Exercise 4.5.29 (Simple groups $G$ of order $|G| < 100$)

Proof. Let $G$ be a non-abelian simple group of order $n<100$. We know that

• if $n=1$, or if $n$ is prime or the square of a prime, then $G$ is abelian.
• If $n=p^k$ for some integer $k>1$, then $G$ is a $p$-group which has a non trivial center (and $Z(G)$ is a proper subgroup, otherwise $G$ is abelian). Thus $G$ is not simple.
• If $n=\mathit{𝑝𝑞}$, with $p<q$, the unique Sylow $q$-subgroup of $G$ is normal, so $G$ is not simple (see Example, “groups of order $\mathit{𝑝𝑞}$” in this section).
• If $n=p^{2}q$, where $p,q$ are prime numbers, then $G$ has a normal Sylow subgroup for either $p$ or $q$ (see Example, “groups of order $p^{2}q$” in this section), so $G$ is not simple.
• If $n=30$, then $G$ has a normal subgroup isomorphic to $Z_{15}$ (see Example, “groups of order $30$” in this section), so $G$ is not simple.

It remains the following values for $n$:

Part 3 of Sylow’s Theorem gives first (see program below in the note):

• If $n=40$, then $n_5=1$.
• If $n=54$, then $n_3=1$.
• If $n=66$, then $n_{11}=1$.
• If $n=70$, then $n_7=1$.
• If $n=78$, then $n_{13}=1$.
• If $n=84$, then $n_7=1$.
• If $n=88$, then $n_{11}=1$.

In every case, $G$ is not simple.

There are $8$ values left to study (excluding $60$):

• If $n=24=2^3\cdot 3$, then $n_2\mid 3$, thus

  $$n_2\in \{1,3\}.$$

  Assume for the sake of contradiction that $n_2\ne 1$. Then $n_2=3$.

  Since $G$ acts by conjugation on the set of $3$ Sylow $2$-subgroups $\{H_1,H_2,H_3\}$, this action affords a homomorphism

  $$\phi :G\to S_3.$$

  This action is not trivial, otherwise $H_1$ would be normal in $G$, therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\ne G$. Since $G$ is simple, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1\}$ or $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=G$, thus

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1\},$$

  so that $\phi$ is injective. This shows that $G$ is isomorphic to a subgroup of $S_3$. Since $|G|=24$ and $|S_3|=6$, this is impossible. This contradiction shows that $n_2=1$, therefore $G$ has a non trivial proper normal subgroup of order $2$, so $G$ is not simple.

• If $n=48=2^4\cdot 3$ or $n=96=2^5\cdot 3$, we obtain similarly $n_2=1$ or $n_2=3$. Since $|G|=48>|S_3|=6$, the same proof shows that $G$ is not simple.

• If $n=36=2^2\cdot 3^2$, then $n_3\in \{1,4\}$.

  Assume for the sake of contradiction that $n_3\ne 1$. Then $n_3=4$.

  Since $G$ acts by conjugation on the set of $4$ Sylow $3$-subgroups $\{H_1,H_2,H_3,H_4\}$, this action affords a homomorphism

  $$\phi :G\to S_4.$$

  This action is not trivial, otherwise $H_1$ would be normal in $G$, therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\ne G$. Since $G$ is simple, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1\}$ or $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=G$, thus

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1\},$$

  so that $\phi$ is injective. This shows that $G$ is isomorphic to a subgroup of $S_4$. Since $|G|=36>|S_4|=24$. This is impossible. This contradiction shows that $n_3=1$, therefore $G$ has a non trivial proper normal subgroup of order $3$, so $G$ is not simple.

• If $n=72$, $n_3\in \{1,4\}$. If $n_3\ne 1$, then $n_3=4$, and similarly $G$ acts faithfully by conjugation on the set of $3$-Sylow, so $G$ is isomorphic to some subgroup of $S_4$. Since $|G|=72\nmid |S_4|=24$, this is impossible.
• If $n=80=2^4\cdot 5$, we obtain similarly $n_5=1$ or $n_5=5$. If $n_5\ne 1$, $G$ acts non trivially by conjugation on the set of $5$ Sylow $5$-subgroups. This affords an injective homomorphism $\phi :G\to S_5$. But $|G|=80\nmid 120=|S_5|$. Therefore $G$ is not simple.

• If $n=56=2^3\cdot 7$, we obtain

  $$n_2\in \{1,7\},n_7\in \{1,8\}.$$

  Suppose for the sake of contradiction that $n_2\ne 1$ and $n_7\ne 1$, so that $n_2=7$ and $n_7=8$. Any pair of distinct Sylow subgroups have trivial intersection, so there are $n_7\cdot 6=48$ elements of order $7$. It remains a set $S$ of $8$ elements, whose order is not $7$. Every Sylow $2$-subgroup $H$ has $8$ elements, whose order is not $7$, so $H\subseteq S$, where $|H|=|S|=8$, so $H=S$. There is only one Sylow $2$-subgroup of $G$, which is normal in $G$. Therefore $G$ is not simple.

• If $n=90=2\cdot 45$, where $45$ is odd, by Exercise 4.2.13, there is a subgroup of $G$ of index 2, which is normal in $G$, so $G$ is not simple.

It remains only the case $n=60$. By Proposition 23, $G\simeq A_5$. □