Note that |G| = 2 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 2. But then g is a generator of G.

Note that |G| = 3 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 3. But then g is a generator of G.

Note that |G| = 5 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 5. But then g is a generator of G.

Note that |G| = 7 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 7. But then g is a generator of G.

Note that |G| = 11 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 11. But then g is a generator of G.

Note that |G| = 13 is a prime. Therefore, G is cyclic. For, by the previous lemma, there is g G of order 13. But then g is a generator of G.

Suppose that the order of G equals 4. If G is not cyclic, then by Lagrange's theorem each element distinct from e has order 2. By the previous lemma, G is commutative. Choose two elements a, b of order 2. Now

G = {e, a, b, ab}.

It is not hard to verify that the map f: Z/2Z × Z/2Z -> G with (i,j) -> aⁱb^j is an isomorphism.

Let G be of order 6. By a previous lemma, there must be an element, say a of order 2 and an element, say b, of order 3.

Consider now the permutation representation G -> Sym(G/<a>) of a previous theorem. If this map is an isomorphism, then G is isomorphic with S₃. Otherwise, the kernel is equal to <a>. But then <a> is a normal subgroup of G and we must have bab^-1 = a, that is ab = ba. The element ab must be a generator of the group: ab e, (ab)² = a²b² = b² e, (ab)³ = a³ b³ = a e. Hence the group is isomorphic with Z/6Z.

Suppose that G has order 8 and is not cyclic. Each element of G has order 1, 2, or 4.

First, suppose all nontrivial elements have order 2. Then G is commutative according to the previous lemma. Choose three distinct elements a, b, c of order 2 in G so that c ab. Now K = {e,a,b,ab} is a subgroup of order 4 and H = {e,c} a subgroup of order 2. It is easily verified that the map Z×Z×Z -> G,   (i,j,k) -> aⁱb^jc^k is an isomorphism.

Suppose now that G has an element b of order 4. Choose, if possible, an element a of order 2 in G\<b>. Note that G = <a,b>. As <b> has index 2 in G, it is a normal subgroup of G. In particular, aba^-1 <b>. The element aba^-1 has order 4, so we can distinguish two cases: aba^-1 = b or aba^-1 = b^-1.
If aba^-1 = b, then G is commutative. But then Z/2Z×Z/4Z -> G,   (i,j) -> aⁱb^j is an isomorphism.

From now on, suppose aba^-1 = b^-1. Consider the permutation representation L : G -> Sym(G/<a>) given in a previous theorem. The 4 cosets of <a> in G are <a>, b<a>, b²<a>, and b³<a>. Numbering these cosets 1, 2, 3, 4, respectively, we find L_b = (1,2,3,4) and L_a = (2,4). Hence, G is isomorphic to H = <(2,4),(1,2,3,4)>. In turn, this group is isomorphic with D₄, the dihedral group of order 4.

Remains the case where each a G\<b> is of order 4. Choose such an a. Then a² = b², the only element of order 2 in G. It follows that ab ba. For, otherwise (ab)² = a²b² = e and we have found a second element of order 2. We conclude that aba^-1 = b^-1. The map f : G -> Q with f(a) = i and f(b) = j now is an isomorphism between G and the quaternion group Q.

Suppose that G has order 9 and is not cyclic. Then each element distinct from e has order 3. Let a be such an element and consider the permutation representation

L : G -> Sym(G/<a>).

By a previous theorem, its kernel is contained in <a>. On the other hand it cannot be trivial (i.e., equal to {e}), for otherwise, the image of G under L would be a subgroup of Sym(G/<a>) of order 9, and so, by Lagrange's theorem, 9 would divide the order of Sym(G/<a>), which is 6.

Hence, the kernel of L is <a>. Consequently, by a previous theorem, <a> is a normal subgroup of G. In particular, the conjugacy class C of a is contained in {e, a, a²}. Clearly, e cannot be conjugate to a. Therefore, C has at most 2 elements. But, by the orbit theorem, the number of elements of C is a divisor of 9, so C = {a}. For each element b G, we have bab^-1 = a, that is, ab = ba. In other words, a lies in the center of G.

Choose now b G\<a>. Since a was chosen arbitrarily amongst the nontrivial elements of G, also b has order 3 and lies in Z(G). It follows that each element of G is of the form aⁱb^j. Letting this element correspond to (i,j) Z/3Z×Z/3Z, we find that G is isomorphic with Z/3Z×Z/3Z.

Let G be a group of order 10. By the previous lemma, there is an element a of order 5 and an element b of order 2 in G. Since <a> has index 2 in G, it is a normal subgroup. In particular, bab^-1 belongs to <a>. This means that bab^-1 = a^k for some k {1,2,3,4}. Rewriting every occurrence of ba by a^kb, we see that

G = {aⁱb^j | i = 0, 1,   j = 0, 1, 2, 3, 4}.

We show that k = 1, 4. For, a = b²ab^-2 = b(bab^-1)b^-1 = b(a^k)b^-1 = (bab^-1)^k = (a^k)^k = a^k2, so k² = 1 mod 5, whence k = 1, 4.

If k = 1, then a and b commute, and Z/5Z×Z/2Z -> G,   (i,j) -> aⁱb^j, is readily seen to be an isomorphism.

If k = 4, the map G -> D₁₀,   determined by a -> (1,2,3,4,5),   b -> (1,5)(2,4) is readily seen to be an isomorphism.