I am reading the first set of examples in Chapter 18 Representation Theory & Character Theory and Dummit and Foote - and cross referencing with examples in James and Liebeck's "Representations and Characters of Groups"

I have a (possible) problem with Example 2 page 844 from D&F and am hopign someone can help!

The text of the problem in D&F is as follows: (see attachement of page 844 D&F)

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Let V = FG and consider this ring as a left module over itself. Then V affords a representation of G of degree equal to |G|. If we take elements of G as a basis for V, then each $\displaystyle g \in G $ permutes these basis elements under the left regular representation:

$\displaystyle g \cdot g_i = g g_i$

[Note/question by Peter: I thought we had to specify an action of a ring element g on an element v of V= FG - but then I suppose if we specify an action on each basis element then this approximates to the same thing??]

With respect to this basis of V, the matrix of the group element g has a 1 in row i and column j if $\displaystyle g g_i $ , and has 0s in alll other positions.

This linear or matrix representation is called the regular representation of G

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Since I understand maths better when I have a concrete example I took an example from James and Liebeck - see attachement pages 56 -57 of J&L)

J&L took V = FG where $\displaystyle G = C_3 $, the cyclic group of order 3 $\displaystyle < a: a^3 = e >$ where the set of elements $\displaystyle \{ g_1, g_2, g_3 \} = \{ e, a, a^2 \} $

Now J&L use a slightly different approach and notation so I tried to follow the D&F approach and notation and work through the problem, thus ....

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Consider $\displaystyle g = g_1 = e $

$\displaystyle g g_1 = ee = e $ so $\displaystyle g g_1 = g_1 $

Thus we have 1 in row 1 and column 1

$\displaystyle g g_2 = ea = a $ so $\displaystyle g g_2 = g_2 $

Thus we have 1 in row 2, column 2

$\displaystyle g g_3 = e a^2 = a^2 $ so $\displaystyle g g_3 = g_3 $

Thus, again following D&F we have a 1 in row 3 and column 2

Now apart from the 1s above the reset of the matrix for $\displaystyle g = g_1 = e $ is the identity matrix

Thus the linear mapping for $\displaystyle g_1 = e$ is as follows:

$\displaystyle e \longrightarrow \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{ar ray}\right) $

Mapping is $\displaystyle e \longrightarrow $

1 0 0

0 1 0

0 0 1

[This squares with J&L page 57 !

Consider now $\displaystyle g = g_2 = a $

$\displaystyle g g_1 = ae = a $ so $\displaystyle g g_1 = g_2$

Thus we have a 1 in row 2 and column 1

$\displaystyle g g_2 = a.a = e $ so $\displaystyle g g_2 = g_1 $

Thus we have a 1 in row 1, column 2

$\displaystyle g g_3 = a a^2 = a $ so $\displaystyle g g_3 = g_2 $

Thus we have a 1 in row 2 and column 3

Thus the linear mapping for $\displaystyle g_2 = a$ is as follows:

Mapping is $\displaystyle a \longrightarrow $

0 1 0

1 0 1

0 0 0

[PROBLEM - this does not sqare with J&L page 57 which gives

Mapping is $\displaystyle a \longrightarrow $

0 1 0

0 0 1

1 0 0 ]

Consider now $\displaystyle g = g_3 = a^2 $

$\displaystyle g g_1 = a^2 e = a^2 $ so $\displaystyle g g_1 = g_3$

Thus we have a 1 in row 3 and column 1

$\displaystyle g g_2 = a^2 a = e $ so $\displaystyle g g_2 = g_1$

Thus we have a 1 in row 1 and column 2

$\displaystyle g g_3 = a^2 a^2 = a $ so $\displaystyle g g_3 = g_2$

Thus we have a 1 in row 2 and column 3

Thus the linear mapping for $\displaystyle g_3 = a^2 $ is as follows:

Mapping is $\displaystyle a^2 \longrightarrow $

0 1 0

0 0 1

1 0 0

[PROBLEM - this does not sqare with J&L page 57 which gives

Mapping is $\displaystyle a^2 \longrightarrow $

0 0 1

1 0 0

0 1 0 ]

Can someone please help by

1. Confirming that my general analysis following D&F is correct (or not!)

2. Resolving the problem between my working and that of J&L pages 56 -57 (please see attachement) [Hmmm ... is there some problem with rows and columns in the two approaches?]

Peter