MHB 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 $$G$$ is finite and has order $$|G| = n$$

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

So, two specific elements could be as follows:

$$\sum \alpha_g g$$

$$ = \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \alpha_{g_n} g_n $$

and

$$\sum \beta_h h$$

$$ = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \beta_{h_n} h_n $$

and the product would be as follows:

$$ ( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n $$
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 $$G$$ is finite and has order $$|G| = n$$

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

So, two specific elements could be as follows:

$$\sum \alpha_g g$$

$$ \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \alpha_{g_n} g_n $$

and

$$\sum \beta_h h$$

$$ = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \beta_{h_n} h_n $$

and the product would be as follows:

$$( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n $$
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

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 $$\sum_{g\in G} k_g\, g$$,where the $k_g \in \Bbb k$. The zero in $\Bbb kG$ is defined to be the sum $$\sum_{g\in G} 0g$$. 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 $$\sum_{g\in G} k_g\, g = 0$$ 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 $$G$$ is finite and has order $$|G| = n$$

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

So, two specific elements could be as follows:

$$\sum \alpha_g g$$

$$ = \alpha_{g_1} g_1 +\alpha_{g_2} g_2 + \ … \ … \ + \alpha_{g_n} g_n $$

and

$$\sum \beta_h h$$

$$ = \beta_{h_1} h_1 + \beta_{h_2} h_2 + \ … \ … \ + \beta_{h_n} h_n $$

and the product would be as follows:

$$ ( \sum \alpha_g g ) ( \sum \beta_h h ) = \alpha_{g_1} \beta_{h_1} g_1 h_1 + \alpha_{g_2} \beta_{h_2} g_2 h_2 + \ … \ … \ + \alpha_{g_n} \beta_{h_n} g_n h_n $$
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