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Section 2.11 An example: the regular representation

As a last example in this section, let us study the regular representation for any finite group, which was introduced in Example 2.1.5. Recall that the regular representation is defined using Cayley's theorem. We think of a finite group \(G\) of order \(n\) as a subgroup of \(S_n\text{,}\) and keep only the matrices of the \(n\)-dimensional permutation representation of \(S_n\) that correspond to the elements of the subgroup to obtain the regular representation of \(G\text{.}\) Thus the regular representation is a \(n\)-dimensional representation.

It turns that for any finite group \(G\text{,}\) the characters of the regular representation are very easy to compute. First, the matrix corresponding to the identity element is the identity matrix, thus its trace is \(n\text{,}\) the dimension of the representation. What is the trace of the matrices corresponding to the other group elements in \(G\text{?}\) Recall from Cayley's theorem that one defines the mapping \(G \to S_n\) by looking at the permutations of the group elements corresponding to the rows of the mulitiplication table. But by the rearrangement theorem Theorem 1.4.2, we know that each row must permute every single element (no element can be left fixed, since they cannot appear twice in the same column in the multiplication table). Thus, the corresponding \(n\)-dimensional permutation matrices will necessarily only have zeroes on the diagonal (otherwise it would leave some elements fixed). Their traces are then necessarily zero!

Therefore we conclude that for any finite group \(G\) of order \(n\text{,}\) its \(n\)-dimensional regular representation \(R\) will have characters \(\chi_1^{(R)} = n\) for the conjugacy class containing the identity element and \(\chi_i^{(R)} = 0\) for all other conjugacy classes \(i > 1\text{.}\)

With this we can show that for all finite groups with \(n \geq 2\) the regular representation is reducible. Indeed, by the reducibility criterion Theorem 2.9.1, we get:

\begin{equation*} \sum_{i=1}^c n_i |\chi_i^{(R)} |^2 = 1 \cdot n^2 + 0 = n^2, \end{equation*}

which is certainly \(> |G| = n\) for all \(n \geq 2\text{.}\)

We can also find its decomposition into a direct sum of irreducible representations for all finite groups! Recall from the decomposition theorem Theorem 2.9.2 that, if we write the decomposition as

\begin{equation*} T^{(S)} = \bigoplus_{\alpha=1}^c m_\alpha T^{(\alpha)}, \end{equation*}

the coefficients are given by

\begin{equation*} m_\alpha = \frac{1}{|G|} \sum_{i=1}^c n_i \chi_i^{(S)} (\chi_i^{(\alpha)})^*. \end{equation*}

In the case of the regular representation of a group of order \(n\text{,}\) since the only non-vanishing character is \(\chi_1^{(S)} = n\text{,}\) for any \(\alpha\) we calculate:

\begin{equation*} m_\alpha = \frac{1}{n} \left( n (\chi_1^{(\alpha)})^* \right)= (\chi_1^{(\alpha)})^*. \end{equation*}

But for any irreducible representation, the character of the identity element is the trace of the identity matrix, which is just the dimension of the representation:

\begin{equation*} \chi_1^{(\alpha)} = d_\alpha. \end{equation*}

Therefore \(m_\alpha = d_\alpha\text{,}\) for all irreducible representations!

Therefore for any finite group the regular representation decomposes as the direct sum of irreducible representations:

\begin{equation*} T^{(S)} = \bigoplus_{\alpha=1}^c d_\alpha T^{(\alpha)}\text{,} \end{equation*}

whose coefficients are precisely the dimensions of the irreducible representations! In other words, in the decomposition each irreducible representation appears exactly the same number of times as its dimension. Isn't that cool?

For instance, the six-dimensional regular representation of \(S_3\) has a direct sum decomposition:

\begin{equation*} T^{(S)} = T^{(1)} \oplus T^{(2)} \oplus T^{(3)} \oplus T^{(3)}, \end{equation*}

where as before \(T^{(1)}\) is the one-dimensional identity representation, \(T^{(2)}\) is the one-dimensional parity representation, and \(T^{(3)}\) is the two-dimensional irreducible representation of \(S_3\) (since the latter is two-dimensional it appears twice in the decomposition of the regular representation).

Remark 2.11.1.

The existence of the regular representation for all finite groups also provide a simple proof of Theorem 2.8.6 about the dimensions of irreducible representations of finite groups. Indeed, the regular representation is \(n\)-dimensional, where \(n = |G|\text{,}\) and is equivalent to a block diagonal matrix with blocks of size \(d_\alpha\text{,}\) where \(d_\alpha\)is the dimension of the irreducible representations, and such that each block appears exactly \(d_\alpha\) times. Thus, simply comparing the dimension of the matrices, we conclude directly that

\begin{equation*} \sum_{\alpha=1}^c d_\alpha^2 = |G|. \end{equation*}
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