non-abelian simple group of order less than 100

On a Saturday morning, I was thinking about sylow theorems.

The question I asked myself is, do I know how to apply sylow theorems?

Only application I was aware about, of sylow theorem, is to assure if a group of finite order is simple or not.

As a first step, I thought to check for groups of order less than 100.

Question is simple :

For which $n<100$, a group of order $n$ will be simple?

A standard result that we (are supposed to) learn in first half a group theory course is that abelian simple groups are boring. Any abelian simple group is cyclic of prime order. So, I rephrased the question to as following question.

For which $n<100$, a non abelian group of order $n$ will be simple?

The first thing to do is list down the numbers from 1 to 100

Sure! Here are the numbers from 1 to 100:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52, 53, 54, 55, 56, 57, 58, 59, 60,
61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80,
81, 82, 83, 84, 85, 86, 87, 88, 89, 90,
91, 92, 93, 94, 95, 96, 97, 98, 99, 100.

If we are looking for non abelian groups, we can safely remove groups of prime order.

Now the question boils down to which numbers are prime numbers. We may not know many things about prime numbers but we know for sure even numbers are not prime numbers (except 2). So, we can safely focus our attention to the odd numbers

Then, multiples of 3 can not be prime numbers (except 3). This is easy to do. Start from 3 and skip two numbers and strike the next number. We get

There is no need to look for multiple of 4 as they are already covered in multiples of 2. Next is to see multiples of 5 (of course ignoring 5). This is simple. Just strike out numbers that ends with 5 and 0. We get

There is no need to look for multiple of 6 as they are already covered in multiples of 2 and 3. Next is to see multiples of 7. One way to do is tell 7 table loudly (seven vanjaar seven, seven tujaar forteen). We get

There is no need to look for multiples of 8 as they are already covered in multiples of 2. There is no need to look for multiples of 9 as they are already covered in multiples of 3. All multiples of 10 are already strike out. If we ignore 1, the remaining numbers are supposed to be primes.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Groups of the above orders are of prime order, so, cyclic, thus abelian. So, we ignore those (along with 1 of course).

We are left with

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100

Another interesting result that assure abelian property of group based on number of elements is the following :

any group of order $p^2$ for a prime $p$ is abelian.

This is very simple to compute.

Just remove $2^2,3^2,5^2,7^2$ that is, $4,9,25,49$. We are left with

6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100

Another interesting result assures that, any $p$-group will have a non-trivial centre. In other words, centre of non abelian $p$-group is always non trivial (apart from being a normal subgroup). So, we can safely ignore all powers of primes.

Just remove $2^2,2^3,2^4,2^5,2^6,3^2,3^3,3^4,5^2,7^2$ that is, $4,8,16,32,64,9,27,81,25,49$. We are left with

6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100

Note that, all the numbers above has more than one prime factor.

We do have an interesting result about groups of order which has exactly two prime factors.

A group $G$ with $|G|=pq$ for distinct primes $p,q$ can not be a simple group.

Sylow theorem says number of sylow $p$-subgroups should be $1$ or $q$. Suppose number of sylow $p$-subgroups is $1$, it is a normal subgroup so, $G$ can not be a simple group. Suppose number of sylow $p$-subgroups is $q$. As they can not have a non-trivial intersection, the number of non-identity elements in these sylow $p$-subgroups is $q(p-1)$ (with $p-1$ non identity elements from each of the $q$ subgroups). So, for sylow $q$-subgroup we are left with only $pq-(pq-q)=q$ elements. With $q$ elements, we can construct exactly one sylow $q$-subgroup of $G$. So, sylow $q$-subgroup is unique, hence normal. In any case $G$ will have either a sylow $p$-subgroup or a sylow $q$-subgroup; hence can not be simple.

With this observation, we can safely remove numbers which are of the form $pq$ for distinct primes $p,q$. They are given by

\[2\times 3, 2\times 5, 2\times 7, 2\times 11, 2\times 13, 2\times 17, 2\times 19, 2\times 23, 2\times 29, 2\times 31, 2\times 37, 2\times 41, 2\times 43, 2\times 47\]
\[3\times 5, 3\times 7, 3\times 11, 3\times 13, 3\times 17, 3\times 19, 3\times 23, 3\times 29, 3\times 31\]
\[5\times 7, 5\times 11, 5\times 13, 5\times 17, 5\times 19\]
\[7\times 11, 7\times 13\]

The numbers coming in above procedure are

6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.

Removing these numbers (some of them are already gone in previous filtering) from the previous sequence gives us,

12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 66, 68, 70, 72, 75, 76, 78, 80, 84, 88, 90, 92, 96, 98, 99, 100

Another interesting result about groups of order which has exactly two prime factors says that,

any group of order $p^2q$ will have either a sylow $p$-subgroup or a sylow $q$-subgroup, hence not simple.

The proof is almost same as that of the other result we saw above. So, it is time to look for numbers of the form $p^2q$ for distinct primes $p,q$. They are

\[2^2\times 3, 2^2\times 5, 2^2\times 7, 2^2\times 11, 2^2\times 13, 2^2\times 17, 2^2\times 19, 2^2\times 23, 2^2\times 29\]
\[3^2\times 2, 3^2 \times 5, 3^2\times 7, 3^2\times 11\]
\[5^2\times 2, 5^2\times 3\]
\[7^2\times 2\]

The numbers coming in above procedure are

12, 18, 20, 28, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99.

Removing these numbers (some of them are already gone in previous filtering) from the previous sequence gives us,

24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100

Do we know anything about groups of order of the form $p^2q^2$ for distinct primes $p,q$?

Some of the numbers in above collection has exactly three prime factors with each prime factor appearing exactly once.

As we saw every group of order $pq$ for distinct primes $p,q$ is not simple, we can also see every group of order $pqr$ for distinct primes $p,q,q$ is also not simple (sylow theorem says one of sylow subgroups is normal).

The numbers of the form $pqr$ for distinct primes $p,q,r$ are

\[30=2\times 3\times 5\]
\[42=2\times 3\times 7\]
\[66=2\times 3\times 11\]
\[70=2\times 5\times 7\]
\[78=2\times 3\times 13\]

Removing these numbers (some of them are already gone in previous filtering) from the previous sequence gives us,

24, 36, 40, 48, 54, 56, 60, 72, 80, 84, 88, 90, 96, 100

Some of the above numbers are of the form $p^3q$ for distinct primes $p,q$; namely $24,40,54,56,88$. Let us see if we can make some quick observations.

For $G$ with $|G|=24=2^3\times 3$, the only interesting case is when $n_2=3$. A trick gives us a group homomorphism $\Phi:G\rightarrow S_3$ with a bound on kernel of $\Phi$, namely $8\leq |Ker(\Phi)|< 24$. As kernel of a group homomorphism is a normal subgroup, we can see such $G$ with $n_2=3$ will have a normal subgroup and hence not simple. So, any group of order $24$ is not simple. One lesson to learn from this example is, it is possible for a group to have a normal subgroup, which is not equal to any of the sylow subgroups.

For $G$ with $|G|=40=2^3\times 5$ it is straight forward to see that only possible value for $n_5$ (number of sylow $5$-subgroups) is $1$ as we need $n_5=1+5k$ to divide $8$ and only possibility is when $n_5=1$. So, any group of order $40$ is not simple.

For $G$ with $|G|=54=2\times 3^3$ it is straight forward to see that only possible value for $n_3$ (number of sylow $3$-subgroups) is $1$ as we need $n_3=1+3k$ to divide $2$ and only possibility is when $n_3=1$. So, any group of order $54$ is not simple.

For $G$ with $|G|=56=2^3\times 7$, only choice for $n_7$ is $1$ or $8$. Suppose $n_7=1$, then it is normal and we are done. Suppose $n_7=8$, due to the order of sylow $7$ subgroup, no two sylow $7$-subgroups will have a non trivial intersection. So, sylow $7$-subgroups constitute $8\times 6=48$ non identity elements of $G$ (with $6$ non-identity elements from each of the $8$ sylow $7$-subgroups). Out of $56$, if $48$ are reserved for sylow $7$-subgroups, we are left with only $8$ elements and with those $8$ elements we can only construct one sylow $2$-subgroup for $G$; which means sylow $2$-subgroup of $G$ is normal. So, any group of order $56$ is not simple.

For $G$ with $|G|=88=2\times 3^11$ it is straight forward to see that only possible value for $n_{11}$ (number of sylow $11$-subgroups) is $1$ as we need $n_{11}=1+11k$ to divide $8$ and only possibility is when $n_{11}=1$. So, any group of order $88$ is not simple.

Removing the above five numbers (some of them are already gone in previous filtering) from the previous sequence gives us,

36, 48, 60, 72, 80, 84, 90, 96, 100

Out of the above numbers, only three of them have three distinct prime factors, namely $60=2^2\times 3\times 5, 84=2^2\times 3\times 7, 90=2\times 3^2\times 5$.

For $G$ with $|G|=84=2^2\times 3\times 7$ it is straight forward to see that only possible value for $n_{7}$ (number of sylow $7$-subgroups) is $1$ as we need $n_{7}=1+7k$ to divide $12$ and only possibility is when $n_{7}=1$. So, any group of order $84$ is not simple.

For $G$ with $|G|=90=2\times 3^2\times 5$ it is not straight forward. It would be fun to try on your own. As in the case of group of order $24$, we need to think slightly differently. In that case we were able to associate a group homomorphism from $G$ and the kernel turned out to be non-trivial normal subgroup of $G$. In this case of $G$ with $|G|=90$, there are many ways to see existence of a normal subgroup, one of which is to involve the notion of "normalizer of a subgroup". With a small effort, we can see any group of order $90$ has a non trivial normal subgroup, thus not a simple group.

A special care needs to be taken when doing similar observation for groups of order $60$, which we will push towards the end of this discussion.

Removing the above five numbers (some of them are already gone in previous filtering) from the previous sequence gives us,

36, 48, 60, 72, 80, 96, 100

For $G$ with $|G|=36$, as in the case of $G$ with $|G|=90$, thinking in terms of normaliser of a subgroup would be useful. Try this on your own.

For $G$ with $|G|=48=2^4\times 3$ only options for $n_3$ are $1$ and $16$. If $n_3=16$, then, we have total $2\times 16=32$ non identity elements (with $2$ non identity elements coming from $16$ sylow $3$-subgroups). As the order of $G$ is $48$, all sylow $2$-subgroups combined can have only $48-32=16$ elements. But, each sylow $2$-subgroup will have exactly $16$ elements. So, there can be exactly one sylow $2$-subgroup. As mentioned before, this means $G$ is not simple.

For $G$ with $|G|=72$, it is not straight forward. As in the case of $G$ with $|G|\in \{24,90\}$, we can associate a homomorphism from $G$ whose kernel gives us a non trivial normal subgroup. Thus, any group of order $72$ is not simple.

For $G$ with $|G|=80=2^4\times 5$, only interesting case for $n_5$ is $16$. If this is the case, we will have contribution of $4\times 16=64$ non identity elements from sylow $5$-subgroups. As the order of $G$ is $80$, all sylow $2$-subgroups combined can have only $80-64=16$ elements. But, each sylow $2$-subgroup will have exactly $16$ elements. So, there can be exactly one sylow $2$-subgroup. As mentioned before, this means $G$ is not simple.

For $G$ with $|G|=96$, as in the case of $G$ with $|G|\in \{36,90\}$, thinking in terms of normaliser of a subgroup would be useful. Try this on your own.

We are left with only two options namely 60 and 100.

We said we are looking for groups of order less than 100. So, we should not really count 100.

We are left with only one possibility, namely 60.

There are examples of groups of order 60 some of which are non-simple. But, there is exactly one simple group (upto isomorphism) of order $60$, namely, the alternating group $A_5$. So, we have the following result

any non abelian simple group of order less than $100$ is isomorphic to $A_5$.

Out of first 100 numbers, if only one number has the possibility of being order of a simple group, does it mean simple groups are interesting?? or uninteresting??