## Abelian groups

Here is the 1st in the series of posts for the Algbera I course I teach now.

An abelian group is a pair $(A,+)$, where $A$ is a set and a binary operation $+$ on $A$ that is

• associative, i.e. $x+(y+z)=(x+y)+z$ for all $x,y,z\in A$,
and thus we can drop brackets $x+(y+z)=x+y+z$ without introducing any ambiguity;
• commitative, i.e. $y+x=x+y$ for all $x,y\in A$,
• there exists $0=0_A\in A$, a special element satisfying $x+0=x$ for all $x\in A.$ It must not be confused with the integer number 0, as in general it has nothing to do with $0_A,$ although it shares similar properties (integers w.r.t. addition are an example of an abelian group, and there these two things are indeed the same). Elements like $0_A$ are called neutral w.r.t. the operation in question.
• for any $x\in A$ there exists $x'\in A$ such that $x+x'=0$. We denote $x'$ by $-x$. We usually abbreviate $x+(-y)$ as $x-y$, etc.

Note that these axioms are essentially the axioms of a vector space that “forgot everything about the field”. The following is also very helpful, although might look evident at first sight.
(But it is not! Later, in ring theory, when we see zero-divisors, we will see that the intuition of “cancelling” may fail you!)

The equation $a+x=b$ for any given $a,b\in A$ has unique solution $x=b-a.$
Thus we can cancel: $x+y=x+z$ iff $y=z$, for all $x,y,z\in A.$

First examples of abelian groups

1. Integers w.r.t. addition as the operation $\mathbb{Z}=(\mathbb{Z},+)$
2. Vector spaces w.r.t. addition of vectors as the operation (such a group is called the additive group of the vector space)
3. additive groups of fields, e.g. the real numbers w.r.t. addition
4. Integers modulo $m$ w.r.t. addition $\mod m$, i.e.
$\mathbb{Z}_m=(\mathbb{Z}_m,+)=\{0,\dots,m-1\}$

The size of a group A is usually called order, and denote |A|. So $|\mathbb{Z}|=\infty,$ while $|\mathbb{Z}_m|=m.$

New abelian groups from old: direct products,, a.k.a. direct sums

Given two abelian groups $(A,+_A)$ and $(B,+_B),$ define the group
$A\oplus B=(A\times B,+)$, the direct sum of $A$ and $B,$ where the operation $+$ is component-wise:
$(a,b)+(a',b')=(a+_A a',b+_B b').$

Similarly, one can define direct sums of an arbitarty number of abelian groups. E.g. the additive groups of n-dimensional vector spaces over a field F can be viewed as direct sums of n copies of the additive group of F.

The next important concept is of homomorphism and isomorphism of abelian groups.

Given two groups $(A,+_A)$ and $(B,+_B)$, a function $\phi:B\to A$ satisfying $\phi(b+_B b')=\phi(b)+_A\phi(b')$ is called a homomorphism from $B$ to $A.$

When $\phi:B\to A$ is a bijection, then it is called isomorphism, and one writes $(A,+_A)\cong (B,+_B)$, or just $A\cong B.$

From the point of view of abstract algebra, two isomorphic groups are hardly distinguishable, although of course they can appear in different disguises.
An example of nontrivial isomorphism is $\mathbb{Z}_6\cong \mathbb{Z}_2\oplus\mathbb{Z}_3.$
(To see the isomorphism, set $\phi(1)=(1,1)$ and try extending it by additivity, i.e. compute $\phi(1+1),$ etc. Eventually you arrive at an isomorphism $\phi:\mathbb{Z}_6\to\mathbb{Z}_2\oplus\mathbb{Z}_3.$
Interestingly, $\mathbb{Z}_4\not\cong \mathbb{Z}_2\oplus\mathbb{Z}_2.$ Can you prove this?)

Subgroups

A subgroup $B$ of $A$ is a subset of $A=(A,+)$ that is closed under the addition in $A.$ (Notation: $B\le A$ or $B\leq A$, the latter does not exlcude $B=A.$)
This means that for any $x,y\in B$ also $-x,x+y\in B$, and immediately implies $0\in B.$

Examples of subgroups:

1. additive group $(U,+)$ of a subspace $U$ of a vectorspace $V$ is a subgroup of $(V,+);$ in shorthand, $(U,+)\leq (V,+);$
2. for any two positive integers $m,n,$ one has $\mathbb{Z}_m\leq\mathbb{Z}_{mn};$ (this is indeed not immediate, but note that $n\in\mathbb{Z}_{nm}$ generates a subgroup $n\mathbb{Z}_{nm}\cong \mathbb{Z}_m.$)
3. $m\mathbb{Z}\leq \mathbb{Z},$ e.g. for $m=2$ we have that the set of all even integers form a subgroup in the integers.
4. for a prime number $p$, define $\mathbb{Z}[\frac{1}{p}]=\{\frac{m}{p^k}\mid m,k\in\mathbb{Z}, k\geq 0\}.$

Generating sets, etc.
First, a bit of terminology: any abelian group $A$ is a $\mathbb{Z}$-module, i.e. in a sense $A$ relates to $\mathbb{Z}$ in a way similar to the way a vectorspace over a field F relates to F. More precisely, this means that for any $a\in A, n\in\mathbb{Z}$, the product $na$ is well-defined, as follows:

1. if $n=0$ then $na=0_A,$
2. if $n<0$ then $na=(-n)(-a),$
3. if $n>0$ then $na={a+\dots+a},$ the sum of $n$ copies of $a.$

Note that, in contrast to vectorspace/field relationship, it can perfectly happen that $na=0_A$ while neither $n=0$ nor $a=0_A$, e.g. $na=0_A$ for any $a\in A=\mathbb{Z}_n.$

Let $S\subseteq A.$ Define $\langle S\rangle,$ the subgroup generated by $S,$ to be $\langle S\rangle=\cap_{S\subseteq B\leq A} B,$ i.e. the intersection of all the subgroups of $A$ containing $S.$

One can prove that $\langle S\rangle=\{\sum_{s\in S} k_s s\mid k_s\in\mathbb{Z}, \text{and only finitely many } k_s\neq 0\}.$ (Note that in algebra infinite sums are, normally speaking, not allowed, as we don’t have a notion of convergence.)

Cyclic subgroups. These are subgroups generated by just one element, $s\in A,$ i.e. one should take $S=\{s\}$ and denote $\langle S\rangle=\langle s\rangle.$
That is as simple as it gets as far as subgroups are concerned. We define the order of $s$ to the be $|\langle s\rangle|,$ the order of the cyclic subgroup it generates.
Similarly, groups generated by one element are called cyclic.
One can show the following classification of cyclic groups.

Either $\langle s\rangle\cong\mathbb{Z}$ (and so $|s|=\infty$), or $\langle s\rangle\cong\mathbb{Z}_m,$ for some $m>0$ (and so $|s|=m$).

We postpond a proof until we learned more about homomorphisms.

Every subgroup $H$ of a cyclic group $A=\langle s\rangle$ is cyclic.

Let $H\leq A,$ and $m$ the minimal positive integer such that $ms\in H.$ Suppose that $ks\in H.$ Without loss of generality, $k\ge 0,$ as $H$ is a subgroup and so $-ks\in H.$ By the integer division with a remainder algorithm, one has $k=nm+\ell,$ for $0\leq \ell\le m.$ On the other hand $\ell s\in H$, as $\ell s=(k-nm)s=ks-nms\in H.$ By the choice of $m,$ we have $\ell =0,$ and so $H=\langle ms\rangle.$

Generally speaking, complete classification of abelian groups is out of question, but groups that are somehow similar to finite-dimensional vector spaces, finitely generated abelian groups, can be completely classified.

An abelian group $A$ is called finitely generated if there exists $S\subset A$ such that $\langle S\rangle=A$ and $|S|<\infty.$

You might like think which of the following groups are finitely generated:

1. the direct sum of a finite number of cyclic groups
2. $(\mathbb{Q},+)$
3. $\mathbb{Z}[\frac{1}{p}],$ for a prime $p$

Finally, note that there are abelian groups that do not admit a countable generating set, leave alone finite. (Hint: to see this, prove first that a group with a countable generating set is countable.)

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### 3 Responses to “Abelian groups”

1. Properties of homomorphisms of abelian groups « Equatorial Mathematics Says:

[…] of homomorphisms of abelian groups Let be a homomorphism of abelian groups and (we denoted operations in both groups by the same symbol – these are […]

2. paradox (shengtao) Says:

is Zm a subgroup of Zmn? aren’t they have different addition operator?

3. Dima Says:

paradox’s comment is valid, to an extent; $n\in\mathbb{Z}_{nm}$ generates a subgroup $n\mathbb{Z}_{nm}\cong \mathbb{Z}_m.$