## Section11.4Finite Abelian Groups

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.

###### Example11.44.

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 $\{ (1,0), (0,1) \}\text{.}$

###### Example11.45.

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.

###### Example11.48.

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.

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

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

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

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

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

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

The proof of the Fundamental Theorem of Finite Abelian Groups depends on several lemmas.

###### 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 $n$ and 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. Then the order of $G$ must be prime. Since $p$ divides $n\text{,}$ we know that $p = n\text{,}$ and $G$ contains an element of order $p$ (in fact, all $p - 1$ non-identity elements have order $p$).

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 11.49 is a special case of Cauchy's Theorem, which states that if $G$ be a finite group and $p$ a prime such that $p$ divides the order of $G\text{,}$ then $G$ contains a subgroup of order $p\text{.}$ This is harder to prove and we will not attempt it here.

###### 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 11.49, $G$ has an element of order $q$ and thus is not a $p$-group.

###### 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

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

If remains for us to determine the possible structure of each $p_i$-group $G_i$ in Lemma 11.51.

###### Proof.

By Lemma 11.50, 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 11.52. Suppose that $G$ is a finite abelian 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. Since $|H| \lt |G|\text{,}$ we can apply mathematical induction.

We now state the more general theorem for all finitely generated abelian groups. The proof of this theorem can be found in any of the references at the end of this chapter.

### ExercisesCollected Homework

###### 1.

Describe all abelian groups of order 200 (up to isomorphism). Explain how you know you have them all.