Classify the groups of order $88$ up to isomorphism.

Question

Classify the groups of order $88$ up to isomorphism.

Here is what I have so far (I'm aware that there are $12$ groups, but I don't know which ones I'm missing as well as why the $3$ groups are abelian and the other $9$ are non-abelian.)

We are asked to classify the groups of order $88$ up to isomorphism. To prove that any group of order $88$ is abelian. So if $|G|=88$, then we can construct all abelian groups of order $88$ by using the Fundamental Theorem of Finitely Generated Abelian groups. We can use the Fundamental Theorem of Finitely Generated Abelian Groups wich states the following:

Let $G$ is a finitely generated abelian group. Then
\begin{equation}G\cong\Bbb Z^r\times\Bbb Z_{n_1}\times \Bbb Z_{n_2}\times\dots\times\Bbb Z_{n_s},\tag1\end{equation} For some integers $r,n_1,n_2,\dots,n_s$ satisfying the following conditions:

1. $r\ge 0$ and $n_j\ge 2$ for all $j$, and
2. $n_{i+1}\mid n_i$ for $1\le i\le s-1$

The expression in $(1)$ is unique: if $G\cong\Bbb Z^t\times\Bbb Z_{m_1}\times \Bbb Z_{m_2}\times\dots\times\Bbb Z_{m_u}$, where $t$ and $m_1,m_2,\dots,m_u$ satisfy 1. and 2. (i.e., $t\ge 0$,$m_j\ge 2$ for all $j$ and $m_{i+1}\mid m_i$ for $1\le i\le u-1$), then $t=r$,$u=s$ and $m_i=n_i$ for all $i$.

This gives us an effective way of listing all finite abelian groups of a given order. Namely, to find (up to isomorphism) all abelian groups of a given order $n$ one must find all finite sequences of integers $n_1, n_2,\dots,n_s$ such that

1. $n_j\ge 2$ for all $j\in\{1,2,\dots,s\}$,
2. $n_{i+1}\mid n_i$,$1\le i\le s-1$, and
3. $n_1 n_2\dots n_s=n$

We can also note that every prime divisior of $n$ must divide the first invariant factor $n_1$. In particular, if $n$ is the product of distinct primes, which are all to the first power, which is called squarefree, we see that $n|n_1$, hence $n=n_1$. This proves that if $n$ is squarefree, there is only one possible list of invariant factors for an abelian group of order $n$. The factorization of $n$ into prime powers is the first step in determining all possible lists of invariant factors for abelian groups of order $n$.

This means that we can break 8$8$ down into its prime factors which would give us the following: $$88=2\cdot 44=2\cdot 2\cdot 22=2\cdot 2\cdot 2\cdot 11$$ So if we say that $n=88=2^3\cdot 11$, as we have stated above we must have that $2×11|n_1$, so possible values of $n_1$ are as follows: $$n_1=2^3\cdot 11~\lor~n_1=2^2\cdot 11~\lor~n_1=2\cdot 11$$ For each of these we need to work out the possible $n_2$’s. For each resulting pair $n_1,n_2$ we need to then work out the possible $n_3$’s and then continue in this manner until all lists satisfying 1. and 3. are obtained. Therefore $88$ can be written as $2^3\cdot 11$. Which would give us the following:

Order $p^\beta$: Partitions of $\beta$ Abelian Groups $$2^3:~3,~\Bbb Z_8;~~~2,1,~\Bbb Z_4\times\Bbb Z_2;~~~1,1,1,~\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$$ $$11^1:~1,~\Bbb Z_{11}$$

We can obtain the abelian groups of order $88$ by taking one abelian group from each of the two lists above and taking their direct product. Doing this in all possible ways gives all isomorphism types: $$\Bbb Z_{88},~\Bbb Z_8\times\Bbb Z_{11},~\Bbb Z_4\times\Bbb Z_2\times\Bbb Z_{11},~\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{11},~\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{22},~\Bbb Z_4\times\Bbb Z_{22},~\Bbb Z_2\times\Bbb Z_{44}$$ When we have completed this we will have $12$ groups. By the Fundamental Theorems above, this is a complete list of all abelian groups of order $88$, every abelian group of this order is isomorphic to precisely one of the groups above and no two of the groups in this list are isomorphic.

We can then define abelian and non-abelian groups. Abelian groups or commutative groups are groups in which the results of applying the group operation to two group elements does not depend on the order in which they are written, in other words these groups are are groups that following the axiom of commutativity.

Abelian groups generalize the arithmetic of addition of integers. Non-abelian groups, also known as non-commutative groups are groups $(G,*)$ in which there exists at least one pair of elements $a$ and $b$ of $G$, such that $a*b\ne b*a$.

Of these $12$ groups $3$ of them are abelian and the other $9$ are non-abelian groups. The three abelian groups are $\Bbb Z_{88}$, $\Bbb Z_4\times\Bbb Z_{22}$, and $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{22}$.

Answer 1

You're going to need to use Sylow's theorems. If $n_{11}$ is the number of Sylow $11$-subgroups of $G$ then $n_{11}\bigm|88$ and $n_{11}\equiv1\pmod{11}$ so $n_{11}=1$. Then $G$ has a normal Sylow $11$-subgroup $P_{11}$. Let $P_2$ be a Sylow $2$-subgroup of $G$. The recognition theorem for semidirect products shows that $G\cong P_{11}\rtimes_\varphi P_2$ for some group homomorphism $\varphi\colon P_2\to\text{Aut}(P_{11})$.

Recall that $\text{Aut}(P_{11})\cong(\mathbb{Z}/11\mathbb{Z})^\times\cong\mathbb{Z}/10\mathbb{Z}$ is cyclic of order $10$ and has a unique subgroup $K$ of order $2$ (i.e. the subgroup $\{\pm1\}\leq(\mathbb{Z}/11\mathbb{Z})^\times$). The image of $\varphi$ must lie in $K$. Let $N=\ker\varphi$. We know that $\varphi$ factors as the composition $$P_2\twoheadrightarrow P_2/N\hookrightarrow K\hookrightarrow\text{Aut}(P_{11}).$$ Since $K$ has order $2$, there is only one choice for the injective homomorphism $P_2/N\hookrightarrow K$. Thus, the choice of $\varphi$ is determined by the choice of a normal subgroup $N$ of $P_2$ of index $1$ or $2$. It remains to go through the possibilities for $P_2$ and the possible normal subgroups of $P_2$ of index $1$ or $2$.

Two notes:

(1) Subgroups of index $1$ or $2$ are necessarily normal, so requiring normality is redundant.

(2) If $N$ and $N^\prime$ differ by an automorphism $\sigma$ of $P_2$ (meaning that $N^\prime=\sigma(N)$) then the resulting groups $G=P_{11}\rtimes_\varphi P_2$ and $G=P_{11}\rtimes_{\varphi^\prime}P_2$ will be isomorphic. Thus, we only need to look at one $N$ from each automorphism-class.

Case 1: $P=C_8$. There are two possibilities for $N$: $C_4$ and $C_8$.

Case 2: $P=C_2\times C_4$. There are three possibilities for $N$: $C_2\times C_2$, $C_4$, and $C_2\times C_4$.

Case 3: $P=C_2\times C_2\times C_2$. There are two possibilities for $N$: $C_2\times C_2$ and $C_2\times C_2\times C_2$.

Case 4: $P=D_4$. There are three possibilities for $N$: $C_2\times C_2$, $C_4$, and $D_4$.

Case 5: $P=Q_8$. There are two possibilities for $N$: $C_4$ and $Q_8$.

This gives $12$ groups.

There are a few points to make though:

(1) You need to check that I really did cover every automorphism-class of index $1$ or $2$ subgroups of $P_2$. This is a pain but it's just a matter of running through what the automorphism groups of each possible $P_2$ do to its subgroups of index $2$.

(2) You need to check that these $12$ groups are pairwise non-isomorphic. This isn't so bad. There is no overlap between the five cases because each case has a different Sylow $2$-subgroup. Also, there is no overlap between picking $N=P_2$ (which results in $G$ having a normal Sylow $2$-subgroup) and picking $N\lneq P_2$ (which results in $G$ not having a normal Sylow $2$-subgroup). Then there are just two pairs of groups which you have to check are not isomorphic (which can be done by counting elements of 2).

(3) We know many of these groups. In the cases where $N=P_2$, we obtain the direct products

$C_{11}\times C_8\cong C_{88}$.

$C_{11}\times C_2\times C_4\cong C_{44}\times C_2$.

$C_{11}\times C_2\times C_2\times C_2\cong C_{22}\times C_2\times C_2$.

$C_{11}\times D_4$.

$C_{11}\times Q_8$.

In the cases where $N\lneq P_2$, we can still identify most of the groups:

$C_{11}\rtimes(C_2\times C_4)$ with $N=C_4$ is isomorphic to $D_{11}\times C_4$.

$C_{11}\rtimes(C_2\times C_2\times C_2)$ with $N=C_2\times C_2$ is isomorphic to $D_{11}\times C_2\times C_2\cong D_{22}\times C_2$.

$C_{11}\rtimes D_4$ with $N=C_4$ is isomorphic to $D_{44}$.

$C_{11}\rtimes(C_2\times C_4)$ with $N=C_2\times C_2$ is isomorphic to $\text{Dic}_{11}\times C_2$.

$C_{22}\rtimes Q_8$ with $N=C_4$ is isomorphic to $\text{Dic}_{22}$.

$C_{11}\rtimes C_8$ is the only one that I can't name.