Fundamental Theorem of Finite Abelian Groups

Theorem
Every finite Abelian group is a direct product of cyclic groups of prime-power order. The number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.

Proof

 * Lemma 1: Given an Abelian group A, any set B of the form $$K= \left\{ x \in A : a^k=e \right\} $$ where $$k \in \mathbb{Z}$$ is a subgroup of A.

Proof: We can demonstrate this because the identity, e, satisfies $$e^k=e$$, and so B is non-empty. Now assume that $$x, y \in B$$. Then $$a^k=b^k=e \in B$$ and

$$(ab^{-1})^k = \underbrace{ab^{-1}ab^{-1}...ab^{-1}}_{k} = a^k (b^{-1})^k = e (b^k)^{-1} = ee^{-1} = e$$

Hence B is a subgroup of A.


 * Lemma 2: Let G be a finite Abelian group of order $$mp^n$$ where p is a prime that does not divide m. Then $$G=H \times K$$ where $$H=  \left\{ x \in G : x^{p^n}=e \right\}  $$ and $$K= \left\{ x \in G : x^m=e \right\}  $$. $$\left|{H}\right| = p^n$$.

Proof: Because G is Abelian, to prove $$G=H\times K$$ we need only show that G = HK and $$H \cap K = \left\{e\right\}$$. Since we have $$\gcd(m,p^n)=1$$, there are integers s and t such that $$1=sm+tp^n$$. For any $$x\in G$$, we have $$x=x^{sm+tp^n}$$, and by Lagrange's Theorem$$\ddagger$$, $$x^{sm} \in H$$ and $$x^{tp^n}\in K$$. Thus, $$G=HK$$.

Now suppose that some $$x \in H \cap K$$. Then $$x^{p^n}=e=x^m$$, and hence $$\left|{x}\right|$$ divides both $$p^n$$ and m. Since p does not divide m, $$\left|{x}\right|=1 \Rightarrow x=e$$.


 * Lemma 3: Let G be an Abelian group of prime-order power and let a be an element of maximal order in G. Then G can be written in the form $$\langle a \rangle \times K$$.

Proof: Denote $$\left|{G}\right|=p^n$$ and induct on n. For n=1, then $$G=\langle a \rangle \times \langle e \rangle$$. Now we assume the lemma is true for all Abelian groups of order $$p^k$$, where $$k < n$$. For an element $$a \in G$$ with maximal order $$p^m, x^{p^m}=e \forall x \in G$$. We can assume $$G \neq \langle a \rangle$$, for then there is nothing to prove. Choose $$b \in G$$ such that $$b \not \in \langle a \rangle$$. The claim is that $$\langle a \rangle \cap \langle b \rangle = \left\{ e \right\}$$. Clearly, this would be established if $$\left|b \right| = p$$. Since $$\left|b^p\right|=\left|b\right|/p$$, we know that $$b^p \in \langle a \rangle$$ by the manner in which b was chosen. Say $$b^p=a^i$$. Note that $$e=b^{p^m}=(b^p)^{p^{m-1}}=(a^i)^{p^{m-1}}$$, so $$\left|a^i\right|\le p^{m-1}$$. Hence $$a^i$$ is not a generator of $$\langle a \rangle$$ and therefore $$gcd(p^m,i) \neq 1$$. This proves that p divides i, so that we can write i=pj. Then $$b^p=a^i=a^{pj}$$. Consider that element $$c=a^{-j}b$$. This element c is not in $$\langle a \rangle$$, for if it were, b would be, too. Also, $$c^p=a^{-jp}b^p=a^{-i}b^p=b^{-p}b^p=e$$. Hence c is an element of order p such that $$c \not \in \langle a \rangle$$. Since b was chosen to have smallest order such that $$b \not \in \langle a \rangle$$, we conclude b also has order p, and the claim is verified.

Now consider the factor group $$\bar{G}=G/\langle b \rangle$$. Let $$\bar{x}$$ denote the coset $$x\langle b \rangle$$ in G. If $$\left|\bar{a}\right| < \left|a\right|=p^m$$, then $$\bar{a}^{p^{m-1}}=\bar{e}$$. This means $$(a\langle b \rangle)^{p^{m-1}}=a^{p^{m-1}}\langle b \rangle = \langle b \rangle$$, so that $$a^{p^{m-1}} \in \langle a \rangle \cap \langle b \rangle = \left\{ e \right\}$$, contradicting the fact that $$\left|a\right| = p^m$$. Thus, $$\left|\bar{a}\right|=\left|a\right|=p^m$$, and therefore $$\bar{a}$$ is an element of maximal order in $$\bar{G}$$. By induction, we know that $$\bar{G}$$ can be written in the form $$\langle \bar{a} \rangle \times \bar{K}$$ for some subgroup $$\bar{K}$$ of $$\bar{G}$$. Let K be the pullback of $$\bar{K}$$ under the natural homomorphism from G to $$\bar{G}$$, specifically, $$K= \left\{ x \in G : \bar{x} \in \bar{K} \right\}$$. The claim is that $$\langle a \rangle \cap K = \left\{ e \right\}$$. For if $$x \in \langle a \rangle \cap K$$, then $$\bar{x} \in \langle \bar{a} \rangle \cap \bar{K} = \left\{ \bar{e} \right\} = \langle b \rangle$$ and $$x \in \langle a \rangle \cap \langle b \rangle =  \left\{ e \right\}$$. It now follows from an order argument that $$G=\langle a \rangle K$$, and therefore $$G=\langle a \rangle \times K$$.


 * Lemma 4: A finite Abelian group of prime-order power is an internal direct product of cyclic groups.

Proof: This follows from Lemma 3 and induction on the order of the group.


 * Lemma 5: Suppose that G is a finite Abelian group of prime-power order. If $$G=H_1 \times H_2 \times ... \times H_m=K_1 \times K_2 \times ... \times K_n$$, where the H's and K's are nontrivial cyclic subgroups with $$\left|H_1\right| \ge \left|H_2\right| \ge ... \ge \left|H_m\right|$$ and $$\left|K_1\right| \ge \left|K_2\right| \ge ... \ge \left|K_n\right|$$.  Then m=n and $$\forall i, \left|H_i\right|=\left|K_i\right|$$.

Proof: We proceed through induction on $$\left|G\right|$$. Clearly, the case where $$\left|G\right|=p$$ is true. Now suppose the lemma is true for all Abelian groups of order less than $$\left|G\right|$$. For any Abelian group L, the set $$L^p=\left\{x^p:x \in L \right\}$$ is a subgroup of L and is a proper subgroup is p divides $$\left|L\right|$$. It follows that $$G^p = H_1^p \times ... \times H_{m'}^p=K_1^p \times ... \times K_{n'}^p$$, where m' is the largest integer i such that $$\left|H_i\right| > p $$ and n' is the largest integer j such that $$\left|K_j\right| > p $$. This ensures the direct product does not have trivial factors. Since $$\left|G^p\right| < \left|G^p\right|$$, we have by induction m'=n' and $$\left|H_i^p\right|=\left|K_i^p\right|$$ for i=1, 2, ..., m'. Since $$\left|H_i\right|=p\left|H_i^p\right|$$, this proves that $$\left|H_i\right|=\left|K_i\right|$$ for all i=1, 2, ..., m'.

Clearly, m-m'=n-n'; this follows directly from the facts that $$\left|H_1\right|\left|H_2\right|...\left|H_{m'}\right|p^{m-m'}=\left|G\right|=\left|K_1\right|\left|K_2\right|...\left|K_{n'}\right|p^{n-n'}$$ and that $$\left|H_i\right|=\left|K_i\right|$$ and m'=n'.


 * Proof of the Fundamental Theorem of Finite Abelian Groups

From Lemma 2, it follows that G is the product of several groups of prime-order power. From Lemma 4, it follows that each such factor is the internal direct product of cyclic groups. Thus, G is an internal direct product of cyclic groups of prime-power order. Lemma 5 demonstrated the uniqueness of those factors. QED.

$$\ddagger$$Specifically, Lagrange's Theorem implies $$\forall a \in G, \left|{a}\right|$$ divides $$\left|{G}\right|$$, which implies $$a^{\left|{G}\right|}=e$$.