## Direct products

If $G_1,\ldots, G_n$ are groups, the direct product $G=G_1\times\cdots\times G_n$ is the group whose elements are $n$-tuples $(g_1,\ldots, g_n)$, with $g_i\in G_i$, and all operations done componentwise.

**Definition:** If $G$ is a group and $H$ and $K$ are normal subgroups of $G$ with $HK=G$ and $H\cap K$ trivial, then the map $(h,k)\to hk$ from $H\times K\to HK=G$ is an isomorphism. $G$ is called the internal direct product of $H$ and $K$.

## Fundamental theorem of finitely generated abelian groups

**Definition:** A group is *finitely generated* if it has a finite generating set.

Finitely generated groups can be finite or infinite. For example $\Z^{n}$ is finitely generated.

**Proposition:** An abelian group $G$ is finitely generated if and only if there is an integer $k$ and a surjective homomorphism \(f:\Z^{k}\to G.\)

**Proof:** If $f$ exists, then $G$ is generated by $f(e_i)$ for $i=1,\ldots,k$, where $e_i$ is the element of $\Z^{k}$ with a $1$ in position $i$ and zeroes elsewhere.

Conversely, if $G$ is generated by $g_1,\ldots, g_k$, define $f:\Z^{k}\to G$ by $f(e_i)=g_i$.

**Theorem:** Let $G$ be a finitely generated abelian group. Then there are integers \(r\ge 0\) and $n_1,\ldots,n_s$, all at least $2$, such that

and such that $n_{i+1}\divides n_{i}$ for $i=1,\ldots, s-1$.

The integer $r$ is called the *free rank* of $G$, and the $n_{i}$ are called the *invariant factors*. Two finitely generated abelian groups are isomorphic if and only if they have the same rank and the same invariant factors.

**Example:** If $G$ has order $20$, then its possible invariant factors are:

- $20$, in which case it’s cyclic and also isomorphic to $\Zn{5}\times\Zn{4}$.
- $10$ and $2$ in which case it’s $\Zn{10}\times\Zn{2}$ or $\Zn{5}\times\Zn{2}\times\Zn{2}$.

Alternatively, suppose that $G$ is finite of order $n$ and $n=p_1^{a_1}p_2^{a_2}\cdots p_{k}^{a_k}$ is the prime factorization of $n$. Then $G$ is a product of its Sylow $p$-subgroups $A_{i}$ which have orders $p_{i}^{a_{i}}$. Each abelian $p$-group $A_{i}$ is a product of cyclic $p$-groups of order $p^{a_{ij}}$ for $j=1,\ldots, t$ where $a_{i1}+a_{i2}+\cdots+a_{is}=a_{ij}$. If you arrange the $a_{ij}$ so that the primes are in increasing order and the exponents are in decreasing (or increasing) order then the $p_{i}^{a_{ij}}$ determine the group up to isomorphism. The powers $p^{a_{ij}}$ are called the *elementary divisors* of $G$.

**Example:** If $G$ has order $20$, then the elementary divisors are either $5,4$ (if the group is cyclic) or $5,2,2$ if not.

## Semidirect products

**Definition:** Let $H$ and $K$ be groups and let $\phi:K\to \Aut(H)$ be a homomorphism. Write $\phi_{k}$ for the map $H\to H$ associated by $\phi$ to $k\in K$. The *semidirect product* $P=H\semi_{\phi} K$ of $H$ and $K$ by $\phi$ is defined as follows:

- As a set, it consists of pairs $(h,k)$.
- The group law is $(h_1,k_1)(h_2,k_2)=(h_1\phi_{k_1}(h_2),k_1k_2).$
- The groups $H=\lbrace (h,1) :h\in H\rbrace$ and $K=\lbrace (1,k): k\in K\rbrace$ are included in $P$ as subgroups.
- $P$ has $\lvert H\rvert\lvert K\rvert$ elements.
- $H$ is a normal subgroup of $P$. In fact $(1,k)(h,1)(1,k^{-1})=(\phi_{k}(h),1)$. So the conjugation action of $K$ on $H$ is given by the automorphism $\phi$.
- The quotient group $P/H$ is isomorphic to $K$.

**Example:** Let $R=\Zn{n}$, where $n$ is odd, and let $S=\Zn{2}$. The cyclic group $\Zn{n}$ (written multiplicatively) has an automorphism $\phi(x)=x^{-1}$ of order $2$. Let $s$ be the generator of $S$ and define $\phi_s(x)=x^{-1}$ for $x\in R$. Then $R\semi_{\phi} S$ is a group of order $2n$. Writing $(x,1)=x$ and $(1,s)=s$, the computation above shows that $sx=(1,s)(x,1)=(x^{-1},s)=(x^{-1},1)(1,s)x^{-1}s$ or, in other words, the group $P$ we have constructed is the dihedral group with $2n$ elements.

**Example:** Let $R=\Zn{11}$. The automorphism group of $R$ is cyclic of order $10$ generated by $2$. Let $U$ be the cyclic group of order $10$ generated by $y$, with $\phi_{y}(x)=x^2$ for $x\in R$.

Then $P=R\semi_{\phi}U$ is a group of order $110$ generated by $x=(1,0)$ and $y=(0,1)$ satisfying $x^{11}=y^{10}=1$ and $yx=x^{2}y.$