PSL(2, 7l). Use the above results to show that PSL(2, il) = (u,l). Show also that u3 = l2 = 1, the identity of PSL(2,71). Suppose, if possible, that some word of the form is equal to I in PSL(2, 7l). Show that there is a word of the fonn that is equal to ±I in SL(2, 7l) and, by considering the trace of w, obtain a contradiction. Finally, show that PSL(2,il) ~ (a,b I a2 28 = b3 = 1).

We have to show that P is cyclic. Now P is a p-subgroup of G so P ::::: P for some Sylow p-subgroup P of G. Since P is cyclic, so then is P. Suppose that PI and P2 are p-subgroups of C with IFII = jP2 1. We have PI ::::: PI and P2 ::::: P 2 where P;, P2 are Sylow p-subgroups of C. But PI is conjugate to P2 , so there exists 9 E G with g-1 Pig = P2 . But now g-IP1g::::: P 2 and [g-IP1gl = 1F1! = iP2 j, so g-IPI g and P2 are su bgroups of the same order in the cyclic grou p P2 . Hence g-I PI 9 = P2 as required.

J(H). We have to show that K ~ H. Suppose that IHI = nand IG/HI = m. Since HK/H ::; G/H we have that IHK/HI divides m. But we know that HK/H ~ K/(H n K), so IH K/ HI divides n. (n, m) = 1 by hypothesis. Hence ]HK/HI = 1, so HK = Hand K::; H as required. 16 If a E F then a-I E F since (g-l ag )-1 = g-l a-l g. But if a,b E F we have ab E F, for g-l abg = g-l agg- 1 bg. Hence F ::; G. Now F is characteristic in G. J E Aut G. J( a). J( a) E F. J t is clearly an additive group morphism which, since t i= 0, is injective.

### Algebra through Practice: A Collection of Problems in Algebra with Solutions. Groups by Byth T.S., Robertson E.F.

by Richard

4.4