Group Algebra - Cohn page 55 - SIMPLE CLARIFICATION

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we read the following on page 55:https://www.physicsforums.com/attachments/3142I am trying to get an idea of what Cohn says and means by a group algebra …

Take a case where \(\displaystyle G\) is finite and has order \(\displaystyle |G| = n\)

Then according to Cohn the elements of the group algebra, \(\displaystyle kG\), are the finite sums \(\displaystyle \sum \alpha_g g\)

So, two specific elements could be as follows:

\(\displaystyle \sum \alpha_g g\)

\(\displaystyle = \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \alpha_{g_n} g_n \)

and

\(\displaystyle \sum \beta_h h\)

\(\displaystyle = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \beta_{h_n} h_n \)

and the product would be as follows:

\(\displaystyle ( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n \)
Now is the above interpretation of the elements and multiplication (dot product?) correct?
Note that in the above text, Cohn writes:

" … … Put more simply, kG has G as a k-basis … …"

What does Cohn mean by a "k-basis"?

Peter

***EDIT*** Apologies to MHB members for the previous version of their post … it had unreadable elements due to an editor problem … problem was resolved thanks to Mark … my thanks to Mark.

 

[Euge]

am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we read the following on page 55:https://www.physicsforums.com/attachments/3142



I am trying to get an idea of what Cohn says and means by a group algebra …

Take a case where \(\displaystyle G\) is finite and has order \(\displaystyle |G| = n\)

Then according to Cohn the elements of the group algebra, \(\displaystyle kG\), are the finite sums \(\displaystyle \sum \alpha_g g\)

So, two specific elements could be as follows:

\(\displaystyle \sum \alpha_g g\)

\(\displaystyle \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \)\(\displaystyle \alpha_{g_n} g_n \)

and

\(\displaystyle \sum \beta_h h\)

\(\displaystyle = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \)\(\displaystyle \beta_{h_n} h_n [/MATH]

and the product would be as follows:

\(\displaystyle ( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n \)
Now is the above interpretation of the elements and multiplication (dot product?) correct?
Note that in the above text, Cohn writes:

" … … Put more simply, kG has G as a k-basis … …"

What does Cohn mean by a "k-basis"?

Peter\)

\(\displaystyle

Given any group $G$ and any field $k$, we can form a $\Bbb k$-algebra such that the elements of $G$ together form a basis. To do so, define a set $\Bbb kG$ to be the set of all formal sums \(\displaystyle \sum_{g\in G} k_g\, g\),where the $k_g \in \Bbb k$. The zero in $\Bbb kG$ is defined to be the sum \(\displaystyle \sum_{g\in G} 0g\). Then $\Bbb kG$ is a vector space over $\Bbb k$. By construction, $G$ spans $\Bbb kG$. Further, $G$ is a linearly independent set in $\Bbb kG$. For if \(\displaystyle \sum_{g\in G} k_g\, g = 0\) is a linear relation in $\Bbb kG$, then $k_g = 0$ for all $g\in G$, again by construction. So $G$ is a $\Bbb k$-basis for $\Bbb kG$, that is, $G$ is a basis for $\Bbb kG$ as a vector space over $\Bbb k$.

Define a multiplication in $kG$ by

$\displaystyle \sum_{g\in G} k_g\, g \cdot \sum_{g\in G} j_g\, g = \sum_{u\in G} \left(\sum_{gh = u} k_g j_h\right) u$.

With this multiplication, $\alpha(xy) = (\alpha x)y = x(\alpha y)$ for all $x, y\in \Bbb kG$ and $\alpha\in \Bbb k$. It follows that $\Bbb kG$ is a $\Bbb k$-algebra, which Cohn calls the group algebra of $G$ over $\Bbb k$.\)

 

[Deveno]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we read the following on page 55:https://www.physicsforums.com/attachments/3142I am trying to get an idea of what Cohn says and means by a group algebra …

Take a case where \(\displaystyle G\) is finite and has order \(\displaystyle |G| = n\)

Then according to Cohn the elements of the group algebra, \(\displaystyle kG\), are the finite sums \(\displaystyle \sum \alpha_g g\)

So, two specific elements could be as follows:

\(\displaystyle \sum \alpha_g g\)

\(\displaystyle = \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \alpha_{g_n} g_n \)

and

\(\displaystyle \sum \beta_h h\)

\(\displaystyle = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \beta_{h_n} h_n \)

and the product would be as follows:

\(\displaystyle ( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n \)
Now is the above interpretation of the elements and multiplication (dot product?) correct?
Note that in the above text, Cohn writes:

" … … Put more simply, kG has G as a k-basis … …"

What does Cohn mean by a "k-basis"?

Peter

***EDIT*** Apologies to MHB members for the previous version of their post … it had unreadable elements due to an editor problem … problem was resolved thanks to Mark … my thanks to Mark.

Let's do an easy example, say:

$G = \{e,a,a^2\}$, a cyclic group of order 3, and $k = \Bbb Q$.

So a typical element of $\Bbb Q[G]$ looks like this:

$q_1e + q_2a + q_3a^2$ where $q_1,q_2,q_3 \in \Bbb Q$.

Let's multiply two such elements together:

$(p_1e + p_2a + p_2a^2)(q_1e + q_2a + q_3a^2)$

$ = p_1q_1(ee) + p_1q_2(ea) + p_1q_3(ea^2) + p_2q_1(ae) + p_2q_2(aa) + p_2q_3(aa^2) + p_3q_1(a^2e) + p_3q_2(a^2a) + p_3q_3(a^2a^2)$

$= (p_1q_1 + p_2q_3 + p_3q_2)e + (p_1q_2 + p_2q_1 + p_3q_3)a + (p_1q_3 + p_2q_2 + p_3q_1)a^2$

(note the similarity with polynomial multiplication, here).

Note that $1e\ (= 1e + 0a + 0a^2)$ serves as a ring identity.

Basically, we use the distributive rule to "expand everything out" and then "collect like terms" after we evaluate the group products $g_ig_j$ (the "sigma" summation notation is more compact, but obscures what is actually happening-one must pay careful attention to the indices).

Sometimes, this product is written:

$\displaystyle \left( \sum_i a_ig_i\right)\left(\sum_j b_j g_j\right) = \sum_{g_ig_j = g_k} a_ib_jg_k$

One can form a similar notion of a "monoid-ring", $k[M]$, in which case we have $k[x] \cong k[\Bbb N]$

(the corresponding notion for $k[\Bbb Z]$ would be the set of expressions:

$\dfrac{a_{-n}}{x^n} +\cdots + \dfrac{a_{-2}}{x^2} + \dfrac{a_{-1}}{x} + a_0 + a_1x + a_2x^2 + \cdots + a_mx^m$

that is to say, a finite Laurent series, or rational function in $x$ over $k$).

For the $\Bbb Q$-algebra $\Bbb Q[G]$ I have outlined above, we have the following multiplicative constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{113} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{123} = 0$
$\gamma_{131} = 0$
$\gamma_{132} = 0$
$\gamma_{133} = 1$
$\gamma_{221} = 0$
$\gamma_{222} = 0$
$\gamma_{223} = 1$
$\gamma_{231} = 1$
$\gamma_{232} = 0$
$\gamma_{233} = 0$
$\gamma_{331} = 0$
$\gamma_{332} = 1$
$\gamma_{333} = 0$

and the ones not listed follow by commutativity of $G$.

It should be clear that the group $G$ itself is a subgroup of the group of units of $k[G]$.

 

[Math Amateur]

Let's do an easy example, say:

$G = \{e,a,a^2\}$, a cyclic group of order 3, and $k = \Bbb Q$.

So a typical element of $\Bbb Q[G]$ looks like this:

$q_1e + q_2a + q_3a^2$ where $q_1,q_2,q_3 \in \Bbb Q$.

Let's multiply two such elements together:

$(p_1e + p_2a + p_2a^2)(q_1e + q_2a + q_3a^2)$

$ = p_1q_1(ee) + p_1q_2(ea) + p_1q_3(ea^2) + p_2q_1(ae) + p_2q_2(aa) + p_2q_3(aa^2) + p_3q_1(a^2e) + p_3q_2(a^2a) + p_3q_3(a^2a^2)$

$= (p_1q_1 + p_2q_3 + p_3q_2)e + (p_1q_2 + p_2q_1 + p_3q_3)a + (p_1q_3 + p_2q_2 + p_3q_1)a^2$

(note the similarity with polynomial multiplication, here).

Note that $1e\ (= 1e + 0a + 0a^2)$ serves as a ring identity.

Basically, we use the distributive rule to "expand everything out" and then "collect like terms" after we evaluate the group products $g_ig_j$ (the "sigma" summation notation is more compact, but obscures what is actually happening-one must pay careful attention to the indices).

Sometimes, this product is written:

$\displaystyle \left( \sum_i a_ig_i\right)\left(\sum_j b_j g_j\right) = \sum_{g_ig_j = g_k} a_ib_jg_k$

One can form a similar notion of a "monoid-ring", $k[M]$, in which case we have $k[x] \cong k[\Bbb N]$

(the corresponding notion for $k[\Bbb Z]$ would be the set of expressions:

$\dfrac{a_{-n}}{x^n} +\cdots + \dfrac{a_{-2}}{x^2} + \dfrac{a_{-1}}{x} + a_0 + a_1x + a_2x^2 + \cdots + a_mx^m$

that is to say, a finite Laurent series, or rational function in $x$ over $k$).

For the $\Bbb Q$-algebra $\Bbb Q[G]$ I have outlined above, we have the following multiplicative constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{113} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{123} = 0$
$\gamma_{131} = 0$
$\gamma_{132} = 0$
$\gamma_{133} = 1$
$\gamma_{221} = 0$
$\gamma_{222} = 0$
$\gamma_{223} = 1$
$\gamma_{231} = 1$
$\gamma_{232} = 0$
$\gamma_{233} = 0$
$\gamma_{331} = 0$
$\gamma_{332} = 1$
$\gamma_{333} = 0$

and the ones not listed follow by commutativity of $G$.

It should be clear that the group $G$ itself is a subgroup of the group of units of $k[G]$.

Thanks Deveno … these examples provide wonderful learning experiences … wish the texts had more like these examples ...

Peter

 

[blue_raver22]

I am not an expert in ring theory, but I can provide a general understanding of Cohn's explanation. A group algebra is a mathematical structure that combines elements from a group (G) with elements from a field (k). In this case, G is a finite group with n elements, and k is a field (such as the real or complex numbers). The elements of the group algebra, denoted as kG, are finite sums where each term includes an element from the field (α) multiplied by an element from the group (g). So, the two specific elements given are examples of these finite sums.

The product of two elements in the group algebra is defined as the sum of all possible products of the elements from the two sums. This is represented by the dot product in the given equation. The first term in the product would be the product of the first term from the first sum and the first term from the second sum, and so on.

When Cohn says that "kG has G as a k-basis," he means that the elements of G form a basis for the group algebra kG. This means that every element in kG can be written as a linear combination of the elements in G using coefficients from the field k. In other words, the elements of G span the entire group algebra kG.

I hope this helps clarify Cohn's explanation of group algebra. If you have further questions, I recommend consulting with a mathematician or referring to other resources on ring theory.