutative ring in its own right.

The addition in $R$ is as you would expect: $\sum_{a \in A} \lambda_{a}a + \sum_{a \in A} \mu_{a}a = \sum_{a \in A} (\lambda_{a}+\mu_{a})a.$ In particular, it is indeed true that $x + x = 0$ for all $x \in R.$

However, it may be causing confusion that here, $A$ itself is considered as a multiplicative Abelian group, not written additively.

I am then saying that $\phi$ may be considered as a (multiplicative) group homomorphism from $H$ into $R^{\times}$, the group of multiplicative units of $R$, which has the multiplicative group $A$ as a subgroup. This may be viewed as a homomorphism from $H \to {\rm GL}(1,R)$, the group of invertible $1 \times 1$ matrices over the commutative ring $R$.

In the usual manner, we may induce this representation to a representation $\psi: G \to {\rm GL}(n,R)$, where $n = [G:H].$ Since $R$ is a commutative ring, there is a well defined determinant map from $M_{n}(R) \to R$ , and ${\rm GL}(n,R)$ is consists of the matrices whose determinants are multiplicative units of $R$. By construction of the induced representation $\psi$, each element of $\psi(G)$ has determinant which is a product of various elements of the form $\phi(h)$ for some $h \in H$, (because $\psi(G)$ has exactly one non-zero entry in each row and column, and each such entry has the from $\phi(x)$ for some $x \in H$).

I hope this helps to clarify things.


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