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Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ respectively.

Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and to a subgroup of ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ (which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$) indeed contains no element of order $12$.

Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ for $p = 5,7$ and $11$ respectively.

Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and to a subgroup of ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ (which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$) indeed contains no element of order $12$. That same result of Blichfeldt also can be used to prove (in the manner alluded to above) that a finite primitive subgroup of ${\rm SL}(2,\mathbb{C})$ has order $24,48$ or $120.$

Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ respectively.

Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$ indeed contains no element of order $12$.

Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ respectively.

Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and to a subgroup of ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ (which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$) indeed contains no element of order $12$.

Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ respectively.

Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).

If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$

Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$

Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.

Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.

Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$

(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ respectively.

Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$ indeed contains no element of order $12$.