In this problem, as you apply Sylow’s Theorem, state precisely which portions you are using.
Prove that there is no simple group of order 30.
Suppose that \(G\) is a simple group of order 60. Determine the number of \(p\)-Sylow subgroups of \(G\) for each prime \(p\) dividing 60, then prove that \(G\) is isomorphic to the alternating group \(A_5\).
Note: in the second part, you needn’t show that \(A_5\) is simple. You need only show that if there is a simple group of order 60, then it must be isomorphic to \(A_5\).
Solution
1. There is no simple group of order \(30\).
Suppose \(|G|=30\) and \(G\) is simple. By Sylow’s third theorem, \[ n_5\equiv1\pmod5,\qquad n_5\mid6, \] so \(n_5\in\{1,6\}\). Simplicity rules out \(n_5=1\), because a unique Sylow subgroup is normal by Sylow’s second theorem. Hence \(n_5=6\).
Similarly, \[ n_3\equiv1\pmod3,\qquad n_3\mid10, \] so \(n_3\in\{1,10\}\), and simplicity forces \(n_3=10\).
Distinct Sylow subgroups of prime order intersect trivially. Thus the six Sylow \(5\)-subgroups contribute \[ 6(5-1)=24 \] nonidentity elements of order \(5\), while the ten Sylow \(3\)-subgroups contribute \[ 10(3-1)=20 \] nonidentity elements of order \(3\). These sets are disjoint, so together with the identity they would give at least \[ 1+24+20=45>30 \] elements, a contradiction.
2. Now suppose \(G\) is simple of order \(60\). Then\[ n_5=6,\qquad n_3=10. \]
For \(p=5\), Sylow’s third theorem gives \[ n_5\equiv1\pmod5,\qquad n_5\mid12, \] so \(n_5\in\{1,6\}\). Simplicity excludes \(1\), hence \(n_5=6\).
For \(p=3\), \[ n_3\equiv1\pmod3,\qquad n_3\mid20, \] so \(n_3\in\{1,4,10\}\). The value \(1\) is excluded by simplicity. If \(n_3=4\), conjugation on the four Sylow \(3\)-subgroups gives a nontrivial homomorphism \[ G\longrightarrow S_4. \] Its kernel is normal. Since \(G\) is simple and the action is nontrivial, the map would be injective, impossible because \(60\nmid24\). Therefore \(n_3=10\).
3. The number \(n_2\) of Sylow \(2\)-subgroups satisfies\[ n_2\in\{5,15\}. \]
Sylow’s third theorem gives \[ n_2\equiv1\pmod2,\qquad n_2\mid15, \] so \(n_2\in\{1,3,5,15\}\). The value \(1\) is excluded by simplicity. If \(n_2=3\), conjugation on the three Sylow \(2\)-subgroups gives a nontrivial homomorphism \(G\to S_3\), which by simplicity would be injective; this is impossible because \(60>6\). Hence only \(5\) and \(15\) remain.
4. If \(n_2=5\), then \(G\) embeds in \(S_5\).
Conjugation gives a transitive action of \(G\) on its five Sylow \(2\)-subgroups, hence a nontrivial homomorphism \[ \varphi:G\longrightarrow S_5. \] Its kernel is normal. Since \(G\) is simple and the action is nontrivial, \(\ker\varphi=1\), so \(\varphi\) is injective.
5. If \(n_2=15\), then \(G\) still has a subgroup of index \(5\), and hence still embeds in \(S_5\).
By 2, \(n_5=6\), so the Sylow \(5\)-subgroups contribute exactly \[ 6(5-1)=24 \] nonidentity elements. Hence only \(60-24=36\) elements, including the identity, remain outside those subgroups.
If the fifteen Sylow \(2\)-subgroups were pairwise disjoint away from the identity, they would contribute \[ 15(4-1)=45 \] nonidentity elements, impossible. Therefore two distinct Sylow \(2\)-subgroups \(P,Q\) intersect nontrivially. Since \(|P|=|Q|=4\), their intersection has order \(2\); let \[ P\cap Q=\langle x\rangle,\qquad x\ne1. \] Every group of order \(4\) is abelian, so both \(P\) and \(Q\) centralize \(x\). Thus \[ P,Q\le C_G(x). \] Because \(P\ne Q\), the centralizer strictly contains \(P\). Its order divides \(60\) and is divisible by \(4\), so \[ |C_G(x)|\in\{12,20,60\}. \] The value \(60\) would put \(x\) in \(Z(G)\), impossible because a nonabelian simple group has trivial center. The value \(20\) would give a subgroup of index \(3\); the coset action on three points would produce a nontrivial homomorphism \(G\to S_3\), hence an injection by simplicity, again impossible. Therefore \[ |C_G(x)|=12, \] so \(C_G(x)\) has index \(5\).
The action of \(G\) on the five left cosets of \(C_G(x)\) gives a nontrivial homomorphism \(G\to S_5\). By simplicity its kernel is trivial, hence \(G\) embeds in \(S_5\).
6. In either case,\[ G\cong A_5. \]
By 4 and 5, there is an injective homomorphism \[ G\hookrightarrow S_5. \] Its image has order \(60\), hence index \(2\) in \(S_5\). Every index-\(2\) subgroup is the kernel of a surjection \(S_5\to C_2\). All transpositions are conjugate and generate \(S_5\), so any nontrivial homomorphism \(S_5\to C_2\) sends every transposition to the nontrivial element; hence it is the sign homomorphism. Therefore the unique index-\(2\) subgroup of \(S_5\) is \[ \ker(\operatorname{sgn})=A_5. \] Thus the image of \(G\) is \(A_5\), and \(G\cong A_5\).
7. Consequently, for a simple group \(G\) of order \(60\),\[ \boxed{n_2=5,\qquad n_3=10,\qquad n_5=6}. \]
2. By 6, \(G\cong A_5\). The Sylow \(2\)-subgroups of \(A_5\) are the five Klein four groups\[ V_i=\{1\}\cup\{(ab)(cd):\{a,b,c,d\}=\{1,2,3,4,5\}\setminus\{i\}\}, \]one for each fixed letter \(i\). Hence \(n_2=5\).