Suppose $\order G = p^n$ with $p$ a prime. We proceed by induction on $n$. === Basis for the induction === For $n = 1$, we have $G = \gen a \times \gen e$, by Prime Group is Cyclic. === Induction Hypothesis === Now we assume the theorem is true for all abelian groups of order $p^k$, where $k < n$. === Induction Step === Let $a$ be an element with maximal order $p^m$ in $G$. For any $x \in G$, we have $\order x \divides \order G$ by Order of Element Divides Order of Finite Group. Thus we have $\order x = p^i$, for some $i \ge 0$. By assumption, $i \le m$. We conclude that $\forall x \in G : x^{p^m} = e$. Now assume $G \ne \gen a$, for the remaining case is trivial. Choose $b \in G$ such that $\order c < \order b$ implies $c \in \gen a$, but $b \notin \gen a$. ==== Lemma ==== We have $\gen a \cap \gen b = \gen e$. <section title="Proof of Lemma" foldable="true" header="4"> Since $\order {b^p} = \dfrac {\order b} p < \order b$, from the definition of $b$ we conclude $b^p \in \gen a$. Thus we have $\exists i \in \Z : b^p = a^i$. Note that $e = b^{p^m} = \paren {b^p}^{p^{m - 1} } = \paren {a^i}^{p^{m - 1} }$, so $\order {a^i} \le p^{m - 1}$. Hence $a^i$ is not a generator of $\gen a$. Therefore $\gcd \set {p^m, i} \ne 1$. It follows that $p$ divides $i$, so that we can write $i = pj$ for a $j \in \Z$. This means that we have $b^p = a^i = a^{pj}$. Now consider the element $c = a^{-j} b$. This element $c$ is not in $\gen a$, for if it were, $b = a^j c$ would be as well. Also, $c^p = a^{-j p} b^p = a^{-i} b^p = b^{-p} b^p = e$. Hence, $\order c = p$, and $c \notin \gen a$. From the definition of $b$, we conclude that also $\order b = p$. It follows from Prime Group is Cyclic that $\gen b$ is generated by all its non-identity elements. Therefore, by Intersection of Subgroups is Subgroup, $\gen a \cap \gen b$ is either $\gen b$ or $\gen e$. However, the first case contradicts $b \notin \gen a$. We conclude that $\gen a \cap \gen b = \gen e$. </section>{{qed|lemma}} Now consider the quotient group $\bar G = G / \gen b$. Let $\bar x$ denote the coset $x \gen b$ in $G$. Assume that $\order {\bar a} < \order a = p^m$. We observe that therefore $\bar a^{p^{m - 1} } = \bar e$. This means: :$\paren {a \gen b}^{p^{m - 1} } = a^{p^{m - 1} } \gen b = \gen b$ so that: :$a^{p^{m - 1} } \in \gen a \cap \gen b = \gen e$ This contradicts $\order a = p^m$. Thus: :$\order {\bar a} = \order a = p^m$ and therefore $\bar a$ is an element of maximal order in $\bar G$. As $\order {\bar G} = p^{n - 1}$ by a corollary of Quotient Group is Group, the induction hypothesis applies. Thus we know that $\bar G$ can be written in the form: :$\gen {\bar a} \times \bar K$ for some subgroup $\bar K$ of $\bar G$. Let $K$ be the preimage of $\bar K$ under the quotient epimorphism from $G$ to $\bar G$: :$K = \set {x \in G : \bar x \in \bar K}$ As this natural homomorphism is $p$-to-$1$ (that is, every element in the image has $p$ preimages), it follows that: :$\order K = p \order {\bar K}$ Also, Pullback is Subgroup ensures that $K \le G$. {{questionable|Pullback is Subgroup should be replaced by something like Preimage is Subgroup}} Now if $x \in \gen a \cap K$, then we have: :$\bar x \in \gen {\bar a} \cap \bar K = \gen {\bar e} = \gen {\bar b}$ It follows that: :$x \in \gen a \cap \gen b = \gen e$ and therefore: :$\gen a \cap K = \gen e$ This means that: :$\order {\gen a K} = \order {\gen a} \order K$ Computing now this order in more detail, we find: :$\order {\gen a} \order K = \order {\gen {\bar a} } \paren {p \order {\bar K } } = p \paren {\order {\gen {\bar a} } \order {\bar K } } = p \order {\bar G} = p^n = \order G$ We conclude that therefore, $G = \gen a K$. Using the Internal Direct Product Theorem, we conclude $G = \gen a \times K$. The result follows by induction. {{qed}} Category:Abelian Groups f0vwmkczwjucimz5se48wnao507edmt