# List: groups up to order 15 with proofs

One of the prime goals after any mathematical structure is defined is to classify all possible structures up to isomorphism. Here, we will do something a little more modest: classify all finite groups up to order fifteen. Despite spending a great amount of time with groups, I’ve never actually done this formally, although I’ve certainly worked with many of them in random examples. So, I thought it might be fun to actually write them down.

## Theorems we need

In this description I will assume some basic facts of group theory such as terminology. However, I have tried to keep this post self-contained by proving elementary results so that beginners won’t have to refer to random texts in case they forgot proofs. I will however assume that the reader has seen the theorem that every finitely-generated abelian group is of the form $\Z^n\times (\Z/p_1^{r_1})\times\cdots\times (\Z/p_k^{r_k})$ for some integer $n$, primes $p_i$ and integers $r_i$. I also will assume the basic Sylow theorem:

Theorem. Let $G$ be a finite group with $|G| = p^em$ where $e$ is a positive natural number and $p$ and $m$ are coprime. Then:

1. There exists a subgroup of $G$ of order $p^e$.
2. Any two subgroups of $G$ of order $p^e$ are conjugate.
3. The number of subgroups of $G$ of order $p$ divides $|G|$ and leaves a remainder $1$ on division by $p$.

A subgroup of order $p^e$ in $G$ as above is called a Sylow $p$-subgroup. Some might think this is already too much technology to assume, but actually this theorem only takes a very short time to prove so the interested reader should not have much trouble learning it. I am going to use another theorem. To explain it, let $G$ be a group and $A$ a $\Z[G]$-module. The semidirect product $A\rtimes G$ is the group whose underlying set is $A\times G$ and whose group multiplication is given by $(a,g)(a’,g’) = (a + ga’,gg’)$. The theorem we will use is:

Abelian Schur-Zassenhaus. If $0\to A\to E\to G\to 1$ is a short exact sequence of finite groups such that the order of $A$ and $G$ are coprime, then $E$ is isomorphic to $A\rtimes G$.

This theorem holds for nonabelian groups also, but then it is a little more complicated to explain the semidirect product and we will not need that here. It takes a bit more work to prove this one but it is an important calculational tool that makes the work of listing finite groups a little easier.

Without further ado, we list the groups.

## The trivial group

We certainly can’t forget the trivial group $G = \{1\}$. Or is it even more trivial to say that there are no groups of order zero? Let’s not go there.

## Groups of prime order

If $G$ is any finite group and $x\in G$ then the order of $x$ divides the order of $G$. Hence, if $p$ is a prime, there is only one group of order $p$ and that is $\Z/p$. That takes care of $p=2,3,5,7,11,13$ in our list and of course all of the infinitely many primes not on our list.

## Groups of order four

We start out with the abelian ones: $\Z/4$ and $\Z/2\times\Z/2$. Are there any others? The answer is no, and in this case it is easy to prove directly. But let’s introduce a general tools that will be helpful later on. Recall that a $p$-group is a group whose order is a power of a prime $p$.

If $G$ is a nontrivial $p$-group then the center $Z(G)$ of $G$ is nontrivial.
Proof. Let $G$ act on itself via conjugation. This action partitions $G$ into conjugacy classes (orbits under conjugation). Each nontrivial orbit has size divisible by $p$, whereas the trivial orbits make up the center $Z(G)$. This must also be divisible by $p$ since $|G|$ is divisible by $p$.

Now, I will leave it as an exercise to the reader to prove this corollary: if $G$ has order $p^2$, then $G$ is abelian. Since we have listed all the abelian groups, there are no other groups of order four.

## Groups of order six

There is only one abelian one: $\Z/6\cong \Z/2\times \Z/3$. What about a nonabelian one. Does such a group exist? Suppose for the sake of argument that such a $G$ exists. Cauchy’s theorem tells us that there exists a $r\in G$ of order three and an $s\in G$ of order two. Let’s start with $r$. This element generates a subgroup $H\cong\Z/3$.

This subgroup $H$ contains half the elements of $G$, and hence this subgroup is normal. Obviously, $s\not\in H$ since it has order two, and the coset $sH$ contains the rest of the elements of $G$. Therefore, $G = \{1,r,r^2,sr,sr^2 \}$. Now, in order to fill out the multiplication table of $G$, we need to know what $rs$ is. Now, $rs\not\in H$ since $s\not\in H$, so there are only three possibilities:

1. $rs = s$: Impossible because this implies $r = 1$.
2. $rs = sr$: Impossible because this implies $G$ is abelian.
3. $rs = sr^2$: The only remaining possibility

Therefore, such a group $G$ would need to be the group generated by two elements $r$ and $s$ and subject to the relations $r^3=1,s^2=1$, and $rs = sr^2$. Does it have six elements? Yes, it does. We can see this by defining a homomorphism $G\to S_3$ by sending $r\mapsto (1 2 3)$ and $s\mapsto (1 3)$. I will leave it to the reader to check that these permutations in $S_3$ do generate a group of six elements and that this homomorphism is well-defined.

In general, we define the dihedral group $D_n$ via generators and relations:
$$\langle~ r,s~|~ r^n, s^2, rs = sr^{n-1}~\rangle.$$ It is not hard to see that $D_n$ is a noncommutative group consisting of $2n$ elements. In fact, $D_n$ is nothing else than the symmetry group of rotations and reflections of the regular $n$-gon.

So there is exactly one noncommutative group of order six and that is the dihedral group $D_3$, the symmetry group of the triangle. These dihedral groups will come in handy later.

## Groups of order eight

We have the three abelian ones: $\Z/2\times\Z/2\times\Z/2$, $\Z/4\times\Z/2$, $\Z/8$. Also, from the discussion on groups of order six, we have the noncommutative dihedral group $D_4$. Are there any other ones?

For example, is there a noncommutative group of order eight in which every element has order two? Suppose such a $G$ existed. Let $r,s\in G$ be two nontrivial elements of order two that do not commute. Now we already know about five elements in our hypothetical group $G$: $\{1,r,s,sr,rs\}$. These five elements must generate $G$ since the next divisor of eight that is greater than five is eight. However we see that $\{1,r,s,sr,rs,rsr,srs\}$ are the only possible nontrivial elements that we could have: if we tried to form an element like $srsr$, that would just be trivial because every element has order two. Therefore, there are at most seven nontrivial elements in this group, which doesn’t make any sense. (Note, the fact that $sr\not=rs$ was required to get this contradiction.)

Therefore, any group $G$ that is nonabelian and has order eight must in fact have at least one element $r$ of order four, which generates a normal subgroup $H = \{1,r,r^2,r^3\}$. One such group is of course the dihedral group. Are there others? Now, we know from Cauchy’s theorem that $G$ contains an element of order two. There are two possibilities for this element: either it is $r^2$, an element already in $H$, or it is an element $s\in G$ that is not in $H$. I will leave it to the reader to check that if there exists an element of order two not in $H$, then the group $G$ must be the dihedral group $D_4$. Therefore, if our mythical group $G$ exists, there exists an element $s\in G$ of order four not in $H$. The elements of $G$ are of course
$$\{1,r,r^2,r^3, s,sr,sr^2,sr^3\}.$$ However, unlike the dihedral group, we have $s^4 = 1$ rather than $s^2 = 1$. But what is $s^2$? We can’t have $s^2 = r$ (why?). Can $s^2 = sr^2$. No, because then $s = r^2$, and then $s\in H$.

In fact, if we play this game, the only possibility that is consistent with our assumptions thus far is that $s^2 = r^2$. What about $rs$? Can we figure out what that is? If we go through the list of possibilities for $rs$, we find that there are only two that are not immediately ruled out: $rs = sr^2$ and $rs = sr^3$. The first possibility implies (because $r^2 =s^2$) that $r=s$, which is not possible. Therefore, that only leaves
$$rs = sr^3.$$ These relations are enough to determine the multiplication table of $G$. In presentation format, we have
$$G := \langle~r,s~|~ r^4, s^2 = r^2, rs = sr^3~\rangle.$$ Does this group have order eight or did these relations, consequences of our reasoning, kill off too many elements? I will leave it to the reader to determine that
\begin{align*}r&\longmapsto\begin{pmatrix}0&i\\i&0\end{pmatrix}\\ s&\longmapsto\begin{pmatrix}0 & 1\\-1&0\end{pmatrix}\end{align*} give an injective map $G\to \GL_2(\C)$ whose image is a group of order eight.

This group $G$ has a name: the quaternions, often denoted by $Q$. Since by our reasoning we have proven that there exists no element of order two not in the subgroup generated by $r$, we have proven that $G\not\cong D_4$. Our reasoning left no other options for other generators and relations so we have found all the groups of order eight.

## Groups of order nine

The abelian ones are $\Z/9$ and $\Z/3\times\Z/3$. This time, we can take a break from the dihedral groups, which have even order! In fact, if we recall our journey into groups of order four, we proved that any group of order $p^2$ where $p$ is a prime is abelian. So actually, there are no other groups of order nine.

## Groups of order ten

The abelian ones is: $\Z/2\times\Z/5$ (notice that there is only one abelian one whenever the prime factorization of the order contains no proper powers of primes).

There is also the dihedral group
$$D_5 = \langle~r^5,s^2 ~|~ rs=sr^{-1}~\rangle.$$ Are there any others? No. This is where we can use the abelian Schur-Zassenhaus theorem and the Sylow theorem. Let $G$ be a group of order ten. Then there exists a $5$-Sylow subgroup and hence an exact sequence $0\to \Z/5\to G\to \Z/2\to 0$ of groups. By Schur-Zassenhaus, $G$ is a semidirect product. The trivial action of $\Z/2$ on $\Z/5$ gives $\Z/2\times\Z/5$. What about nontrivial actions?

In this case, $\Z/5^\times$ is the automorphism group of $\Z/5$. (Note, when writing $\Z/5^times$, we are referring to $\Z/5$ as a ring, the corresponding elements of $\Z/5^\times$ act via multiplication on the abelian group $\Z/5$.) The only element of order two is $4$ as $4^2 = 16$, so this is the other semidirect product, which must be $D_5$.

Note: in general, if a group has order $2p$ it is either $\Z/2p$ or $D_p$. Can you prove that?

## Groups of order twelve

Abelian ones: $\Z/2\times\Z/2\times\Z/3$ and $\Z/4\times\Z/3$. We also have the dihedral group $D_6$.

Let $H$ be a $3$-Sylow and $K$ be a $2$-Sylow (it has four elements). We claim at least one of $H$ or $K$ is normal. If $H\cong\Z/3$ is not normal, then it must have exactly four conjugates via the Sylow theorem. Each of these has pairwise trivial intersection. This accounts for nine elements, with three to spare. Since the $2$-Sylow $K$ has trivial intersection with each of the conjugates of $H$, we see that these remaining three elements make up the $2$-Sylow and therefore $K$ is normal.

If $H$ is normal, we get one of two possible short exact sequences:
$$0\to \Z/3\to G\to \Z/4\to 0\\ 0\to \Z/3\to G\to \Z/2\times\Z/2\to 0.$$ Now we invoke Schur-Zassenhaus: $G$ must be a semidirect product. To find all such products we just need to list all such actions of $\Z/4$ on $\Z/3$ and all actions of $\Z/2\times\Z/2$ on $\Z/3$ (in general of course, this procedure might list duplicates so we have to be careful).

Let us note that the only automorphism of $\Z/3$ is multiplication by $2$.

Let’s start with the case where the quotient of $G$ by $\Z/3$ is $\Z/2\times \Z/2$. There are at most two nonisomorphic possibilities: either $(1,0)$ acts nontrivially and $(0,1)$ acts trivially, or they both act nontrivially. These two actions give isomorphic groups; the matrix
$$\begin{pmatrix}1 & 1\\1 & 0\end{pmatrix}$$ is an automorphism of $\Z/2\times\Z/2$ that takes on action to the other. In that case, we obtain the group $D_3\times\Z/2$. I will leave it to the reader to prove that this is actually isomorphic to $D_6$.

The other case is a new group which we will leave as $\Z/3\rtimes\Z/4$ where $1\in\Z/4$ acts nontrivially. It is not isomorphic to any of the groups we have seen so far.

Now we suppose that $H$ is not normal. As we have seen, this means that $K$ is normal and there are two short exact sequences
$$0\to \Z/4\to G\to\Z/3\to 0.\\ 0\to \Z/2\times \Z/2 \to G\to\Z/3\to 0.$$ In the first case, $\Z/4^\times \cong \Z/2$ so $\Z/3$ can only act trivially on $\Z/4$. The second case is more interesting. In that case, the automorphim group of $\Z/2\times\Z/2$ is ${\rm GL}_2(\Z/2)\cong D_6$.

The elements of $D_6$ are $\{1,r,r^2,s,sr,sr^2 \}$. Without loss of generality, if $\Z/3$ acts on $\Z/2\times\Z/2$ it must act via $r$, or more concretely, via the matrix
$$\begin{pmatrix}1 & 1\\1 & 0\end{pmatrix}.$$ Let us recall the symmetric group $S_4$, the permutation group on four letters. It has as a subgroup $A_4$: the alternating group of even permutations. These are the permutations that can be written as a product of an even number of transpositions (a.k.a. 2-cycles). Alternatively, we could define it has the subgroup of permutations with determinant one, when we realize the permutation group as a group of permutation matrices.

Now $A_4$ has twelve elements and is nonabelian because $(1 3 2)$ and $(2 3 4)$ are even permutations that do not commute. Is $A_4$ isomorphic to $D_6$? It is not. There are probably many ways to observe this. One way is to check that in the dihedral group $D_6$ generated by $r$ and $s$ as before, the only elements of order three are $r^2$ and $r^4$. Whereas, in $A_4$, we see that $(1 3 2), (1 2 3), (2 3 4)$ are just some of the elements of order three.

However, another way to realize that $A_4$ is not isomorphic to $D_6$ is to observe that it is in fact isomorphic to our new group $\Z/2\times \Z/2\rtimes\Z/3$. Why is this?

Let us list all the elements of $A_4$:
$$\{ e, (12)(34), (13)(24), (14)(23), (123),(132),(124),(142), (134),(143) (234),(243) \}.$$ We see that $A_4$ has a subgroup $P$ isomorphic to $\Z/2\times\Z/2$ consisting of the first four elements in this list. We can pick any representative for coset representatives: for example, $P, (123)P, (132)P$ are a complete list of cosets and these cosets form a group isomorphic to $\Z/3$. We know that the action of $\Z/3$ is nontrivial because $A_4$ is not abelian, and we know there is only one nontrivial action so $A_4$ must be isomorphic to the group $(\Z/2\times\Z/2)\rtimes\Z/3$ be constructed above. (It might be instructive to check directly that $(123)$ acts via conjugation through the matrix we wrote down above or possibly its square.)

## Groups of order fourteen

We already mentioned that a group of order $2p$ for an odd prime $p$ is either cyclic or dihedral. So in this case $\Z/2\times\Z/7\cong\Z/14$ or $D_7$. Note we did not actually prove this theorem, but the same argument we used for groups of order ten works here.

## Groups of order fifteen

We claim that the only group of order fifteen is $\Z/15\cong\Z/5\times\Z/3$.

Let’s prove it. Let $G$ be an arbitrary group of order fifteen. Let $H$ be a $5$-Sylow subgroup and $K$ be a $3$-Sylow subgroup. From the Sylow theorem we see that $H$ is normal and that $K$ is normal. Thus we have an exact sequence
$$0\to \Z/5\to G\to \Z/3\to 0.$$ However, all the automorphisms of $\Z/5$ have order $1$ or $2$ or $4$. Therefore, $\Z/3$ acts trivially on $\Z/5$ and thus $G\cong \Z/3\times\Z/5$.

## The complete list

1. Order 1 (1): The trivial group $\{1\}$
2. Order 2 (1): $\Z/2$.
3. Order 3 (1): $\Z/3$
4. Order 4 (2): $\Z/4$ and $\Z/2\times \Z/2$
5. Order 5 (1): $\Z/5$
6. Order 6 (2): $\Z/2\times\Z/3$ and $D_3$.
7. Order 7 (1): $\Z/7$
8. Order 8 (5): $\Z/2\times\Z/2\times\Z/2$ and $\Z/4\times\Z/2$ and $\Z/8$ and $Q$ (quaternions)
9. Order 9 (2): $\Z/9$ and $\Z/3\times\Z/3$
10. Order 10 (2): $\Z/2\times\Z/5$ and $D_5$
11. Order 11 (1): $\Z/11$
12. Order 12 (5): $\Z/2\times\Z/2\times\Z/3$ and $\Z/4\times\Z/3$ and $\Z/3\rtimes \Z/4$ and $D_6$ and $A_4$ and $\Z/3\rtimes\Z/4$
13. Order 13 (1): $\Z/13$
14. Order 14 (2): $\Z/2\times\Z/7$ and $D_7$
15. Order 15 (1): $\Z/3\times\Z/5$

Here is a larger list of the number of subgroups of each order starting at order one:

1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2, 2, 1, 231, 1, 5, 2, 16, 1, 4, 1, 14, 2, 2, 1, 45, 1, 6, 2, 43, 1, 6, 1, 5, 4, 2, 1, 47, 2, 2, 1, 4, 5, 16, 1, 2328, 2, 4, 1, 10, 1, 2, 5, 15, 1, 4, 1, 11, 1, 2, 1, 197, 1, 2, 6, 5, 1, 13, 1, 12, 2, 4, 2, 18, 1, 2, 1, 238, 1, 55, 1, 5, 2, 2, 1, 57, 2, 4, 5, 4, 1, 4, 2, 42, 1, 2, 1, 37, 1, 4, 2, 12, 1, 6, 1, 4, 13, 4, 1, 1543, 1, 2, 2, 12, 1, 10, 1, 52, 2, 2, 2, 12, 2, 2, 2, 51, 1, 12, 1, 5, 1, 2, 1, 177, 1, 2, 2, 15, 1, 6, 1, 197, 6, 2, 1, 15, 1, 4, 2, 14, 1, 16, 1, 4, 2, 4, 1, 208, 1, 5, 67, 5, 2, 4, 1, 12, 1, 15, 1, 46, 2, 2, 1, 56092, 1, 6, 1, 15, 2, 2, 1, 39, 1, 4, 1, 4, 1, 30, 1, 54, 5, 2, 4, 10, 1, 2, 4, 40, 1, 4, 1, 4, 2, 4, 1, 1045, 2, 4, 2, 5, 1, 23, 1, 14, 5, 2, 1, 49, 2, 2, 1, 42, 2, 10, 1, 9, 2, 6, 1, 61, 1, 2, 4, 4, 1, 4, 1, 1640, 1, 4, 1, 176, 2, 2, 2, 15, 1, 12, 1, 4, 5, 2, 1, 228, 1, 5, 1, 15, 1, 18, 5, 12, 1, 2, 1, 12, 1, 10, 14, 195, 1, 4, 2, 5, 2, 2, 1, 162, …