Finite Groups, Groups of Order p² and p³

Groups of Order p² and p³

Let G have order p², and let H be its center. If H = G then G is abelian. If not then H = Z_p. Let x be an element not in H. Now x has order p in G/H, hence x^p lies in H. The powers of x represent all the cosets of H in G, and they commute with each other. They also commute with H, since H is the center. Therefore G is abelian. It is either Z_pZ_p or Z_p².

Next let G have order p³ and let H be the center. If H = G then G is abelian, and is either Z_pZ_pZ_p, Z_p²Z_p, or Z_p³.

If H has order p² then select x not in H and argue as above, whence G is abelian once again.

Let H have order p and let K = G/H. Since K has order p² it is abelian. If K is cyclic, let x generate K, whence x and H generate G, x and H commute, and G is abelian. So let K = Z_pZ_p, and let x and y generate K. If x and y commute than G is abelian, so assume they don't commute.

At this point I must apologize for the inconsistent notation that is coming up. I'm going to use additive notation for H, which is cyclic, and multiplicative notation for the rest of G, which is not abelian. Think of H as Z_p, i.e. the integers mod p. Thus 0 is the additive identity in H, and the identity element for G, while 1 is the generator of H, 1 2 3 4 etc, and is not the identity for G. I hope this isn't too confusing.

Suppose x^p = j for some nonzero j in H. If j is not 1, let v be the inverse of j mod p and replace x with x^v. Raise the new x to the j power and combine with something from H to get the old x back again, hence G is still spanned. Furthermore, x^p = 1.

Do the same for y. Hereinafter we may assume both generators, when raised to the p power, produce 0 or 1 in H.

If both generators produce 0, and yx = xyw, let w generate H. In other words, yx = xy1. This is the first nonabelian group of order p³, up to isomorphism.

This group is isomorphic to the 3×3 upper triangular matrices over Z_p with ones down the main diagonal. When 1 is in the upper right we have w, the generator for the center of G. Place 1 in the top middle and 1 in the middle right to get y and x respectively. Verify that w commutes with x and y, and yx = xyw. Also, each of x y and w, raised to the p, gives the identity matrix.

We can show that every element in this group has order p. Raise a generic element xⁱy^jw^k to the p power and push the instances of y forward past the instances of x. The number of swaps is i×j×(1+2+3+…+p-1), which is divisible by p when p is odd. (We'll deal with p = 2 later.) Hence the commuting factor drops out. The original instances of x y and w are raised to the p, and drop out as well. The result is 0.

What about p = 2? The construction of the group is still valid, including its representation as 3×3 upper triangular matrices, but this time it contains an element of order 4. In fact this group is equal to the dihedral group D₄, i.e. the reflections and rotations of the square. The 180° rotation generates the center of the group, 1 in the above notation. The vertical reflection is x and the reflection through the main diagonal is y. Note that yx = xy1, and x² and y² are 0.

Next case; let both x and y lead to 1 in H. Rescale y again, so that y^p = -1. Now use xy as a generator, instead of y. Expand (xy)^p, and push the instances of y forward. Every time y and x commute we introduce a factor w from H. Again, this happens 1+2+3+…+p-1 times, which drops to 0 when p is odd. This leaves x^py^p = 1-1 = 0. Therefore the new generator produces 0 in H.

We may now assume x^p = 1 and y^p = 0. Thus |x| = p². This is definitely a different group than we've seen before.

With p still odd, renormalize y as follows. If yx = xyj, and v is the inverse of j mod p, replace y with y^v. Now y^p is still 0, x^p is still 1, and yx = xy1. This is the second nonabelian group of order p³.

Return to p = 2. Suppose x² = 1 and y² = 0. Replace x with xy, and both generators lead to 0 in H. We've already handled this case.

Finally both x² and y² are 1, and yx = xy1. This is isomorphic to the quaternion group Q₈. To see the isomorphism, consider the quaternions mod 2. These are the elements ±1, ±i, ±j, and ±k, where i j and k squared give -1. Also, ji = -ij etc. Think of i as x, and j as y, and k as xy. The multiplicative identity 1 corresponds to 0 in H, and -1 corresponds to 1, the involution in H. These are indeed the same group, hence the nonabelian groups of order 8 are D₄ and Q₈.

Thanks to somebody's computer program, the number of groups of order 2ⁱ, for i = 1 to 7, is 1, 2, 5, 14, 51, 267, 2328. We just verified this for i = 3, 3 abelian groups and 2 nonabelian groups.