 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2020/3/a/solution

The componentwise product is closed in

G_{1} \times G_{2}

. Associativity follows componentwise from associativity in each group:

((g_{1}, g_{2}) (h_{1}, h_{2})) (k_{1}, k_{2}) = ((g_{1} h_{1}) k_{1}, (g_{2} h_{2}) k_{2}) = (g_{1} (h_{1} k_{1}), g_{2} (h_{2} k_{2})) .

The identity is

(e_{1}, e_{2})

, and

(g_{1}, g_{2})^{- 1} = (g_{1}^{- 1}, g_{2}^{- 1})

. These verify all the group axioms, so this is the direct product of groups.

\times

\otimes

. Its mixed-product identity and the representation laws of

R_{1}, R_{2}

give

R (g_{1}, g_{2}) R (h_{1}, h_{2}) = (R_{1} (g_{1}) \otimes R_{2} (g_{2})) (R_{1} (h_{1}) \otimes R_{2} (h_{2})) = R_{1} (g_{1}) R_{1} (h_{1}) \otimes R_{2} (g_{2}) R_{2} (h_{2}) = R_{1} (g_{1} h_{1}) \otimes R_{2} (g_{2} h_{2}) = R ((g_{1}, g_{2}) (h_{1}, h_{2})) .

Also

R (e_{1}, e_{2}) = I_{n_{1}} \otimes I_{n_{2}} = I_{n_{1} n_{2}}

, and

R (g_{1}, g_{2})^{- 1} = R_{1} (g_{1})^{- 1} \otimes R_{2} (g_{2})^{- 1} .

n_{1} n_{2}

, called the external tensor product of group representations. No irreducibility assumption is needed for this construction.

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