The group SL2(F3).

We shall study the group SL2(F3) of 2 × 2-matrices over a field F3 of three elements, and of determinant 1. There are alltogether 34 = 81 2 × 2-matrices over F3, but only 24 have determinant 1. In this exercise we shall determine the structure of this group. The field F3 has some nice properties. First of all, we write F3 = {0, 1, −1} and compute modulo 3, i.e. −1 ≡ 2. It is easy to see that for a 6= 0, we have a2 = 1, and consequently a3 = a. This is in fact Fermats little theorem in this setting. a) Show that the 24 elements of SL2(F3) are the matrices  a b d(bc + 1) b U = , b 6= 0 and V = , d 6= 0 a,b −b 0 b,c,d c d There are 6 of the first type and 18 of the second type, and the two families are disjoint. −1 0  b) Show that the only element in SL ( ) of order 2 is −e = . 2 F3 0 −1 c) Show that in the U-family we have order |Ua,b| = 4 if and only if a = 0, and in the V -family we have |Vb,c,d| = 4 if and only if bc = 1. Thus we have 6 elements of order 4 in the group.

The 6 elements of order 4 are U0,1, U0,−1, V1,1,1, V−1,−1,1, V1,1,−1, V−1,−1,−1.

 0 b −d b U = ,V = 0,b −b 0 b,b,d b d

d) Let H = {e, −e, U0,1,U0,−1,V1,1,1,V−1,−1,1,V1,1,−1,V−1,−1,−1} be the set consisting of the 6 elements of order 4, together with ±e. Show that H is a subgroup of SL2(F3). Use Sylow theory to prove that H is normal. e) The quaternion group Q has 8 elements, Q = {±1, ±i, ±j, ±k}, where i2 = j2 = k2 = −1, and ij = k = −ji, jk = i = −kj and ki = j = −ik. We can write Q = hi, j | i4 = e, ji = i3j, j2 = i2i Show that H ' Q. 24 The factor group SL2(F3)/H has 8 = 3 elements. There is only one group of three elements, the cyclic group Z3. Using multiplicative notation, we write 2 Z3 = {H, tH, t H}, where t can be taken to be any element in SL2(F3), outside of H. f) For your choice of t, list the elements of the two cosets tH and t2H. g) In exercise e) you identified H with the quaternion group, i.e. you identified the quaternion elements i, j and k with some matrices in SL2(F3). We know that H is a normal subgroup of SL2(F3), thus we have tH = Ht, and we can find elements qi, qj, qk ∈ Q ' H such that ti = qit, tj = qjt and tk = qkt. Determine these elements in Q. h) Conclude that SL2(F3) is isomorphic to the group generated by i, j and t and write up the defining relations. (We can drop k since k = ij)

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