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Let $R$ be a finite ring, and $G$ be a finite group. I need to compute the number of elements of the group ring $R(G)$ in terms of $|R|$ and $|G|$ (where $|R|$ is the number of elements in the ring $R$, and $|G|$ is the number of elements in the group $G$).

Recall that a group ring $R(G)$ is the set of equivalence classes of all formal sums

$r_{1}g_{1}+r_{2}g_{2}+\cdots + r_{k}g_{k}$

where the $r_{i} \in R$, $g_{i} \in G$.

We say that two formal sums are said to be equivalent if they have the same reduced form.

In each sum, we have $k$ terms, and $|R|$ choices for each coefficient, so if we didn't care about repeats or whether we should choose only one representative from each equivalence class, I'd say the group ring contained $k|R|$ elements.

However, I'm assuming that in calculating the size of a group ring, we count each representative of an equivalence class only once. So, how do I factor this in when counting the number of elements in $R(G)$? Also, how does $|G|$ come into play?

Thank you in advance!

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    @MichaelBurr what about taking into account elements with the same reduced form? – 2017-02-27
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    The elements of $G$ form a basis, so two elements are equal if and only if they have the same coefficients. Can you give an example of what you mean by the same reduced form? – 2017-02-27
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    @MichaelBurr I suppose it would depend on the ring the coefficients come from. In $R(G)$, "same reduced form" is an equivalence relation, so take any group ring, and divide it into equivalence classes. I wish I could explain it better. – 2017-02-27
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    Then, perhaps you could define the equivalence relation that you're using. – 2017-02-27
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    @MichaelBurr you know what, I don't think I need to worry about that anymore, because I noticed it my notes something I hadn't noticed before. It says "clearly, each formal sum has the unique reduced form", so I think that's been taken care of for me in the definition of formal sums in the group ring. So never mind. I think what you told me is okay and I am unnecessarily complicating things. – 2017-02-27
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    @MichaelBurr just one more thing though. You say that $G$ has a basis. $G$ is finite, but nowhere did I say that $G$ is abelian. I thought only finite abelian groups had bases. – 2017-02-27
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    No, it's not that $G$ has a basis, it's that the space you construct has the elements of $G$ as a basis. – 2017-02-27
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    @ALannister As Michael says, the elements of $G$ form a basis for $R(G)$, not $G$ itself. Regarding your comment to the answer: you presumably know what a vector space over a field is. An $R$-module is basically just a vector space where the base "field" is no longer assumed to be a field: it's only assumed to be a ring $R$. Unlike vector spaces, not all modules have bases, but the answer says that $R(G)$ is a free module, which means exactly that it does have a basis, namely $G$. Then, as Michael says, each element has a *unique* representation as $r_1 g_1 + \cdots + r_k g_k$. – 2017-02-27
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    @MichaelBurr is it too early in the morning for me or would what you said not give you $|R|^{|G|}$ elements? – 2017-02-27
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    @AlexMathers Yes, you're right, $|R|^{|G|}$ is right. – 2017-02-27

2 Answers 2

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Suppose that $G=\{g_1,\cdots,g_n\}$. Then, I like to write the group ring $R[G]$ as the collection of all sums of the form $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n} $$ where $r_i\in R$. This differs from your notation in the question by replacing $g_i$ with $e_{g_i}$. I like this notation because it's harder to confuse the group elements with the ring elements.

The equality is just like for a basis: $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n}=s_1e_{g_1}+s_2e_{g_2}+\cdots+s_ne_{g_n} $$
iff $r_i=s_i$ for all $i$. To turn this into a ring, we use the product $$ (r_ie_{g_i})(r_je_{g_j}):=r_ir_je_{g_ig_j}. $$ In other words, multiply the ring and group elements separately. This product is then extended by linearity to sums.

Since elements are of the form $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n}, $$ there are $|G|$ different $e_{g_i}$'s and each $e_{g_i}$ has $|R|$ different coefficients. This results in $|R|^{|G|}$ different ring elements.

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    thank you so much for this - it makes everything so much clearer! – 2017-02-27
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Hint: as an $R$-module, $R(G)$ is isomorphic to $R^{\oplus |G|}$.

Edit: in simpler terms, $R(G)$ is isomorphic as an abelian group to $R\oplus R\oplus\cdots\oplus R$, where there are $|G|$ copies of $R$ in the direct sum. This in particular as a set is equal to $R\times R\times\cdots\times R$, the cartesian product of $|G|$ copies of $R$.

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    I don't know what an $R$-module is. I also have no idea what $R^{\oplus|G|}$ is. So, while this answer would be undoubtedly very useful to someone studying at a more advanced level, it is not terribly useful to me. I of course have nothing against learning new things, but immediate goals first. – 2017-02-27
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    @ALannister My edit may help (or may not). – 2017-02-27