Abelian Groups and Fields

The spaces dealt with in linear algebra are vector spaces, a concept generalizing the coordinate spaces we learned in high school. A vector space is, literally, a space of vectors, and addition between these vectors is well-defined. We can also multiply a vector by a constant to obtain another vector; when working with vector spaces, these constants are called scalars. The set of scalars must form a field, which we will define shortly.[^1]

Fields

Definition 1 A set \(G\), together with a binary operation \(+:G\times G\rightarrow G\) defined on \(G\), is called an abelian group if the following conditions are satisfied.

1. \(+\) satisfies the associative law. That is, for all \(a,b,c\in G\), we have \((a+b)+c=a+(b+c)\).
2. There exists an identity element for \(+\). That is, there exists some \(0\in G\) such that \(a+0=0+a=a\) holds for all \(a\in G\).
3. An inverse for \(+\) always exists. That is, whenever \(a\in G\) is given, there exists \(-a\in G\) such that \(a+(-a)=(-a)+a=0\).
4. \(+\) satisfies the commutative law. That is, for all \(a,b\in G\), we have \(a+b=b+a\).

If the last condition is not required, we call \(G\) a group.

In any case, we can use the definition to prove a simple proposition.

Proposition 2 Let \(G\) be an abelian group. Then

1. the identity element of \(G\) is unique.
2. for every \(a\in G\), the inverse \(-a\) of \(a\) is also unique.
3. if \(a+b=a+c\) holds, then \(b=c\).

Proof

1. Let \(0'\) be another element satisfying the second condition of Definition 1. Applying the second condition with \(a=0\) and the identity \(0'\), we get

   \[0+0'=0'+0=0\]

   On the other hand, applying the second condition with \(a=0'\) and the identity \(0\), we likewise obtain

   \[0+0'=0'+0=0'\]

   Therefore \(0=0'\), so the identity element is unique.

2. Proceed as in the first part. Let \((-a)'\) be another element satisfying the third condition of Definition 1. Then

   \[(-a)=(-a)+0=(-a)+(a+(-a)')=((-a)+a)+(-a)'=0+(-a)'=(-a)'\]

   so \((-a)=(-a)'\) holds.

3. Add \((-a)\) to both sides.

In an abelian group, thanks to the fourth condition \(a+b=b+a\), it suffices to state the identity and inverse conditions as the single equations \(a+0=a\) and \(a+(-a)=0\), respectively. However, we did not use this reasoning in the proof above, so the uniqueness of the identity and inverses holds for general groups as well.

The following corollary follows immediately from the uniqueness of the identity and inverses; its first statement also holds for general groups.

Corollary 3 For an abelian group \((G,+)\) and arbitrary elements \(a,b\), the following hold.

1. \(-(-a)=a\),
2. \(-(a+b)=-a+(-b)=-a-b\).

Proof

1. We must show that the inverse \(-(-a)\) of \(-a\) equals \(a\). Since inverses are unique, if the equation

   \[(-a)+x=x+(-a)=0\]

   holds for some \(x\in G\), then \(x\) must be \(-(-a)\). But when \(x=a\), the above equation holds because \(-a\) is the inverse of \(a\). Hence \(a=-(-a)\).

2. As in the previous proof, it suffices to show that both \(x=-a+(-b)\) and \(x=-a-b\) satisfy

   \[(a+b)+x=x+(a+b)=0\]

   For example, when \(x=-a+(-b)\),

   \[\begin{aligned}(a+b)+x&=(a+b)+((-a)+(-b))=(b+a)+((-a)+(-b))=b+(a+(-a))+(-b)=b+(-b)\\&=0\end{aligned}\]

   and similarly (or by the commutative law) one can also show that \(x+(a+b)=0\).

In Definition 1 we assumed that the operation of \(G\) is addition, but in fact the operation need not be addition. Even if the operation of \(G\) is written as multiplication, \(G\) is still called an abelian group as long as it satisfies all four conditions 1 through 4. Of course, in that case it is natural to write the identity as \(1\) instead of \(0\), and the inverse as \(a^{-1}\) instead of \(-a\).

Example 4 The set of real numbers \(\mathbb{R}\) is an abelian group under addition, and the same is true of the set of complex numbers \(\mathbb{C}\).

Multiplication is also defined on \(\mathbb{R}\) and \(\mathbb{C}\), but they are not abelian groups under multiplication. This is because for \(0\in\mathbb{R}\) (or \(\mathbb{C}\)), there is no real (or complex) number \(a\) satisfying \(0a=a0=1\). Instead, \(\mathbb{R}^\times=\mathbb{R}\setminus\{0\}\) (or \(\mathbb{C}^\times=\mathbb{C}\setminus\{0\}\)) is an abelian group under multiplication.

We define the situation of the example above as follows.

Definition 5 Suppose two binary operations \(+\) and \(\times\) are defined on a set \(\mathbb{K}\). We say that \(\mathbb{K}\) is a field if the following three conditions are satisfied:

1. \(\mathbb{K}\) is an abelian group under \(+\).
2. \(\mathbb{K}^\times=\mathbb{K}\setminus\{0\}\) is an abelian group under \(\times\).

3. \(\times\) satisfies the distributive law over \(+\). That is, for all \(a,b,c\in\mathbb{K}\),

   \[a\times (b+c)=(a\times b)+(a\times c),\quad (b+c)\times a=(b\times a)+(c\times a)\]

   holds.

In the definition above, 0 and 1 must be distinct elements, since \(1\in\mathbb{K}^\times=\mathbb{K}\setminus\{0\}\).

Since high school, we have preferred to abbreviate the operation \(\times\) as \(a\cdot b\) rather than writing it out explicitly, and in fact we prefer to omit the operation altogether and simply write \(ab\). From now on we follow this convention and write \(ab\) instead of \(a\times b\).

Meanwhile, applying Proposition 2 to \(\mathbb{K}\) and \(\mathbb{K}^\times\) respectively, we can also easily check that the following hold:

1. \(0\) and \(1\) are unique.
2. For every \(a\in\mathbb{K}\), \(-a\) is unique, and if moreover \(a\neq 0\), then \(a^{-1}\) also exists uniquely.
3. If \(a+b=a+c\), then \(b=c\). Also, if \(a,b,c\) are all nonzero and \(ab=ac\), then \(b=c\).

Proposition 6 For every element \(a\in\mathbb{K}\) of a field \(\mathbb{K}\), we have \(0a=0\) and \((-1)a=-a\).

Proof

First we should carefully unpack what the proposition means: saying \(0a=0\) means that multiplying \(0\) and \(a\) yields the additive identity \(0\), and saying \((-1)a=-a\) means that multiplying \((-1)\) and \(a\) yields the inverse of \(a\). For this, just as we proved several properties right after Proposition 2, it seems we should use the uniqueness of inverses and the identity. To prove the first equation, we would need to show that \(0a+b=b+0a=b\) holds for arbitrary \(b\), but no way of simplifying the expression \(0a+b\) comes to mind. We need to find some other approach. Let us use the third statement of Proposition 2. If we can just show that \(0a+0a=0a\), then since \(0a=0a+0\), we get \(0a=0\) from \(0a+0a=0a+0\). So it suffices to show that \(0a+0a=0a\), and this is clear from

\[0a+0a=(0+0)a=0a\]

Once this is proved, the second part is even easier. We need only show that \((-1)a\) satisfies the condition for being the inverse of \(a\); since

\[(-1)a+a=(-1)a+1a=((-1)+1)a=0a=0\]

we have \((-1)a=-a\).

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References

[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.

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