Abstract Algebra/Group Theory/Products and Free Groups

During the preliminary sections we introduced two important constructions on sets: the direct product and the disjoint union. In this section we will construct the analogous constructions for groups.

Product Groups

Definition 1: Let $G$ and $H$ be groups. Then we can define a group structure on the direct product $G\times H$ of the sets $G$ and $H$ as follows. Let $(g_1,h_1),(g_2,h_2)\in G\times H$. Then we define the multiplication componentwise: $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$. This structure is called the direct product of $G$ and $H$.

Remark 2: The product group is a group, with identity $(e_G,e_H)$ and inverses $(g,h)^{-1}=(g^{-1},h^{-1})$. The order of $G\times H$ is $|G\times H|=|G||H|$.

Theorem 3: Let $G$ and $H$ be groups. Then we have homomorphisms $\pi_1\,:\, G\times H \rightarrow G$ and $\pi_2\,:\, G\times H \rightarrow H$ such that $\pi_1(g,h)=g$ and $\pi_2(g,h)=h$ for all $(g,h)\in G\times H$. These are called the projections on the first and second factor, respectively.

Proof: The projections are obviously homomorphisms since they are the identity on one factor and the trivial homomorphism on the other. ∎

Corollary 4: Let $G$ and $H$ be groups. Then $\frac{G\times H}{H}\approx G$ and $\frac{G\times H}{G}\approx H$.

Proof: This follows immediately from plying the first isomorphism theorem to Theorem 3 and using that $G\times\{e_H\}\approx G$ and $\{e_G\}\times H\approx H$. ∎

Theorem 5: Let $G$ and $H$ be groups. Then $G\times\{e_H\}$ and $\{e_G\}\times H$ are normal subgroups of $G\times H$.

Proof: We prove the theorem for $G\times\{e_H\}$. The case for $\{e_G\}\times H$ is similar. Let $g,g^\prime\in G$ and $h\in H$. Then $(g,h)(g^\prime,e_H)(g,h)^{-1}=(gg^\prime g^{-1},hh^{-1})=(gg^\prime g^{-1},e_H)\in G\times\{e_H\}$. ∎

Commutative diagram showing the universal property satisfied by the direct product.

We stated that this is an analogous construction to the direct product of sets. By that we mean that it satiesfies the same universal property as the direct product. Indeed, to be called a "product", a construction should have to satisfy this universal property.

Theorem 6: Let $G$ and $H$ be groups. Then if $K$ is a group with homomorphisms $\phi_1\,:\, K\rightarrow G$ and $\phi_2\,:\, K\rightarrow H$, then there exists a unique homomorphism $u\,:\,K\rightarrow G\times H$ such that $\phi_1=\pi_1\circ u$ and $\phi_2=\pi_2\circ u$.

Proof: By the construction of the direct product, $u\,:\, K\rightarrow G\times H$ is a homomorphism if and only if $\pi_1\circ u$ and $\pi_2\circ u$ are homomorphisms. Thus $u\,:\, K\rightarrow G\times H$ defined by $u((g,h))=(\phi_1(g),\phi_2(h))$ is one homomorphism satisfying the theorem, proving existence. By the commutativity condition this is the only such homomorphism, proving uniqueness. ∎

Products of Cyclic Groups

Theorem 7: The order of an element $(a,b)\in\mathbb{Z}_m\times\mathbb{Z}_n$ is $|(a,b)|=\mathrm{lcm}(|a|,|b|)$.

Proof: The lowest positive number $c$ such that $(a,b)^c=(ac,bc)=(0,0)$ is the smallest number such that $ac=rm$ and $bc=sn$ for integers $r,s$. It follows that $c$ divides both $|a|$ and $|b|$ and is the smallest such number. This is the definition of the least common divider. ∎

Theorem 8: $\mathbb{Z}_m\times \mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_{mn}$ if and only if $m$ and $n$ are relatively prime.

Proof: We begin with the left implication. Assume $\mathbb{Z}_m\times\mathbb{Z}_n\approx \mathbb{Z}_{mn}$. Then $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic, and so there must exist an element with order $mn$. By Theorem 7 we there must then exist a generator $(a,b)\neq (0,0)$ in $\mathbb{Z}_m\times\mathbb{Z}_n$ such that $\mathrm{lcm}(|a|,|b|)=mn$. Since each factor of the generator must generate its group, this implies $\mathrm{lcm}(m,n)=mn$, and so $\gcd(m,n)=1$, meaning that $m$ and $n$ are relatively prime. Now assume that $m$ and $n$ are relatively prime and that we have generators $a$ of $\mathbb{Z}_m$ and $b$ of $\mathbb{Z}_n$. Then since $\gcd(m,n)=1$, we have $\mathrm{lcm}(m,n)=mn$ and so $|(a,b)|=mn$. this implies that $(a,b)$ generates $\mathbb{Z}_m\times\mathbb{Z}_n$, which must then be isomorphic to a cyclic group of order $mn$, im particular $\mathbb{Z}_{mn}$. ∎

Theorem 9 (Characterization of finite abelian groups): Let $G$ be an abelian group. Then there exists prime numbers $p_1,...,p_n$ and positive integers $r_1,...,r_n$, unique up to order, such that

$G\approx \mathbb{Z}_{p_1^{r_1}}\times ... \times \mathbb{Z}_{p_n^{r_n}}$

Proof: A proof of this theorem is currenly beyond our reach. However, we will address it during the chapter on modules. ∎

Subdirect Products and Fibered Products

Definition 10: A subdirect product of two groups $G$ and $H$ is a proper subgroup $K$ of $G\times H$ such that the projection homomorphisms are surjective. That is, $\pi_1(K)=G$ and $\pi_2(K)=H$.

Example 11: Let $G$ be a group. Then the diagonal $\Delta=\{(g,g)\mid g\in G\}\subseteq G\times G$ is a subdirect product of $G$ with itself.

Definition 12: Let $G$, $H$ and $Q$ be groups, and let the homomorphisms $\phi\,:\, G\rightarrow Q$ and $\psi\,:\, H\rightarrow Q$ be epimorphisms. The fiber product of $G$ and $H$ over $Q$, denoted $G\times_Q H$, is the subgroup of $G\times H$ given by $G\times_Q H=\{(g,h)\in G\times H\mid \phi(g)=\psi(h)\}$.

In this subsection, we will prove the equivalence between subdirect products and fiber products. Specifically, every subdirect product is a fiber product and vice versa. For this we need Goursat's lemma.

Theorem 13 (Goursat's lemma): Let $G$ and $G^\prime$ be groups, and $H\subseteq G\times G^\prime$ a subdirect product of $G$ and $G^\prime$. Now let $N=\ker \,\pi_2$ and $N^\prime=\ker\,\pi_1$. Then $N$ can be identified with a normal subgroup of $G$, and $N^\prime$ with a normal subgroup of $G^\prime$, and the image of $H$ when projecting on $G/N\times G^\prime /N^\prime$ is the graph of an isomorphism $G/N\approx G^\prime/N^\prime$.

Proof:

Further Reading

More on the automorphism groups of finite abelian groups. Some results require theory of group actions and ring theory, which is developed in a later section.

http://arxiv.org/pdf/math/0605185v1.pdf

Free Groups

In order to properly define the free group, and thereafter the free product, we need some preliminary definitions.

Definition 10: Let $A$ be a set. Then a word of elements in $A$ is a finite sequence $a_1a_2...a_n$ of elements of $A$, where the positive integer $n$ is the word length.

Definition 11: Let $x=a_1...a_n$ and $y=a_{n+1}...a_{n+k}$ be two words of elements in $A$. Define the concatenation of the two words as the word $xy=a_1...a_na_{n+1}...a_{n+k}$.

Now, we want to make a group consisting of the words of a given set $A$, and we want this group to be the most general group of this kind. However, if we are to use the concatenation operation, which is the only obvious operation on two words, we are immediately faced with a problem. Namely, deciding when two words are equal. According to the above, the length of a product is the sum of the lengths of the factors. In other words, the length cannot decrease. Thus, a word of length $n$ multiplied with its inverse has length at least $n$, while the identity word, which is the empty word, has length $0$. The solution is an algorithm to reduce words into irreducible ones. These terms are defined below.

Definition 12: Let $A$ be any set. Define the set $W(A)$ as the set of words of powers of elements of $A$. That is, if $a_1,...,a_n\in A$ and $r_1,...,r_n\in\mathbb{Z}$, then $a_1^{r_1}...a_n^{r_n}\in F(A)$.

Definition 13: Let $x=a_1^{r_1}...a_n^{r_n}\in W(A)$. Then we define a reduction of $x$ as follows. Scan the word from the left until the first pair of indices $j,j+1$ such that $a_j=a_{j+1}$ is encountered, if such a pair exists. Then replace $a_j^{r_j}a_{j+1}^{r_{j+1}}$ with $a_j^{r_j+r_{j+1}}$. Thus, the resulting word is $x_{(1)}=a_1^{r_1}...a_{j-1}^{r_{j-1}}a_j^{r_j+r_{j+1}}a_{j+2}^{r_{j+2}}...a_n^{r_n}$. If no such pair exists, then $x=x_{(1)}$ and the word is called irreducible.

It should be obvious if $x\in W(A)$ with length $n$, then $x_{(n)}$ will be irreducible. The details of the proof is left to the reader.

Definition 14: Define the free group $F(A)$ on a set $A$ as follows. For each word $x\in W(A)$ of length $n$, let the reduced word $x_{(n)}\in F(A)$. Thus $F(A)\subseteq W(A)$ is the subset of irreducible words. As for the binary operation on $F(A)$, if $x,y\in F(A)$ have lengths $n$ and $m$ respectively, define $x*y$ as the completely reduced concatenation $(xy)_{(n+m)}$.

Theorem 15: $F(A)$ is a group.

Proof:

Example 16: We will concider free groups on 1 and 2 letters. Let $A_1=\{a\}$ and $A_2=\{a,b\}$. Then

$F(A_1)=\{a^n\mid n\in \mathbb{Z}\}$ with $a^na^m=a^{n+m}$.

$F(A_2)=\{\prod_{i=1}^n a_{i}b_{i}\mid a_{i}\in F(\{a\}),b_{i}\in F(\{b\})\}$ such that $a_i\neq e$ for any $i>1$ and $b_i\neq e$ for any $i<n$. Example product: $(a^2b^{-3}a)(a^{-1}ba)=a^2b^{-3}aa^{-1}ba=a^2b^{-3}ba=a^2b^{-2}a$.

Group Presentations

In this subsection we will breifly introduce another method used for defining groups. This is by prescribing a group presentation.

Definition 17: Let $G$ be a group and $H$ a subgroup. Then define the normal closure of $H$ in $G$ as the intersection of all normal subgroups in $G$ containing H. That is, if $N$ is the normal closure of $H$, then

$N=\bigcap_{\stackrel{K\trianglelefteq G}{H\subseteq K}}K$.

Definition 18: Let $S$ be a set and $R\subseteq F(S)$. Let $N$ be the normal closure of $R$ in $F(S)$ and define the group $\langle S\mid R\rangle=F(S)/N$. The elements of $S$ are called generators and the elements of $R$ are called relators. If $G$ is a group such that $G\approx \langle S\mid R\rangle$, then $\langle S\mid R \rangle$ is said to be a presentation of $G$.

The Free Product

Using the previously defined notion of a group presentation, we can now define another type of group product.

Defintion : Let $G$ and $G^\prime$ be groups with presentations $\langle S\mid R\rangle$ and $\langle S^\prime \mid R^\prime\rangle$. Define the free product of $G$ and $G^\prime$, denoted $G*G^\prime$, as the group with the presentation $\langle S\cup S^\prime \mid R\cup R^\prime\rangle$.

Remark : Depending on the context, spesifically if we only deal with abelian groups, we may require the free product of abelian groups to be abelian. In that case, the free product equals the direct product. This is another example of abelian groups being better behaved than nonabelian groups.

Lemma : The free product includes the component groups as subgroups.

Remark : The free product is not a product in the sense discussed previously. It does not satifsy the universal property other products do. Instead, it satisfies the "opposide", or dual property, obtained by reversing the direction of all the arrows in the commutative diagram. We usually call a construction satisfying this universal property a coproduct.

Problems

Problem 1: Let $H$ and $K$ be groups of relatively prime orders. Show that any subgroup of $H\times K$ is the product of a subgroup of $H$ with a subgroup of $K$.

Answer

Coming soon. ∎