# Download PDF by Miller G. A.: An Overlooked Infinite System of Groups of Order pq2 By Miller G. A.

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20 EXERCISES 1. Show that the left and right representations f and r of a group G (in any LP(G) space) are equivalent. ) 2. Let it be a unitary representation of a group G. Show that 7r and is are equivalent. '---s (v f . ) 3. If it and rr' are two representations of the same group G (acting in respective Hilbert spaces H and H'), show that the matrix coefficients of rr ® rC' and (e. ) of H' 3 of H 0 H' ) are products of matrix coefficients of it and rr' (Kronecker product of matrices). 4. Let 1n denote the identity representation of a group G in dimension n ( the space of this identity representation is thus an and In (s) = idan for every s E G) .

Proof. 3) above. To see that ii) ==> iii) it is enough to remember that a closed subgroup of a Lie group is a Lie group, and to apply this result to the real Lie group Un((E) (also observe that since G is compact, any continuous injective map G - Un(T) is a homeomorphism into). In this case, one could even see that G is an algebraic group. Finally, the implication iii) - i) is known classically. In our case, we can use the exponential map Mn(]R) -- Gln(R) , A *-* exp(A) = L An/n! n>0 This map is a local diffeomorphism in the neighbourhood of 0 E Mn(R): there exists a neighbourhood V of the zero matrix of Dn(]R) for which e x p : V - - + exp(V) (this is a neighbourhood of In in Gln(R) ).

In which case G is a profinite group. The reader will certainly have understood that we can obtain many variations on this theme... 2) above. Start with a continuous function f e C(G) and a positive e > 0. Choose an open symmetric neighbourhood U of the neutral element of G such that y1x EU )f (Y) - f(x)I < E (f is uniformly continuous on the compact group G). Take then a continuous function Y on G such that '=4 0 , Supp (p) C U , , dy = SG 1 T (Y) Thus we can write f (x) f f (x) = y (y lx) dy G and coming back to the operator K L2(G) - L2(G) with kernel k(x,y) : = 4 (ylx), we see that Kf(x) = fG f(y) T(ylx) dy .