Group Theory

10 Groups of small order

Let’s try to classify groups of small order. Here \(p,q\) will be primes. The first thing to note is that a group of order \(p\) for must be \(C_p\) as any nontrivial element is order \(p\). Next, we can try to classify groups of order \(p^2\). The center cannot have prime index, but also cannot be trivial as it is a \(p\)-group, so the group is abelian and is \(C_{p^2}\) or \(C_p\times C_p\).

Next, let’s try groups of order \(pq, p > q\). If it is abelian, it is \(C_{pq}\). If not, \(n_p=1\) by the Sylow theorems since \(p \geq q\). Thus there is a normal subgroup of order \(p\). Any \(q\)-Sylow will split the corresponding exact sequence, so we get that the group is a semidirect product \(C_p \rtimes C_q\). The automorphisms of \(C_p\) are given by \(C_{p-1}\), so unless \(p \equiv 1 \pmod q\), there are no nontrivial maps and \(C_pq\) is the only group. If \(p \equiv 1 \pmod {q}\), then there is only one homomorphism up to an automorphism of \(C_q\), so we get one non-abelian \(C_p\rtimes C_q\). Note that when \(q = 2\), this is \(D_n\).

Now, lets try groups of order \(p^3\). Here, the abelian ones are \(C_{p^3},C_{p^2}\times C_p, C_p\times C_p\times C_p\). A non-abelian one must have center that is order \(p\), and is also the commutator subgroup by our previous classification. Say that the quotient is \(C_p^2\). Then there must be an element of order \(p^2\) reducing to a generator, so this exact sequence splits. There are no homomorphisms \(C_{p^2}\to \Aut (C_p)\), so this case cannot happen (it is actually abelian). Thus the quotient is \(C_p\times C_p\).

Suppose there is an element of order \(p^2\). The subgroup it generates must be its centralizer. There must be another element not in the center of order \(p\). Otherwise, the group can be partitioned into the center and copies of \(C_{p^2}-C_p\), giving that \(p + k(p^2-p) = p^3\) for some \(k\). Looking mod \(p^2\), \(k \equiv 1 \pmod p\), but also \(1<k<p\), a contradiction. Now the element of order \(p\) is nontrivial in the abelianization, hence it splits an exact sequence of the form \(1 \to C_{p^2} \to G \to C_p \to 1\). There is an essentially unique map \(C_p \to \Aut (C_{p^2})\), so in this case our group is \(C_{p^2} \rtimes C_p\).

Now suppose that there is no element of order \(p^2\). Then all elements are order \(p\). Once again, any such element not in the center splits an exact sequence \(1 \to C_p \times C_p \to G \to C_p \to 1\), so our group is determined by a homomorphism \(C_p \to Aut(C_p\times C_p) = \GL _2(\FF _p)\). The image of a generator is a nontrivial element whose minimal polynomial divides \(X^p-1 = (X-1)^p\), and since it is \(2\times 2\) and nontrivial, its minimal polynomial is exactly \((X-1)^2 = 0\). By rational canonical form, up to conjugation it is the matrix \(\begin {pmatrix}0 & -1\\ 1 & 2 \end {pmatrix}\). In particular it is unique, so our group is \((C_p \times C_p) \rtimes C_p\). When \(p = 2\), this group is called \(Q_8\), and has the presentation \(\langle \bar {e},i,j,k|\bar {e}i^2,\bar {e}j^2,\bar {e}k^2,\bar {e}ijk\rangle \).

Finally, let’s try groups of order \(p^2q\). The abelian ones are \(C_{p^2q}\) and \(C_{pq}\times C_p\). If \(C_q\) is the center, \(G/Z(G)\) will be order \(p^2\), hence the group will be nilpotent, hence abelian, a contradiction. Now suppose that \(C_p\) is the center. The quotient by the center is a non-abelian group of order \(pq\) since \(G\) is not nilpotent, so there is a further quotient onto a group of order \(p\) or \(q\). If it is order \(q\), \(p \equiv 1 \pmod q\), and a \(q\)-Sylow will split the sequence yielding a semidirect product with a \(p\)-Sylow. If the \(p\)-Sylow is \(C_p\times C_p\), since \(C_q\) must act trivially on the center, this group will be \(C_p \times (C_p \rtimes C_q)\). If the \(p\)-Sylow was \(C_{p^2}\), \(\Aut (C_{p^2}) = C_{p(p-1)}\), and the \(C_{p-1}\) inside this group fixes the subgroup \(C_p\), so our group is \(C_{p^2} \rtimes C_q\).

If the quotient was order \(p\), then \(q \equiv 1 \pmod p\). That sequence will split iff there is an element of order \(p\) not in the center. In this case, we will get an action of \(C_p\) on \(C_p\times C_q\). Since \(C_p\) acts trivially on \(C_p\), this is \(C_p\times (C_q \rtimes C_p)\).

If there are no elements of order \(p\) not in the center, a \(p\)-Sylow must be \(C_{p^2}\), and \(n_p \equiv 1 \pmod p, n_p |q\). Were it \(1\), it would be normal and give \(C_{p^2} \rtimes C_q\) again, so it is \(q\). \(n_q \equiv 1 \pmod q\) and divides \(p^2\).