## A few more comments on semidirect products.

**Proposition:** The 8-element quaternion group $Q$ is not a semidirect product.

**Proof:** The only normal subgroup of $Q$ is the 2-element subgroup $Z=\lbrace \pm 1\rbrace$. The quotient group $Q/Z$ is isomorphic to $\Zn{2}\times\Zn{2}$. So if $Q$ *were* a semidirect product it would have to be a semidirect product of $Z$ with $\Zn{2}\times\Z{2}$. But that would mean $Q$ would have a subgroup isomorphic to $\Zn{2}\times\Zn{2}$, so it would have $3$ elements of order $2$ *other than* $\pm 1$. That’s not the case.

**Proposition:** If $H$ and $K$ are two groups, and $\phi:K\to \Aut(H)$ is a homomorphism, you can construct the semi-direct product $T=H\rtimes_{\phi}K$. If you have an automorphism $\tau\in\Aut(K)$, you can modify $\phi$ by taking $\phi’=\tau\phi$. The resulting semidirect product $T’=H\rtimes_{\phi’}K$ is isomorphic to $T$.

**Proof:** Indeed, the map \(f:T'\to T\) defined by $f(hk)=h\tau(k)$ is an homomorphism: \(f(t_1t_2)=f(h_1 k_1 h_2 k_2)=f(h_1\phi'_{k_1}(h_2)k_1k_2)=h_{1}\phi'_{k_1}(h_2)\tau(k_1)\tau(k_2)=h_{1}\phi_{\tau(k_1)}(h_2)\tau(k_1)\tau(k_2).\) On the other hand \(f(t_1)=h_1\tau(k_1)\) and \(f(t_2)=h_2\tau(k_2).\) Then \(f(t_1)f(t_2)=h_1\phi_{\tau(k_1)}(h_2)\tau(k_1)\tau(k_2).\) Since it’s bijective, it’s an isomorphism.

## The fundamental theorem of finitely generated abelian groups

**Theorem:** Let $G$ be a finitely generated abelian group. Then there is an integer $r\ge 0$ and integers $d_1,\ldots, d_k$ all at least two such that \(G\isom \mathbb{Z}^{r}\times \mathbb{Z}/d_{1}\mathbb{Z}\times\cdots\times\mathbb{Z}/d_{k}\mathbb{Z}.\)

**Remark:** The integer $r$ is determined by $G$ up to isomorphism. There are various ways to standardize the $d_{i}$ so that they determine $G$ up to isomorphism but we take that up separately.

We will give a more-or-less constructive proof of this theorem.

Let $g_1,\ldots, g_s$ be a generating set for $G$. Consider the surjective homomorphism \(\pi: \overbrace{\Z\times\cdots\times\Z}^{s} \to G\)

that sends the element $e_{i}$ having a one in position $i$ and zeros elsewhere to $g_{i}$.

By the isomorphism theorems, if $N$ is the kernel of this map, then $G$ is isomorphic to $\Z^{s}/N$. We will study $N$.

Let’s write the elements of $\Z^{s}$ as column vectors using the basis $\lbrace e_{i}\rbrace_{i=1}^{s}$. At first we know very little about $N$ except that it is generated by a (potentially infinite) set of elements of $\Z^{s}$. Let’s choose generators $n_{1},n_{2},\ldots$ for $N$ and organize them in a matrix with integer entries (and potentially infinitely many columns):

\[\mathbf{N}=\left[\begin{matrix} n_{11} & n_{12} & n_{13} & \ldots \cr \vdots & \vdots & \vdots & \vdots \cr n_{s1} & n_{s2} & n_{s3} & \ldots \cr\end{matrix}\right]\]First we are going to show that in fact $N$ is finitely generated, so we can assume $N$ has only finitely many columns.

### Subgroups of finitely generated abelian groups are finitely generated.

**Lemma:** We show first that any subgroup of $\Z^{s}$ is finitely generated.

**Proof:** We will proceed by induction on $s$. If $s=1$, $N$ is a subgroup of $\Z$, and we know that any such subgroup is of the form $d\Z$ and is therefore generated by one element.

Suppose that any subgroup of $\Z^{s-1}$ is finitely generated and that $n_{1},\ldots$ generate $N\subset \Z^{s}$. Consider the last row of the associated matrix $\mathbf{N}$. The integers $n_{s1},n_{s2},\ldots$ generate a subgroup of $\Z$ that is generated by the smallest positive element $d$ of this subgroup, which is the greatest common divisor of all of these $n_{sj}$. Then there are finitely many indices $i_1,i_2,\ldots,i_p$ and integers $a_{1},\ldots, a_{p}$ such that

\[d=\sum_{j=1}^{p} a_{j}n_{si_{j}}.\]The vector $n=\sum_{j=1}^{p}a_{j}n_{i_{j}}\in N$ has $d$ as its last element. For each column $n_{i}$ of $\mathbf{N}$, there is an integer $k_{i}$ so that $n_{i}’=n_{i}-k_{i}n$ has a zero in its last entry. The set consisting of $n$ and the $n_{i}’$ generate $N$. But the $n_{i}’$ all belong to a copy $\Z^{s-1}$ since their last entry is zero, so by induction the subgroup of $\Z^{s-1}$ they generate is finitely generated.

Therefore $N$ is finitely generated.

In fact we’ve shown that any subgroup of $\Z^{s}$ is generated by at most $s$ elements because we add at most one generator at each step in the induction.

**Corollary:** Any subgroup of a finitely generated abelian group $G$ is finitely generated.

**Proof:** A finitely generated abelian group is a quotient of $\Z^{s}$. Given a subgroup $H$ of $G$, it is the image of a subgroup $\tilde{H}$ of $\Z^{s}$, which is finitely generated. The images of the generators of $\tilde{H}$ in $H$ generate $H$.

**Note:** THIS IS FALSE for nonabelian groups.

### Proof of the fundamental theorem

As above, given as finitely generated abelian group, choose a surjective map

\[\pi: \Z^{s}\to G.\]The kernel $N$ of this map is generated by a set $\mathbf{n}$ of at most $s$ elements, and we can arrange the generators of the kernel into a $s\times s$ matrix $\mathbf{N}$, adding zero columns if necessary.

The matrix $\mathbf{N}$ defines a homomorphism (we use the sa)

\[\mathbf{N}:\Z^{s}\to\Z^{s}\]sending an element of the first $\Z^{s}$ viewed as a column vector $v$ to the second by matrix multiplication $\mathbf{n}v$. .

The image of this homomorphism is exactly the kernel $N$ of the map $\pi$. This is because $N$ is generated by the columns of the matrix $\mathbf{N}$, and if

\(v=\left[\begin{matrix} v_1\cr \vdots \cr v_s \cr\end{matrix}\right]\) is a column vector,

then

\[\mathbf{N}v=\sum_{i=1}^{s} v_{i}n_{i}\]where the $n_{i}$ are the columns of $\mathbf{N}$, so the collection of products $\mathbf{N}v$ is exactly the subgroup of $\Z^{s}$ generated by the columns of $N$.

We showed the following in our discussion of automorphisms.

**Lemma:** Suppose $f:\Z^{s}\to\Z^{s}$ is an automorphism. Then there is an invertible $s\times s$ matrix $\mathbf{F}$ with integer entries so that $f(v)=\mathbf{F}v$ where we write $v\in\Z^{s}$ as a column vector.

The composition $\mathbf{N}\circ f$ is given by the matrix $\mathbf{N}\mathbf{F}$. Since $f$ is an automorphism, the image of $\mathbf{N}\mathbf{F}$ in $\Z^{s}$ is the same as the image of $\mathbf{N}$, namely $N$.

Now suppose we apply an automorphism $k$ of $\Z^{s}$ on the right side of the map $\mathbf{N}$. In that case, the image $k(N)$ need not be $N$, but it is still the case that $\Z^{s}/g(N)$ and $\Z^{s}/N$ are isomorphic, and therefore $\Z^{s}/g(N)$ is isomorphic to our original group $G$.

Therefore up to isomorphism we can modify our matrix $\mathbf{N}$ by multiplying it on either side by invertible $s\times s$ integer matrices without changing the quotient $G$ we are trying to compute.

In particular we can:

- swap rows and columns of $\mathbf{N}$
- multiply any row or column of $\mathbf{N}$ by $\pm 1$
- do elementary row and column operations on $\mathbf{N}$ – that is, replace a column $n_{i}$ by $n_{i}-an_{j}$ for any integer $a$, and similarly for rows.

Using these operations, proceed as follows:

- If the matrix is zero, the $G$ is $\Z^{s}$ and we’re done. Otherwise, swap rows so the first row isn’t zero.
- Make every element in the first row positive, and then swap columns so the smallest element in the first row is in the upper left corner of $\mathbf{N}$. Call that element $a$.
- Use elementary column operations to reduce the other elements in the first row to be less than $a$.
- Repeat steps $1$ and $2$ until the upper left entry is the only nonzero entry.
- Now follow the same process using row operations until the only nonzero entry in the first column is in the upper left corner.
- Each round of this makes the entry in the upper left corner smaller, so eventually you must reach a point where the first row and column of the matrix are zero, except for the upper left entry, which could be zero or nonzero.
- Now continue this reduction process on the $(s-1)\times (s-1)$ submatrix.
- Eventually you reach a diagonal matrix.

When you have a diagonal matrix, you can see that

\[G\isom \Z^{r}\times \Z/d_{1}\Z\times \cdots \Z/d_{s-r}\Z\]where $r$ is the number of zero diagonal elements and the $d_{i}$ are the nonzero ones.

**Corollary:** If the matrix $\mathbf{N}$ is invertible, its determinant is the order of $G$.