What is the minimal test to decide whether a permutation π ∈ S24 lies in our preferred copy of M24? The space C is 12 dimensional and so if we choose a basis of 12 codewords of C, apply π to each codeword in the basis and verify that the image is also in C then π ∈ M24. The 12-dimensional subspace C is self-orthogonal with respect to the usual inner product, and so C = C⊥. Thus a vector is in C if, and only if, it is orthogonal to every codeword in a basis of C. Now one codeword in our basis may be chosen to be the all 1s vector that is clearly fixed by any permutation; the other 11 can be chosen to be octads. In this chapter we show that we can do much better than this. In fact we show that we can choose 8 octads that are contained in one, and only one, copy of C, but that any set of 7 octads is contained in no copy of C or in more than one. To this set of 8 octads we add a further 3 to form a basis together with the all 1s codeword. We now have a minimal test for membership of M24: apply π to each of the 8 octads; if the image in each case intersects each of the 11 octads in the basis evenly, then π is in M24, otherwise it is not. When working with M24 we often require an element possessing certain properties. In this chapter we show how to construct elements of shape 18.28, 212 and 16.36. We also reproduce a diagram due to Todd and Conway showing the orbits of M24 on the subsets of Ω.