Alt(5) is the only finite simple group of order less than 100

If G is a nonabelian simple group of order < 100, prove that G \cong A_5. [Hint: Eliminate all orders but 60.]


In the following table, we list the integers 2 \leq n \leq 99, with a note justifying the assertion that no group of order n is simple.

n Factorization Note
2 (prime) \blacksquare
3 (prime) \blacksquare
4 2^2 \heartsuit
5 (prime) \blacksquare
6 2 \cdot 3 \triangle
7 (prime) \blacksquare
8 2^3 \heartsuit
9 3^2 \heartsuit
10 2 \cdot 5 \triangle
11 (prime) \blacksquare
12 2^2 \cdot 3 \clubsuit
13 (prime) \blacksquare
14 2 \cdot 7 \triangle
15 3 \cdot 5 \triangle
16 2^4 \heartsuit
17 (prime) \blacksquare
18 2 \cdot 3^2 \clubsuit
19 (prime) \blacksquare
20 2^2 \cdot 5 \clubsuit
21 3 \cdot 7 \triangle
22 2 \cdot 11 \triangle
23 (prime) \blacksquare
24 2^3 \cdot 3 Lemma 3
25 5^2 \heartsuit
26 2 \cdot 13 \triangle
27 3^3 \heartsuit
28 2^2 \cdot 7 \clubsuit
29 (prime) \blacksquare
{30} 2 \cdot 3 \cdot 5 \Diamond
31 (prime) \blacksquare
{32} 2^5 \heartsuit
33 3 \cdot 11 \triangle
34 2 \cdot 17 \triangle
35 5 \cdot 7 \triangle
{36} 2^2 \cdot 3^2 Lemma 8
{{37}} (prime) \blacksquare
38 2 \cdot 19 \triangle
39 3 \cdot 13 \triangle
{{40}} 2^3 \cdot 5 Lemma 4
{41} (prime) \blacksquare
42 2 \cdot 3 \cdot 7 \Diamond
43 (prime) \blacksquare
{44} 2^2 \cdot 11 \clubsuit
45 3^2 \cdot 5 \clubsuit
{46} 2 \cdot 23 \triangle
{47} (prime) \blacksquare
48 2^4 \cdot 3 Lemma 3
49 7^2 \heartsuit
50 2 \cdot 5^2 \clubsuit
51 3 \cdot 17 \triangle
52 2^2 \cdot 13 \clubsuit
53 (prime) \blacksquare
54 2 \cdot 3^3 Lemma 1
55 5 \cdot 11 \triangle
{56} 2^3 \cdot 7 Lemma 6
57 3 \cdot 19 \triangle
58 2 \cdot 29 \triangle
59 (prime) \blacksquare
60 2^2 \cdot 3 \cdot 5 NA
61 (prime) \blacksquare
62 2 \cdot 31 \triangle
63 3^2 \cdot 7 \clubsuit
64 2^6 \heartsuit
65 5 \cdot 13 \triangle
{66} 2 \cdot 3 \cdot 11 \Diamond
67 (prime) \blacksquare
68 2^2 \cdot 17 \clubsuit
69 3 \cdot 23 \triangle
70 2 \cdot 5 \cdot 7 \Diamond
71 (prime) \blacksquare
72 2^3 \cdot 3^2 Lemma 9
73 (prime) \blacksquare
74 2 \cdot 37 \triangle
75 3 \cdot 5^2 \clubsuit
{{76}} 2^2 \cdot 19 \clubsuit
77 7 \cdot 11 \triangle
78 2 \cdot 3 \cdot 13 \Diamond
79 (prime) \blacksquare
80 2^4 \cdot 5 Lemma 5
81 3^4 \heartsuit
82 2 \cdot 41 \triangle
83 (prime) \blacksquare
84 2^2 \cdot 3 \cdot 7 Lemma 7
85 5 \cdot 17 \triangle
{86} 2 \cdot 43 \triangle
87 3 \cdot 29 \triangle
88 2^3 \cdot 11 Lemma 2
89 (prime) \blacksquare
90 2 \cdot 3^2 \cdot 5 Lemma 10
91 7 \cdot 13 \triangle
92 2^2 \cdot 23 \clubsuit
93 3 \cdot 31 \triangle
94 2 \cdot 47 \triangle
95 5 \cdot 19 \triangle
96 2^5 \cdot 3 Lemma 3
97 (prime) \blacksquare
98 2 \cdot 7^2 \clubsuit
99 3^2 \cdot 11 \clubsuit
  • \blacksquare: All such groups are cyclic, hence abelian.
  • \triangle: No group of order pq, p, q distinct primes, is simple.
  • \clubsuit: No group of order p^2q, p and q distinct primes, is simple.
  • \Diamond: No group of order pqr, p, q, and r distinct primes, is simple.
  • \heartsuit: Every p-group has a nontrivial center, which is normal.

Lemma 1: If G is a finite group of order pq^a, where p < q, then G is not simple. Proof: By Sylow's Theorem, the number of Sylow q-subgroups divides p and is congruent to 1 mod q. Thus n_q = 1, and the (unique) Sylow q-subgroup is normal. Hence G is not simple. \square

Lemma 2: If G is a finite group of order pq^a, where q^a < p, then G is not simple. Proof: By Sylow's Theorem, n_p divides q^a and is congruent to 1 mod p. Thus n_p = 1, and the (unique) Sylow p-subgroup is normal. Hence G is not simple. \square

Lemma 3: If G is a finite group of order 2^k \cdot 3, with k \geq 2, then G is not simple. Proof: Suppose otherwise. The number n_2 of Sylow 2-subgroups divides 3 and is not 1 since G is simple. Thus n_3 = 3. Note that G acts on its Sylow 2-subgroups by conjugation, yielding a permutation representation G \rightarrow S_3. Let K be the kernel of this action; note that for every Sylow 2-subgroup P_2, we have K \leq N_G(P_2). Since K is normal in G and G is simple, there are two possibilities. If K = G, we have N_G(P_2) = G for each Sylow 2-subgroup, thus a Sylow 2-subgroup is normal in G, a contradiction. If K = 1, then the permutation representation is injective. But then |G| \leq 6, a contradiction. Thus G is not simple. \square

Lemma 4: No group of order 40 = 2^3 \cdot 5 is simple. Proof: By Sylow's Theorem, n_5 divides 8 and is congruent to 1 mod 5. Thus n_5 = 1, and the (unique) Sylow 5-subgroup of G is normal. Hence G is not simple. \square

Lemma 5: No group of order 80 = 2^4 \cdot 5 is simple. Proof: Suppose otherwise that G is a simple group of order 80. By Sylow's Theorem, n_2 \in \{ 1,5 \} and n_5 \in \{ 1, 16 \}. Since G is simple, n_2 = 5 and n_5 = 16. Now the Sylow 5-subgroups have prime order, and thus intersect trivially. Thus G has 16 \cdot 4 = 64 elements of order 5. Let P_2 be a Sylow 2-subgroup. There exists a second Sylow 2-subgroup Q_2, and there exists an element x \in Q_2 not in P_2; then G has at least 16 + 1 = 17$ elements in Sylow 2-subgroups. So G has at least 81 elements, a contradiction. So no simple group of order 80 exists. \square

Lemma 6: No group of order 56 = 2^3 \cdot 7 is simple. Proof: Suppose to the contrary that G is a simple group of order 56. Sylow's Theorem and the simplicity of G force n_7 = 8 and n_2 = 7. Since all Sylow 7-subgroups of G intersect trivially, G has at least 6 \cdot 8 = 48 elements of order 7. Let P_2 be a Sylow 2-subgroup, and let Q_2 be a Sylow 2-subgroup distinct from P_2. Q_2 contains an element x not in P_2; so G contains at least 8 + 1 = 9 elements in Sylow 2-subgroups. Hence |G| \geq 57, a contradiction. Thus no such group exists. \square

Lemma 7: No group of order 84 = 2^2 \cdot 3 \cdot 7 is simple. Proof: By Sylow’s Theorem, n_7 divides 12 and is congruent to 1 mod 7; thus n_7 = 1. The (unique) Sylow 7-subgroup is then normal. \square

Lemma 8: No group of order 2^2 \cdot 3^k is simple for k \geq 2. Proof: Suppose to the contrary that G is a simple group of this order. By Sylow’s Theorem, n_3 \in \{ 1,4 \}. If n_3 = 4, then G acts by conjugation on its (four) Sylow 3-subgroups, yielding a permutation representation G \rightarrow S_4 with kernel K. Note that K \leq N_G(P_3) for each Sylow 3-subgroup P_3. If K = G, then N_G(P_3) = G, so that P_3 is normal and n_3 = 1, a contradiction. If K = 1, then |G| divides S_4. But 9 divides |G| and does not divide 24, a contradiction. Thus no simple group of this order exists. \square

Lemma 9: No group of order 2^3 \cdot 3^k is simple for k \geq 2. Proof: Suppose to the contrary that G is a simple group of this order. By Sylow’s Theorem, n_3 \in \{ 1,4 \}. If n_3 = 4, then G acts by conjugation on its (four) Sylow 3-subgroups, yielding a permutation representation G \rightarrow S_4 with kernel K. Note that K \leq N_G(P_3) for each Sylow 3-subgroup P_3. If K = G, then N_G(P_3) = G, so that P_3 is normal in G and n_3 = 1, a contradiction. If K = 1, then G \leq S_4; but |G| \geq 72 > 24 = |S_4|, a contradiction. So no simple group of this order exists. \square

Lemma 10: No group of order 90 = 2 \cdot 3^2 \cdot 5 is simple. Proof: Suppose to the contrary that G is a simple group of order 90. Sylow’s Theorem forces n_5 \in \{ 1,6 \}; we may assume n_5 = 6 since G is simple. Now G acts on its (six) Sylow 5-subgroups by conjugation, yielding a permutation representation G \rightarrow S_6 with kernel K. Note that K \leq N_G(P_5) for each Sylow 5-subgroup P_5. If K = G, then N_G(P_5) = G, so that P_5 is normal in G, a contradiction. If K = 1, we have G \leq S_6. Note that every element of order 5 in G is represented by a 5-cycle in S_6; thus A_6 \cap G \neq 1. Clearly G \neq A_6 by order considerations. Now A_6 \cap G is a nontrivial normal subgroup of G, a contradiction. Thus no simple group of order 90 exists. \square

Finally, note that if G is a finite group of order n < 100 with n \neq 60, then G is not simple. If G is a simple group of order 60, we saw in the text that G \cong A_5.

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Comments

  • nhanttruong  On June 29, 2011 at 6:45 pm

    Typo in the last paragraph: n not 60.

    Why do you have 2r 2x 2y 2W 2X . . . in the leftmost column of your table?

    • nbloomf  On June 30, 2011 at 2:11 pm

      Thanks!

      As for the letters in the left column- I’m not sure what’s going on. It’s correct in the page source. Weird…

      • nbloomf  On June 30, 2011 at 2:20 pm

        Okay, the numbers show up correctly now.

        To fix it, I put the messed up numbers in a pair of braces. Compare: (no braces) 32 and (braces) {32}.

        Strangely, 37, 40, and 76 required two sets of braces. Compare: (no braces) 40, (one pair) {40}, (two pairs) {{40}}.

        Strange.

        • nhanttruong  On June 30, 2011 at 7:12 pm

          > Strangely, 37, 40, and 76 required two sets of braces.

          First Dummit and Foote, now wordpress. When will it end?!

  • Gobi Ree  On December 21, 2011 at 12:43 am

    I don’t understand “Note that every element in G is represented by a 5-cycle in S_6” in Lemma 10.

    • nbloomf  On January 18, 2012 at 1:59 pm

      That should read ‘every element of order 5 in G‘.

      For some reason, WordPress isn’t letting me fix the typo at the moment.

      • nbloomf  On January 19, 2012 at 10:03 am

        Okay- it’s fixed now.

  • Zach  On March 20, 2012 at 11:26 pm

    Just a thought, isn’t Z/pZ simple for prime p because there do not exist any nontrivial proper subgroups? If there are no nontrivial proper subgroups, there cannot possibly be any nontrivial proper normal subgroups.

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