# Why is an abelian group an algebra over $GF(2)$?

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?

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 $$x \in H$$).