To understand the definition of induced representation, let us work through a simple example in detail.

Let $G$ be the group of permutations of three objects and let $H$ be the subgroup of even permutations. We have $$G = \{ e, (ab), (ac), (bc), (abc), (acb) \}$$ $$H = \{ e, (abc), (acb) \}$$

Let $V$ be the one dimensional representation of $H$ . Being one-dimensional, $V$ is spanned by a single basis vector $v$ . The action of $H$ on $V$ is given as $$e v = v$$ $$(abc) v = \exp (2 \pi i / 3) v$$ $$(acb) v = \exp (4 \pi i / 3) v$$

Since $H$ has half as many elements as $G$ , there are exactly two cosets, $\sigma_1$ and $\sigma_2$ in $G/H$ where $$\sigma_1 = \{ e, (abc), (acb) \}$$ $$\sigma_2 = \{ (ab), (ac), (bc) \}$$

Since there are two cosets, the vector space of the induced representation consists of the direct sum of two formal translates of $V$ . A basis for this space is $\{ \sigma_1 v, \sigma_2 v \}$ .

We will now compute the action of $G$ on this vector space. To do this, we need a choice of coset representatives. Let us choose $g_1 = e$ as a representative of $\sigma_1$ and $g_2 = (ab)$ as a representative of $\sigma_2$ . As a preliminary step, we shall express the product of every element of $G$ with a coset representative as the product of a coset representative and an element of $H$ . $$e \cdot g_1 = e = g_1 \cdot e$$ $$e \cdot g_2 = (ab) = g_2 \cdot e$$ $$(ab) \cdot g_1 = (ab) = g_2 \cdot e$$ $$(ab) \cdot g_2 = e = g_1 \cdot e$$ $$(bc) \cdot g_1 = (bc) = g_2 \cdot (acb)$$ $$(bc) \cdot g_2 = (abc) = g_1 \cdot (abc)$$ $$(ac) \cdot g_1 = (ac) = g_2 \cdot (abc)$$ $$(ac) \cdot g_2 = (acb) = g_1 \cdot (acb)$$ $$(abc) \cdot g_1 = (abc) = g_1 \cdot (abc)$$ $$(abc) \cdot g_2 = (bc) = g_2 \cdot (acb)$$ $$(acb) \cdot g_1 = (acb) = g_1 \cdot (acb)$$ $$(acb) \cdot g_2 = (ac) = g_2 \cdot (abc)$$ We will now compute of the action of $G$ using the formula $g(\sigma v) = \tau (hv)$ given in the definition. $$e (\sigma_1 v) = [e \cdot g_1] (e v) = \sigma_1 v$$ $$e (\sigma_2 v) = [e \cdot g_2] (e v) = \sigma_2 v$$ $$(ab) (\sigma_1 v) = [(ab) \cdot g_1] (e v) = \sigma_2 v$$ $$(ab) (\sigma_2 v) = [(ab) \cdot g_2] (e v) = \sigma_1 v$$ $$(bc) (\sigma_1 v) = [(bc) \cdot g_1] ((acb) v) = \exp(4 \pi i / 3) \sigma_2 v$$ $$(bc) (\sigma_2 v) = [(bc) \cdot g_2] ((abc) v) = \exp(2 \pi i / 3) \sigma_1 v$$ $$(ac) (\sigma_1 v) = [(ac) \cdot g_1] ((abc) v) = \exp(2 \pi i / 3) \sigma_2 v$$ $$(ac) (\sigma_2 v) = [(ac) \cdot g_2] ((acb) v) = \exp(4 \pi i / 3) \sigma_1 v$$ $$(abc) (\sigma_1 v) = [(abc) \cdot g_1] ((abc) v) = \exp(2 \pi i / 3) (\sigma_1 v)$$ $$(abc) (\sigma_2 v) = [(abc) \cdot g_2] ((acb) v) = \exp(4 \pi i / 3) (\sigma_2 v)$$ $$(acb) (\sigma_1 v) = [(acb) \cdot g_1] ((acb) v) = \exp(4 \pi i / 3) (\sigma_1 v)$$ $$(acb) (\sigma_2 v) = [(acb) \cdot g_2] ((abc) v) = \exp(2 \pi i / 3) (\sigma_2 v)$$ Here the square brackets indicate the coset to which the group element inside the brackets belongs. For instance, $[(ac) \cdot g_2] = [(ac) \cdot (ab)] = [(acb)] = \sigma_1$ since $(acb) \in \sigma_1$ .

The results of the calculation may be easier understood when expressed in matrix form $$e \qquad \to \qquad \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix}$$ $$(ab) \qquad \to \qquad \begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}$$ $$(bc) \qquad \to \qquad \begin{pmatrix} 0 & \exp(2 \pi i / 3) \cr \exp(4 \pi i / 3) & 0 \end{pmatrix}$$ $$(ac) \qquad \to \qquad \begin{pmatrix} 0 & \exp(4 \pi i / 3) \cr \exp(2 \pi i / 3) & 0 \end{pmatrix}$$ $$(abc) \qquad \to \qquad \begin{pmatrix} \exp(2 \pi i / 3) & 0 \cr 0 & \exp(4 \pi i / 3) \end{pmatrix}$$ $$(acb) \qquad \to \qquad \begin{pmatrix} \exp(4 \pi i / 3) & 0 \cr 0 & \exp(2 \pi i / 3) \end{pmatrix}$$ Having expressed the answer thus, it is not hard to verify that this is indeed a representation of $G$ . For instance, $(acb) \cdot (ac) = (bc)$ and $$\begin{pmatrix} \exp(4 \pi i / 3) & 0 \cr 0 & \exp(2 \pi i / 3) \end{pmatrix} \begin{pmatrix} 0 & \exp(4 \pi i / 3) \cr \exp(2 \pi i / 3) & 0 \end{pmatrix} = \begin{pmatrix} 0 & \exp(2 \pi i / 3) \cr \exp(4 \pi i / 3) & 0 \end{pmatrix}$$