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The professor put an example of a ring on the board today:

Let $G$ be a finite group and define $RG = (a_1g_1+a_2g_2 + ... a_ig_i |a_i \in \mathbb{Z})$.

She then gave an example of multiplaction of "monomials" in this ring: $2g_i5g_j=10g_k$ where $g_ig_j=g_k$. This example was confusing to me becuase a ring is one set with 2 operations. But here we're using 2 sets ($\mathbb{Z}$ and $G$). So What rules of the ring are allowing me to commute $5$ and $g_i$?

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2 Answers 2

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The elements of $RG$ are linear combinations of elements of $G$ with coefficients in $\mathbb Z$. More compactly, this is called $\mathbb Z$-linear combinations of elements of $G$. The construction as a whole is known as the group ring of $G$. To be clear, usually $R$ is a specific ring, and so the construction you're talking about would typically be called $\mathbb Z[G]$, but it generalizes to an arbitrary ring $R$, which would be called $R[G]$ or $RG$. I'll use $\mathbb Z[G]$ below instead of $RG$ to reflect the fact that we're considering the $R=\mathbb Z$ case, though everything I say below can be immediately generalized to any ring $R$ simply by replacing $\mathbb Z$ with $R$ everywhere.

$5g_i$ is not a combination of an element $5$ of $\mathbb Z[G]$ and an element $g_i$ of $\mathbb Z[G]$, it is merely itself an element. Similarly, $3g_1+2g_7$ is another element. To be pedantic, $5$ and $g_i$ are not elements of $\mathbb Z[G]$, though we could identify them with $5e$ and $1g_i$ where $e$ is the unit of $G$. The multiplication rule on these elements is defined in terms of the original multiplication rule of $G$ as well as rules for manipulating linear combinations in general. Concretely, call the multiplication operation $\cdot$, then the rule is $$\left(\sum_{g\in G} a_g g\right)\cdot\left(\sum_{g\in G} b_g g\right) = \sum_{g\in G}\left(\sum_{g=hk}a_hb_k\right)g$$ where $a_g,b_g\in\mathbb Z$ for each $g\in G$. These sums are formal linear combinations. Indeed, the function $g\mapsto a_g : G \to \mathbb Z$ completely describes the element $\sum_{g\in G} a_g g$, so we could say that the elements of $\mathbb Z[G]$ are these functions. Write $\bar a, \bar b : G\to\mathbb Z$ for two such functions. Then we can describe the multiplication equivalently as $\bar a\cdot\bar b = g\mapsto \sum_{g=hk}\bar a(h)\bar b(k)$. Both of these descriptions are implicitly relying on $G$ being finite, though they can easily be adapted to infinite $G$ with a bit of care.

At any rate, the main thing to get here is that the elements of a ring don't have to be "simple" things, and that we're building a new ring with its own notion of element and multiplication from the original group $G$. The new multiplication is defined in terms of the multiplication on $G$, but is not the same as it. There is no question of "commuting" any elements of $G$.

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  • $\begingroup$ Hi -- thanks, I'm almost there! 1. What does it mean to be "formal" -- like you write it out "in the spirit" of something that looks familiar but is not yet defined? 2. What do you mean by "the function $g\mapsto a_g : G \to \mathbb Z$ completely describes the element $\sum_{g\in G} a_g g$" and why would such a characterization give us all the elements of $\mathbb{Z}[G]$? For instance, if I associate $g \mapsto a_g$ then the $\sum a_gg \mapsto \sum gg$ so I got confused. $\endgroup$
    – yoshi
    Commented Apr 18, 2018 at 1:18
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    $\begingroup$ Answering your second question: $g\mapsto a_g$ means the function $f$ where $f(g)=a_g$. I used the name $\bar a$ rather than $f$ in the answer. I don't know how you get $\sum a_g g \mapsto\sum gg$. The function isn't defined for formal linear combinations, and even if it was it maps $g$ to $a_g$ not the other way around. $\endgroup$ Commented Apr 18, 2018 at 4:24
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    $\begingroup$ As for "formal", it's a relatively simple concept but hard to explain concisely since people are so thoroughly trained to identify syntax with its interpretation. I may write a blog post providing such an explanation as there doesn't seem to be any handily accessible, short, introductory explanation. The technical definition is that we are considering a quotient of a term algebra, but that probably doesn't mean much to you. As an oversimple example, when we say $2+3i$ for $\mathbb C$, we really just mean the pair $(2,3)$, we don't actually have an interpretation for how to add $2$ and $3i$. $\endgroup$ Commented Apr 18, 2018 at 4:26
  • $\begingroup$ ah okay I think I get it, $\mathbb{Z}[G]$ is the ring of maps (examples of which are $\overline{a}$ and $\overline{b}$), with multiplication and addition appropriately defined? $\endgroup$
    – yoshi
    Commented Apr 18, 2018 at 5:17
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    $\begingroup$ @yoshi More or less, yes. $\mathbb Z[G]$ is a ring of maps $G\to\mathbb Z$ where $\bar a$ and $\bar b$ are (generic) elements. Addition is defined point-wise, but multiplication is defined with the convolution formula I gave, but this convolution only makes sense when $G$ is finite, or at least we restrict to maps with finite support. We could also define multiplication point-wise, but that would give a different ring than $\mathbb Z[G]$, and in that case it wouldn't matter if $G$ was a group or finite. $\endgroup$ Commented Apr 18, 2018 at 5:24
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You should be thinking of an element $a_i g_i$ (where $a_i \in \mathbb{Z}$) as the sum of $g_i$ with itself $a_i$ times. So this element is really $$ 2g_i 5g_j = (g_i + g_i)(g_j + g_j + g_j + g_j + g_j) $$ and using the basic ring axioms, you can see that this expands to $g_{i}g_j$ added to itself $10$ times.

In a more "categorical" sense, and perhaps this won't make sense to you now, but what this says is that for every ring $R$ there is a unique ring morphism $\mathbb{Z} \to R$ which lands in the center of $R$. Namely, I must send $1 \in \mathbb{Z}$ to $1 \in R$, and this uniquely determines the ring morphism. Since this lands in the center of $R$, this tells you why the $5$ and the $g_i$ commute.

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    $\begingroup$ This is misleading. The construction being discussed, though the OP probably isn't aware of it, is the group ring construction. In general, the coefficients could be elements of some arbitrary ring for which your description would not work. This suggests something else is going on, i.e. that we're not relying on identifying $2g_i$ with $g_i+g_i$. $\endgroup$ Commented Apr 17, 2018 at 4:18
  • $\begingroup$ Well yes, I suppose you mean the general principle is of a ring morphism $R \to S$ where the image lands in the center of $S$? I'm not really sure why it's misleading, since I'm simply saying you can just think of this ring $RG$ as the free $\mathbb{Z}$-algebra (free from the category of groups) generated by $G$, i.e. the group ring. $\endgroup$
    – MT_
    Commented Apr 17, 2018 at 4:23
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    $\begingroup$ My point is that the professor could have used $\mathbb R$ instead of $\mathbb Z$ and we could not describe $\pi g_i$ as the sum of $\pi$-many copies of $g_i$. What I find misleading is that your description strongly suggests that the coefficients being integers is important, but it is only important to a coincidence of the particular example. For the general group ring construction, you should not "be thinking of an element $a_ig_i$ [...] as the sum of $g_i$ with itself $a_i$ times" as this view fails as soon as we consider other coefficient rings. $\endgroup$ Commented Apr 17, 2018 at 4:39
  • $\begingroup$ I was just giving an ad hoc way to do it in this case, and offered insight to the general case at hand. And my post is more emphasizing the fact that every ring has a canonical map from $\mathbb{Z}$, while yours is more about the formal construction which works for an $R$-algebra. I think it's fine, but okay. $\endgroup$
    – MT_
    Commented Apr 17, 2018 at 12:52

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