By Ben Ayed M., El Mehdi K., Pacella F.
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4) holds when i = k − 1: xk−1 xk = yk−1 x + yj x ∈ Uk−1 Uk−1 since yk−1 x ∈ Uk−1 Uk−1 and yj x ∈ Uk−1 Uk−1 . 4) holds in this case. • Suppose that the above two possibilities do not occur. So we have that Uk−1 U = Uk−1 Uk−1 and there exist x y ∈ U \ Uk−1 such that x y ∈ Uk−1 Uk−1 . Suppose further that yk−2 yk−1 + yk−2 x ∈ Uk−2 Uk−2 This implies that yk−2 x ∈ Uk−2 Uk−2 . Set xi = yi for all i ∈ 1 2 k − 2 , set xk−1 = x and xk = y. So Vi = Ui for 1 i k − 2. 4) clearly holds when i < k − 2, and also in the case i = k − 2 since yk−2 x ∈ Uk−2 Uk−2 .
6) follows. 4 A small set of relations In this section we show that we may always find a presentation for a pgroup G of a restricted type. Our eventual enumeration in the next section will depend on the fact that there are comparatively few possibilities for presentations of this sort. We begin by choosing our generating set for G and finding a collection of relations this set satisfies. We then prove that we have a presentation for G. Let G be a group of order pm . Let G = G1 G2 · · · Gc Gc+1 = 1 be the lower central series of G.
3 Let N1 N2 Gr . Then Gr /N1 exists ∈ Aut Gr such that N1 = N2 . Gr /N2 if and only if there Proof: Note that quotients by N1 and N2 make sense, since Gr is contained in the centre of Gr . An element ∈ Aut Gr mapping N1 to N2 induces an isomorphism from Gr /N1 to Gr /N2 . We need to show the converse. Let Gr /N1 → Gr /N2 be an isomorphism. Let y1 y2 yr ∈ Gr be such that xi N1 = yi N2 . 1 implies that there exists a homomorphism Gr → Gr such that xi = yi . Now, since is an isomorphism, y1 y2 yr and N2 together generate Gr .