In our investigation of cyclic groups we found that every group of prime order was isomorphic to \({\mathbb Z}_p\text{,}\) where \(p\) was a prime number. We also determined that \({\mathbb Z}_{mn} \cong {\mathbb Z}_m \times {\mathbb Z}_n\) when \(\gcd(m, n) =1\text{.}\) 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{equation*}
{\mathbb Z}_{p_1^{\alpha_1}} \times \cdots \times {\mathbb Z}_{p_n^{\alpha_n}}\text{,}
\end{equation*}
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
\(G\text{,}\) 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
\(G\text{,}\) then
\(G\) is generated by the set
\(\{g_i : i \in I \}\text{.}\) In this case the
\(g_i\)’s are said to be the
generators of
\(G\text{.}\) If there is a finite set
\(\{ g_i : i \in I \}\) that generates
\(G\text{,}\) then
\(G\) is
finitely generated.
Example 13.1.2.
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 \(p_1/q_1, \ldots,
p_n/q_n\text{,}\) 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 \(q_1, \ldots, q_n\text{.}\) We claim that \(1/p\) cannot be in the subgroup of \({\mathbb Q}\) that is generated by \(p_1/q_1, \ldots,
p_n/q_n\text{,}\) 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{equation*}
p_i / q_i + p_j / q_j = (p_i q_j + p_j q_i)/(q_i q_j)\text{.}
\end{equation*}
Proof.
Let
\(K\) be the set of all products of the form
\(g_{i_1}^{\alpha_1} \cdots g_{i_n}^{\alpha_n}\text{,}\) where the
\(g_{i_k}\)s are not necessarily distinct. Certainly
\(K\) is a subset of
\(H\text{.}\) We need only show that
\(K\) is a subgroup of
\(G\text{.}\) If this is the case, then
\(K=H\text{,}\) since
\(H\) is the smallest subgroup containing all the
\(g_i\)s.
Clearly, the set \(K\) is closed under the group operation. Since \(g_i^0 = 1\text{,}\) the identity is in \(K\text{.}\) It remains to show that the inverse of an element \(g =g_{i_1}^{k_1} \cdots g_{i_n}^{k_n}\) in \(K\) must also be in \(K\text{.}\) However,
\begin{equation*}
g^{-1} = (g_{i_1}^{k_{1}} \cdots g_{i_n}^{k_n})^{-1} = (g_{i_n}^{-k_n} \cdots g_{i_{1}}^{-k_{1}})\text{.}
\end{equation*}
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
\(a^4 b^5\)).
Now let us restrict our attention to finite abelian groups. We can express any finite abelian group as a finite direct product of cyclic groups. More specifically, letting
\(p\) be prime, we define a group
\(G\) to be a
\(p\)-group if every element in
\(G\) has as its order a power of
\(p\text{.}\) For example, both
\({\mathbb Z}_2 \times {\mathbb Z}_2\) and
\({\mathbb Z}_4\) are
\(2\)-groups, whereas
\({\mathbb Z}_{27}\) is a
\(3\)-group. We shall prove the Fundamental Theorem of Finite Abelian Groups which tells us that every finite abelian group is isomorphic to a direct product of cyclic
\(p\)-groups.
Proof.
We will prove this lemma by induction. If
\(n = 1\text{,}\) then there is nothing to show. Now suppose that the lemma is true for all groups of order
\(k\text{,}\) where
\(k \lt n\text{.}\) Furthermore, let
\(p\) be a prime that divides
\(n\text{.}\)
If
\(G\) has no proper nontrivial subgroups, then
\(G = \langle a \rangle\text{,}\) where
\(a\) is any element other than the identity. By
Exercise 4.5.39, the order of
\(G\) must be prime. Since
\(p\) divides
\(n\text{,}\) we know that
\(p = n\text{,}\) and
\(G\) contains
\(p - 1\) elements of order
\(p\text{.}\)
Now suppose that \(G\) contains a nontrivial proper subgroup \(H\text{.}\) Then \(1 \lt |H| \lt n\text{.}\) If \(p \mid |H|\text{,}\) then \(H\) contains an element of order \(p\) by induction and the lemma is true. Suppose that \(p\) does not divide the order of \(H\text{.}\) Since \(G\) is abelian, it must be the case that \(H\) is a normal subgroup of \(G\text{,}\) and \(|G| = |H| \cdot |G/H|\text{.}\) Consequently, \(p\) must divide \(|G/H|\text{.}\) Since \(|G/H| \lt |G| = n\text{,}\) we know that \(G/H\) contains an element \(aH\) of order \(p\) by the induction hypothesis. Thus,
\begin{equation*}
H = (aH)^p = a^pH\text{,}
\end{equation*}
and \(a^p \in H\) but \(a \notin H\text{.}\) If \(|H| = r\text{,}\) then \(p\) and \(r\) are relatively prime, and there exist integers \(s\) and \(t\) such that \(sp + tr = 1\text{.}\) Furthermore, the order of \(a^p\) must divide \(r\text{,}\) and \((a^p)^r = (a^r)^p = 1\text{.}\)
We claim that \(a^r\) has order \(p\text{.}\) We must show that \(a^r \neq 1\text{.}\) Suppose \(a^r = 1\text{.}\) Then
\begin{align*}
a & = a^{sp + tr}\\
& = a^{sp} a^{tr}\\
& = (a^p)^s (a^r)^t\\
& = (a^p)^s 1\\
& = (a^p)^s\text{.}
\end{align*}
Since \(a^p \in H\text{,}\) it must be the case that \(a= (a^p)^s \in H\text{,}\) which is a contradiction. Therefore, \(a^r \neq 1\) is an element of order \(p\) in \(G\text{.}\)
Lemma 13.1.6 is a special case of Cauchy’s Theorem (
Theorem 15.1.1), which states that if
\(G\) is a finite group and
\(p\) a prime such that
\(p\) divides the order of
\(G\text{,}\) then
\(G\) contains a subgroup of order
\(p\text{.}\) We will prove Cauchy’s Theorem in
Chapter 15.
Proof.
Since \(G\) is an abelian group, we are guaranteed that \(G_i\) is a subgroup of \(G\) for \(i = 1, \ldots, k\text{.}\) Since the identity has order \(p_i^0 = 1\text{,}\) we know that \(1 \in G_i\text{.}\) If \(g \in G_i\) has order \(p_i^r\text{,}\) then \(g^{-1}\) must also have order \(p_i^r\text{.}\) Finally, if \(h \in G_i\) has order \(p_i^s\text{,}\) then
\begin{equation*}
(gh)^{p_i^t} = g^{p_i^t} h^{p_i^t} = 1 \cdot 1 = 1\text{,}
\end{equation*}
where \(t\) is the maximum of \(r\) and \(s\text{.}\)
We must show that
\begin{equation*}
G = G_1 G_2 \cdots G_k
\end{equation*}
and \(G_i \cap G_j = \{1 \}\) for \(i \neq j\text{.}\) Suppose that \(g_1 \in G_1\) is in the subgroup generated by \(G_2, G_3, \ldots, G_k\text{.}\) Then \(g_1 = g_2 g_3 \cdots g_k\) for \(g_i \in G_i\text{.}\) Since \(g_i\) has order \(p_i^{\alpha_i}\text{,}\) we know that \(g_i^{p^{\alpha_i}} = 1\) for \(i = 2, 3, \ldots, k\text{,}\) and \(g_1^{p_2^{\alpha_2} \cdots p_k^{\alpha_k}} = 1\text{.}\) Since the order of \(g_1\) is a power of \(p_1\) and \(\gcd(p_1, p_2^{\alpha_2} \cdots p_k^{\alpha_k}) = 1\text{,}\) it must be the case that \(g_1 = 1\) and the intersection of \(G_1\) with any of the subgroups \(G_2, G_3, \ldots, G_k\) is the identity. A similar argument shows that \(G_i \cap G_j = \{1 \}\) for \(i \neq j\text{.}\)
Next, we must show that it possible to write every \(g \in G\) as a product \(g_1 \cdots g_k\text{,}\) where \(g_i \in G_i\text{.}\) Since the order of \(g\) divides the order of \(G\text{,}\) we know that
\begin{equation*}
|g| = p_1^{\beta_1} p_2^{\beta_2} \cdots p_k^{\beta_k}
\end{equation*}
for some integers \(\beta_1, \ldots, \beta_k\text{.}\) Letting \(a_i = |g| / p_i^{\beta_i}\text{,}\) the \(a_i\)’s are relatively prime; hence, there exist integers \(b_1, \ldots,
b_k\) such that \(a_1 b_1 + \cdots + a_k b_k = 1\text{.}\) Consequently,
\begin{equation*}
g = g^{a_1 b_1 + \cdots + a_k b_k} = g^{a_1 b_1} \cdots g^{a_k b_k}\text{.}
\end{equation*}
Since
\begin{equation*}
g^{(a_i b_i ) p_i^{\beta_i}} = g^{b_i |g|} = e\text{,}
\end{equation*}
it follows that \(g^{a_i b_i}\) must be in \(G_{i}\text{.}\) Let \(g_i = g^{a_i b_i}\text{.}\) Then \(g = g_1 \cdots g_k \in G_1 G_2 \cdots G_k\text{.}\) Therefore, \(G = G_1 G_2 \cdots G_k\) is an internal direct product of subgroups.
Proof.
By
Lemma 13.1.7, we may assume that the order of
\(G\) is
\(p^n\text{.}\) We shall induct on
\(n\text{.}\) If
\(n= 1\text{,}\) then
\(G\) is cyclic of order
\(p\) and must be generated by
\(g\text{.}\) Suppose now that the statement of the lemma holds for all integers
\(k\) with
\(1 \leq k \lt n\) and let
\(g\) be of maximal order in
\(G\text{,}\) say
\(|g| = p^{m}\text{.}\) Then
\(a^{p^m} = e\) for all
\(a \in G\text{.}\) Now choose
\(h\) in
\(G\) such that
\(h \notin \langle g \rangle\text{,}\) where
\(h\) has the smallest possible order. Certainly such an
\(h\) exists; otherwise,
\(G = \langle g \rangle\) and we are done. Let
\(H = \langle h \rangle\text{.}\)
We claim that \(\langle g \rangle \cap H = \{ e \}\text{.}\) It suffices to show that \(|H|=p\text{.}\) Since \(|h^p| = |h| / p\text{,}\) the order of \(h^p\) is smaller than the order of \(h\) and must be in \(\langle g \rangle\) by the minimality of \(h\text{;}\) that is, \(h^p = g^r\) for some number \(r\text{.}\) Hence,
\begin{equation*}
(g^r)^{p^{m - 1}} = (h^p)^{p^{m - 1}} = h^{p^{m}} = e\text{,}
\end{equation*}
and the order of \(g^r\) must be less than or equal to \(p^{m-1}\text{.}\) Therefore, \(g^r\) cannot generate \(\langle g \rangle\text{.}\) Notice that \(p\) must occur as a factor of \(r\text{,}\) say \(r = ps\text{,}\) and \(h^p = g^r = g^{ps}\text{.}\) Define \(a\) to be \(g^{-s}h\text{.}\) Then \(a\) cannot be in \(\langle g \rangle\text{;}\) otherwise, \(h\) would also have to be in \(\langle g \rangle\text{.}\) Also,
\begin{equation*}
a^p = g^{-sp} h^p = g^{-r} h^p = h^{-p} h^p = e\text{.}
\end{equation*}
We have now formed an element \(a\) with order \(p\) such that \(a \notin \langle g \rangle\text{.}\) Since \(h\) was chosen to have the smallest order of all of the elements that are not in \(\langle g\rangle\text{,}\) \(|H| = p\text{.}\)
Now we will show that the order of \(gH\) in the factor group \(G/H\) must be the same as the order of \(g\) in \(G\text{.}\) If \(|gH| \lt |g| = p^m\text{,}\) then
\begin{equation*}
H = (gH)^{p^{m-1}} = g^{p^{m-1}} H;
\end{equation*}
hence, \(g^{p^{m-1}}\) must be in \(\langle g \rangle \cap H = \{ e \}\text{,}\) which contradicts the fact that the order of \(g\) is \(p^m\text{.}\) Therefore, \(gH\) must have maximal order in \(G/H\text{.}\) By the Correspondence Theorem and our induction hypothesis,
\begin{equation*}
G/H \cong \langle gH \rangle \times K/H
\end{equation*}
for some subgroup \(K\) of \(G\) containing \(H\text{.}\) We claim that \(\langle g \rangle \cap K = \{ e \}\text{.}\) If \(b \in \langle g \rangle \cap K\text{,}\) then \(bH \in \langle gH \rangle \cap K/H = \{ H \}\) and \(b \in \langle g \rangle \cap H = \{ e \}\text{.}\) It follows that \(G = \langle g \rangle K\) implies that \(G \cong \langle g \rangle \times K\text{.}\)
The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly from
Lemma 13.1.8 and
Lemma 13.1.9. By
Lemma 13.1.8,
\(G\) is a product of
\(p\)-groups. Suppose
\(G\) is a
\(p\)-group and let
\(g\) be an element of maximal order in
\(G\text{.}\) If
\(\langle g \rangle = G\text{,}\) then we are done; otherwise,
\(G \cong {\mathbb Z}_{|g|} \times H\) for some subgroup
\(H\) contained in
\(G\) by the
Lemma 13.1.9. Since
\(|H| \lt |G|\text{,}\) we can apply mathematical induction.
Theorem 13.1.10. The Fundamental Theorem of Finitely Generated Abelian Groups.
Every finitely generated abelian group \(G\) is isomorphic to a direct product of cyclic groups of the form
\begin{equation*}
{\mathbb Z}_{p_1^{ \alpha_1 }} \times {\mathbb Z}_{p_2^{ \alpha_2 }} \times \cdots \times {\mathbb Z}_{p_n^{ \alpha_n }} \times {\mathbb Z} \times \cdots \times {\mathbb Z}\text{,}
\end{equation*}
where the \(p_i\)’s are primes (not necessarily distinct).
