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Finite Abelian Groups

In our investigation of cyclic groups we found that every group of prime order was isomorphic to {\mathbb Z}_p, where p was a prime number.

Finite Abelian Groups

In our investigation of cyclic groups we found that every group of prime order was isomorphic to \({\mathbb Z}_p\), where \(p\) was a prime number. We also determined that \({\mathbb Z}_{mn} \cong {\mathbb Z}_m \times {\mathbb Z}_n\) when \(\gcd(m, n) =1\). In fact, much more is true. Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order; that is, every finite abelian group is isomorphic to a group of the type \[\begin{aligned}\end{aligned}\], where each \(p_k\) is prime (not necessarily distinct).

First, let us examine a slight generalization of finite abelian groups. Suppose that \(G\) is a group and let \(\{ g_i\}\) be a set of elements in \(G\), where \(i\) is in some index set \(I\) (not necessarily finite). The smallest subgroup of \(G\) containing all of the \(g_i\)'s is the subgroup of \(G\) generated by the \(g_i\)'s. If this subgroup of \(G\) is in fact all of \(G\), then \(G\) is generated by the set \(\{g_i : i \in I \}\). In this case the \(g_i\)'s are said to be the generators of \(G\). If there is a finite set \(\{ g_i : i \in I \}\) that generates \(G\), then \(G\) is finitely generated.

Example

Obviously, all finite groups are finitely generated. For example, the group \(S_3\) is generated by the permutations \((1 \, 2)\) and \((1 \, 2 \,3)\). The group \({\mathbb Z} \times {\mathbb Z}_n\) is an infinite group but is finitely generated by \(\{ (1,0), (0,1) \}\).

Example

Not all groups are finitely generated. Consider the rational numbers \({\mathbb Q}\) under the operation of addition. Suppose that \({\mathbb Q}\) is finitely generated with generators \(p_1/q_1, \ldots, p_n/q_n\), where each \(p_i/q_i\) is a fraction expressed in its lowest terms. Let \(p\) be some prime that does not divide any of the denominators \(q_1, \ldots, q_n\). We claim that \(1/p\) cannot be in the subgroup of \({\mathbb Q}\) that is generated by \(p_1/q_1, \ldots, p_n/q_n\), since \(p\) does not divide the denominator of any element in this subgroup. This fact is easy to see since the sum of any two generators is \[\begin{aligned}\end{aligned}\].

The reason that powers of a fixed \(g_i\) may occur several times in the product is that we may have a nonabelian group. However, if the group is abelian, then the \(g_i\)s need occur only once. For example, a product such as \(a^{-3} b^5 a^7\) in an abelian group could always be simplified (in this case, to \(a^4 b^5\)).

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

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Imibuzo abantu bebuza

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

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IiNkqubo Abstract Algebra