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xy is injective (by the cancellation hw) we deduce that IxXI = IXI and hence, since X is finite, that xX = X. Consequently, x = xe for some e E X. The cancellation hw gives e = 1, and so we have that 1 E X.

Elements. Thus there are two conj ugacy classes of cycles of length n - 1 in An each containing ~n(n - 2)! elements. 97 x E G commutes with a E G if and only if it commutes with a-I. Hence Nc(a) = Nc(a- l ). The number of conjugates of a, being the index of Nc(a) in G, must therefore be the same as the number of conjugates of a-I. Suppose now that IGI is even and that 1 is the only element of G that is conjugate to its inverse. For each conjugacy class A; let B; be that containing the inverses.

If A2 = /2 then it is easily seen that either A = /2 or A = -h, and in either case :A is the identity of PSL(2, F) so is not an element of order 2. 20 Every element of C 2 x C2 has order 2 so, since IC2 x C2 = 4, it follows that C2 x C2 cannot be cyclic. Suppose now that G is non-cyclic and of order 4. Then every element of G has order 2 and the multiplication table is uniquely determined. Since this is the same as that of C2 x C2 the result follows. 22 Groups Let K be a normal subgroup of G x H with K =f {(I, I)}.