## 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 $G \cong A_5$.

• 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.

• nbloomf  On August 21, 2012 at 10:31 am

True.