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I have just started studying group representations with the book Representations of Groups by Lux and Pahlings (published by Cambridge). I have tried to solve some exercises to understand the concept of group algebras, but I must admit that it seems I don't get the subtleties of this concept. Here is the problem I can't solve:

Let $$g= \left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right) \in GL_2(\mathbb{F}_2)$$ It is easy to check that $G:=\langle g \rangle\cong C_3$. Now the exercise asks to prove that the group algebra $\mathbb{F}_2G$ is not isomorphic to $$S:=\{\sum^{3}_{i=1}a_ig^i \mid a_i \in \mathbb{F}_2\} \subseteq \mathbb{F}_2^{2 \times 2}.$$

However, - and this is where I confess my ignorance - I thought that $\mathbb{F}_2G$ would be isomorphic to $S$. Now, since the 'not' in the exercise is emphasized in the book, I suppose that this exercise warns about a common mistake beginners make.

Could you help me on this and explain me the subtlety that I do not get?

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In the group algebra $\mathbb{F}_{2}G$, the elements $1_{G},g$ and $g^{2}$ are linearly independent. In the matrix algebra $\mathbb{F}_{2}^{2 \times 2}$, for the given matrix $g$ of order $3$, we do have $I + g + g^{2} = [0]_{2 \times 2}$. This is quite closely related to what Carl says below. –  Geoff Robinson Jul 28 '11 at 13:09
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4 Answers

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Someone else will write a longer answer, but the key here is that the element $g + g^2$ of the group algebra is not the same as the element $g^3$ (which is the identity of the algebra). A quick calculation shows that in $\mathbb{F}^{2\times 2}_2$ the matrix $g + g^2$ is the same as the matrix $g^3$. In other words $S$ does not have $9$ distinct elements, but the group algebra does. The elements of the group algebra are formal sums, while the elements in $\mathbb{F}^{2\times 2}_2$ are actual sums.

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OK, I get it. However, I guess you mean 8 elements instead of 9, isn't it? –  Thomas Connor Jul 28 '11 at 13:06
    
I didn't do all the other calculations to see how many elements $S$ actually has, and I don't know off the top of my head how many there will be. But there are definitely no more than 8. –  Carl Mummert Jul 28 '11 at 13:15
    
Well, I see your point. Thank you for this answer. –  Thomas Connor Jul 28 '11 at 13:18
    
@Thomas Right, there's an $\mathbf F_2$-basis $\{1, g, g^2\}$ for $\mathbf F_2[G]$. –  Dylan Moreland Jul 28 '11 at 14:00
    
... and $\{I_2, g\}$ appears to be an $\mathbf F_2$-basis for $S$. –  Dylan Moreland Jul 28 '11 at 14:26
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You may perhaps be less surprised about the same result for a different representation. For $g=\mathrm e^{2\pi\mathrm i/3}$, we also have $G:=\langle g \rangle\cong C_3$, but it would probably not have occurred to you that

$$S:=\{\sum^{3}_{i=1}a_ig^i \mid a_i \in \mathbb C\} =\mathbb C$$

should be isomorphic to the group algebra $\mathbb CG$.

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Thanks! This helps very much. –  Thomas Connor Jul 28 '11 at 14:00
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The point is that the group algebra is not a sub-thing of the group itself, nor of matrices made from the group elements! So the scalar multiples $a_ig^i$ are not scalar multiples of the matrices $g^i$! The sum is not a sum of matrices!

"What is it, then?!?" is the reasonable question. Some people say "oh, it's just symbols, or marks-on-the-page", but this evades the question. A reasonable interpretation/definition/characterization is that the group algebra is _functions_on_the_group_.

(Edit: and, yes, as in comments, after we're used to the idea that the "group algebra" of a matrix group is not literally a ring of matrices, refinements/corrections/improvements are appropriate or necessary: in any case, the multiplication is certainly not pointwise, but/and can be "discovered" through the idea that repn spaces for the group should naturally be modules over the group algebra... dictating the product! For discrete groups, often _finite_support_ is required, and possibly _compact_support_ more generally. The covariant/contravariant issue is similarly understood by looking at which way arrows ought to go.)

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What I don't like about functions on the group, though, is that it makes me think of a contravariant gadget, while it should be thought of as covariant in the group (and the functions should be finitely supported if the group is infinite). But that's probably too nitpick-y. –  t.b. Jul 28 '11 at 13:02
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I think this is an important nitpick. In addition to this contravariant/covariant issue, functions on the group and the group algebra have different products (pointwise vs. the natural product on formal sums) and different coproducts. Instead you should think of the group ring as dual to the ring of functions. Then you get everything right (since dual is contravariant). –  Noah Snyder Jul 28 '11 at 14:32
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IOW, the group algebra is the algebra of distribution on the group :) –  Mariano Suárez-Alvarez Jul 28 '11 at 15:27
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Definitions are just definitions, so you may want to ask: even if a group ring is defined differently, doesn't the ring I defined make sense? What property does it lack? I'll use the same way of defining a ring in a few cases to show the answer depends on how you write down the group, but that there is a "best" way to write down the group that handles all the others. The best way is called both the regular representation and the group ring.

The small ring S

Your original ring S can be defined using G: $$g = \left[\begin{array}{rr} . & 1 \\ -1 & -1 \end{array}\right] \qquad G = \langle g \rangle \qquad S = F[g] = \left\{ a g^0 + b g^1 : a,b \in F \right\}$$

The ring S has an F-basis { g0, g1 }, since g2 + g1 + g0 = 0. In other words, we have an F-algebra generated by a single element subject to a relation. That means SF[x]/(x2+x+1).

This S is actually a field whenever x2+x+1 is irreducible, so when F is the field of two elements, S is the field of four elements. Fields are very nice, so S is a nice ring, but it turns out it is too small to really capture all cyclic groups of order three.

A bigger ring T

The original matrices of G can be cleverly expanded using the Kronecker product to form a group H, by defining: $$ h = \left[\begin{array}{rr|rr} . & . & . & 1 \\ . & . & -1 & -1 \\ \hline . & -1 & . & -1 \\ 1 & 1 & 1 & 1 \end{array}\right] = \left[\begin{array}{rr} 0g & 1g \\ -1g & -1g \end{array}\right] \qquad H = \langle h \rangle \qquad T = F[h]$$

Now one quickly verifies that while h3 = h0, the set { h0, h1, h2 } is linearly independent over F. Putting these two together we have that { h0, h1, h2 } is a basis of T. As before T is an F-algebra generated by a single element subject to a relation, and TF[x]/(x3−1) ≅ S × F.

It is still obvious that H ≅ C3, so we'd expect our rings for G and for H to be isomorphic. However S has F-dimension 2, and T has F-dimension 3, so they are not isomorphic over any field F.

Even worse, there is no (unital) ring homomorphism from S to T. This is particularly easy to see when F is the field of two elements, since then S is a field, and so the only (nonzero, unital) ring homomorphisms are isomorphisms. Another way to phrase this is to say that T is not even an S-module; this is true over all fields F, but is particularly easy to see when S is a field.

Simply put, the ring we defined using G, namely S, is too small; it does not capture properties of C3 that are clearly evident using H and T.

The just-right ring R

Ok, so maybe S is too small. Why don't we use T? Well, T is three-dimensional, but h is a 4×4 matrix. If we used the basis { h0, h1, h2 } then we can write each power of h as a 3×3 matrix actng as multiplication on that basis. We would get a group K, defined as: $$k = \begin{bmatrix} . & 1 & . \\ . & . & 1 \\ 1 & . & . \end{bmatrix} \qquad K=\langle k\rangle \qquad R = F[k]$$

Now again K ≅ C3. Coincidentally, RT, taking the basis { k0, k1, k2 } in the obvious way to the basis { h0, h1, h2 }. This matrix representation is called the regular representation and is used in the typical definition of the group ring. It represents the group of order n as n × n matrices acting as permutations on a basis indexed by the underlying group.

Would we ever need a bigger ring? I mean sure H looked bigger while the smaller K still covered it, but maybe there are versions of C3 that are even bigger! Luckily K is always big enough. Any ring we define for C3 should at the very least be spanned by the group, and so we can write down the group as a quotient of the polynomial ring in the variable x, subject of course to x3 = 1 so that x generates a group X isomorphic to C3. Thus we have a quotient of a three-dimensional ring, F[x]/(x3−1) = R.

In other words, every ring defined by a representation of C3 over the field F will be a quotient of the one-true ring R. Hence R is universal, and is called the group ring.

Other finite groups

Notice how S, T, and R were direct products of extension fields of F. This holds not just for C3, but for any finite abelian group and any base field whose characteristic is coprime to the order of the group. For any finite abelian group, the regular representation is the smallest dimensional representation that still generates the group ring. The fields involved are defined as the splitting fields of the irreducible factors of xn−1, where n is the exponent of the group. If the characteristic of the field divides the order of the group, then xn−1 is no longer separable and you get mildly more complicated rings.

For other finite groups, the rings involved cannot all be commutative, otherwise the group inside the group ring would still be commutative. As long as the characteristic of the field does not divide the group order, the rings involved are all matrix rings over division F-algebras. If xn−1 already splits over F, then you just get a direct product of matrix rings over F. When the characteristic of the field divides the order of the group, things are significantly more complicated. The regular representation is usually not the smallest representation generating the group ring, but it is still the easiest to understand.

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OK, I think I'll need some more time to catch everything, but I'm getting a better understanding on the subject! Thank you very much for this answer. –  Thomas Connor Jul 28 '11 at 14:42
    
@Thomas: No problem. My answer is basically the same as joriki's. His says, "if you use your definition then you get really small rings!" Mine says "there are big rings too, but actually there is a biggest ring, and it is called the group ring." –  Jack Schmidt Jul 28 '11 at 14:52
    
@Thomas: I rewrote it to be a little easier to read piece by piece. I also wrote it so that it doesn't depend on the field very much, though when I said TF × S, I was assuming F had characteristic not 3 (when F has char 3, then $S\cong F[x]/(x^2)$ and $T \cong F[x]/(x^3)$ are no longer direct products of fields). –  Jack Schmidt Jul 28 '11 at 16:31
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