Example13.1

Obviously, all finite groups are finitely generated. For example, the group $S_3$ is generated by the permutations $(12)$ and $(123)\text{.}$ The group ${\mathbb Z} \times {\mathbb Z}_n$ is an infinite group but is finitely generated by $\{ (1,0), (0,1) \}\text{.}$

Example13.2

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

Proposition13.3

Let $H$ be the subgroup of a group $G$ that is generated by $\{ g_i \in G : i \in I \}\text{.}$ Then $h \in H$ exactly when it is a product of the form

where the $g_{i_k}$s are not necessarily distinct.

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,

Theorem13.4Fundamental Theorem of Finite Abelian Groups

Every finite abelian group $G$ is isomorphic to a direct product of cyclic groups of the form

here the $p_i$'s are primes (not necessarily distinct).

Example13.5

Suppose that we wish to classify all abelian groups of order $540=2^2 \cdot 3^3 \cdot 5\text{.}$ The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities.

• ${\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_5\text{;}$

• ${\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5\text{;}$

• ${\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_{27} \times {\mathbb Z}_5\text{;}$

• ${\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_5\text{;}$

• ${\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5\text{;}$

• ${\mathbb Z}_4 \times {\mathbb Z}_{27} \times {\mathbb Z}_5\text{.}$

Lemma13.6

Let $G$ be a finite abelian group of order $n\text{.}$ If $p$ is a prime that divides $n\text{,}$ then $G$ contains an element of order $p\text{.}$

Proof

We will prove this lemma by induction. If $n = 1\text{,}$ then there is nothing to show. Now suppose that the order of $G$ is $n$ 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 4.4.39, 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,

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

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 $G\text{.}$

Lemma13.7

A finite abelian group is a $p$-group if and only if its order is a power of $p\text{.}$

Proof

If $|G| = p^n$ then by Lagrange’s theorem, then the order of any $g \in G$ must divide $p^n\text{,}$ and therefore must be a power of $p\text{.}$ Conversely, if $|G|$ is not a power of $p\text{,}$ then it has some other prime divisor $q\text{,}$ so by Lemma 13.6, $G$ has an element of order $q$ and thus is not a $p$-group.

Lemma13.8

Let $G$ be a finite abelian group of order $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}\text{,}$ where where $p_1, \ldots, p_k$ are distinct primes and $\alpha_1, \alpha_2, \ldots, \alpha_k$ are positive integers. Then $G$ is the internal direct product of subgroups $G_1, G_2, \ldots, G_k\text{,}$ where $G_i$ is the subgroup of $G$ consisting of all elements of order $p_i^k$ for some integer $k\text{.}$

Proof

Since $G$ is an abelian group, we are guaranteed that $G_i$ is a subgroup of $G$ for $i = 1, \ldots, n\text{.}$ 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

where $t$ is the maximum of $r$ and $s\text{.}$

We must show that

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 $p^{\alpha_i}\text{,}$ 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

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,

Since

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.

Lemma13.9

Let $G$ be a finite abelian $p$-group and suppose that $g \in G$ has maximal order. Then $G$ is isomorphic to $\langle g \rangle \times H$ for some subgroup $H$ of $G\text{.}$

Proof

By Lemma 13.7, 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,

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,

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

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,

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 $G \cong \langle g \rangle \times K\text{.}$

Theorem13.10The 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

where the $p_i$'s are primes (not necessarily distinct).