The character tables of the irreducible representations of $T_d$ and $C_{3v}$ are linked. In the notation on those pages, $A_1$ and $A_2$ are irreducible representations of degree 1, $E$ is degree 2 and $T$ is degree 3. All the representation are in characteristic 0. Refer only to the character tables, not the product tables.

$T_d$ is isomorphic to the symmetric group $S_4$ . In the representations of $T_d$, identify the conjugacy class $E$ with identity element, the conjugacy class $8C_3$ with $8$ elements which are $3-$cycles, the class $3C_2$ as the $3$ elements which are product of $2$ $2-$cycles, the class $6\sigma_d$ with 6 elements which are $2-$cycles, and the class $6S_4$ with 6 elements which are $4-$cycles.

$C_{3v}$ is isomorphic to $S_3$. In the representations of $C_{3v}$, identify the conjugacy class $E$ with identity element, the conjugacy class $2C_3$ with $2$ elements which are $3-$cycles and the class $3\sigma_v$ with 3 elements which are $2-$cycles.

If $V_N=\{(), (1 2)(3 4), (13)(24),(14)(23)\}$ is the normal Kelin-4 subgroup of $S_4$, then $S_4=S_3\ltimes V_N$. In the isomorphism $$S_4/V_N\cong S_3,$$ it is readily verified that the $3-$cycles in $S_4$ map to $3-$cycles in $S_3$, the $2-$cycles as well as the $4-$cycles in $S_4$ map to $2-$cycles in $S_3$, whereas the elements which are products of 2 $2-$cycles in $S_4$ map to the identity in $S_3$.

The point that comes across is that the characters of all the classes in $T_d$ in representations which are identified by the same symbol in $C_{3v}$ are exactly equal to the characters of the classes in $C_{3v}$ to which those classes in $T_d$ are mapped. For example, in the representations $E$, the characters of $6\sigma_d$ and the $6S_4$ classes in the group $T_d$ are the same as that of the class $3\sigma_v$ of the group $C_{3v}$. Similarly for other classes and characters.

I know how to calculate characters of irreducible representations of such a group which is a direct products of its subgroups (whose irreducible characters are known). Is there a general way to do this for groups which are semi-direct products as above? Can the observation made in the previous paragraph be generalized in some way, or these observations are specific to this particular example?


1 Answer 1


What you want to read is Serre's book on Linear Representation of Finite Groups. Section 8.2 of chapter 8 covers the case of a semi-direct product by an abelian group. For the more general case, you want to use the theory of induced characters in a more subtle way. This is not as trivial as it sounds though ; make sure you have the mathematical background to read it. If some words sound unfamiliar, feel free to comment here to ask what they mean so that you know where to look.

Hope that helps,

  • 1
    $\begingroup$ The material in the first 5 chapters, especially at the end of $5^{th}$ nearly solves my problem and actually by the same example. My familiarity with semisimple modules (which seems essential for further reading) is limited, but would read on them when time permits. Thanks. $\endgroup$
    – vnd
    Apr 30, 2016 at 5:35
  • $\begingroup$ You don't need to be "familiar" with semisimple modules to read on ; the book explains the necessary module theory that it uses. Don't be scared of the word "semisimple" ; if $G$ is a finite group, any finitely generated $\mathbb C[G]$-module is semisimple by Maschke's theorem. $\endgroup$ May 1, 2016 at 20:32

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