Question

Prove that there is no simple nonabelian group of order less than 60. [Hint : Exercise 18 may be helpful.]

Short answer

It is proved that, there is no simple nonabelian group of the order less than 60.

Step-by-step solution

Referring to the results of other Exercises from Section 9.3

• If order $n=p^k$for some prime p, then group G is not simple.
• If order$n=pq$for distinct prime p and q, then group G is not simple
• If order$n=p^{k}m$for some prime p, and $m<p$, then group G is not simple
• If order $n=p^{2}q$ for distinct prime p, and q, then group G is not simple
• If order $n=pqr$ for pqr distinct prime, then group G is not simple

Proving that there is no simple nonabelian group of the order less than 60

Let G be a group of order n, $1\le n<60$.

To prove that there is no simple group of order 60, we will use the result from other exercises in section 9.3.

• If order $n=p^k$ for some prime p, then group G is not simple.

According to this, groups with orders 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 27, 32,and 49 are not simple. So, they cannot be a simple nonabelian group.

• If order $n=pq$for distinct prime p and q, then group G is not simple.

According to this, groups with orders 6, 10, 14, 15, 21, 39, 55, and 57 are not simple. So, they cannot be simple nonabelian groups.

• If order $n=p^{k}m$ for some prime p, and $m<p$ then group G is not simple. According to this, groups with orders 18 and 50 are not simple.

So, they cannot be simple nonabelian groups.

• If order $n=p^{2}q$ for distinct prime p, and q, then group G is not simple. According to this, groups with orders 12, 20, 28, 44, and 52 are not simple. So they cannot be simple nonabelian groups.

• If order $n=pqr$ for pqr distinct prime, then group G is not simple.

According to this, groups with orders 30 and 42 are not simple. So, they cannot be a simple nonabelian group.

Therefore, the only groups which we need to check are groups of orders 24, 36, 40, 48, 54, and 56.

• $n=24=2^3·3$

Since G has a Sylow 2-subgroup of index 3, it may be a simple group.

But 24|3.

Therefore, it cannot be a simple group.

• $n=36=2^2·3^3$

Since G has a Sylow 3-subgroup of index 4, it may be a simple group.

But 36|4, therefore, it cannot be a simple group.

• $n=40=2^3·5$

Since G has a unique Sylow 5-subgroup, so it is a normal group that is not simple.

• $n=48=2^4·3$

Since G has a Sylow 2-subgroup of index 3, it may be a simple group.

But 48|3, therefore, it cannot be a simple group.

• $n=54=2·3^3$

Since G has a unique Sylow 2-subgroup, it is a normal group that is not simple.

• $n=58=2^3·7$

Since G has a unique Sylow 7-subgroup, it is a normal group that is not simple.

Since there is no group of the order 60 that is simple, it is proved that there is no simple nonabelian group of the order less than 60.