# Fundamental theorem of finite abelian groups proof

Can someone give a proof of the "technical part" of the following proof?

One direction seems pretty trivial, but the other I am having trouble with. Why does it matter that the subgroup $H$ we choose is maximal?

• If you are familiar with modules, it is much simpler and I find it more intuitive to prove it from the decomposition of finitely generated modules. Apr 29 '17 at 9:25

I'm not sure whether this approach works, but you can try it:

Suppose $\left\langle g\right\rangle H\neq G$. Then there exists an $x\in G\setminus \left\langle g\right\rangle H$. Let $H'$ denote the smallest subgroup of $G$ containing $H$ and $x$. By maximality of $H$, $H'\cap \left\langle g\right\rangle\neq \left\{e\right\}$. Hence there exists a non-trivial $y\in G$ such that $y\in H'\cap \left\langle g\right\rangle$. Since $G$ is commutative, this implies that we can write $y=hx^i$ for some $h\in H$ and some $i\neq 0$ (otherwise $y\in H\cap \left\langle g\right\rangle=\left\{e\right\}$), and $y=g^n$ for some $n$. Hence $g^n=hx^i$ and thus $x^i=g^nh^{-1}\in \left\langle g\right\rangle H$.

Can you conclude from this that $x\in \left\langle g\right\rangle H$? That would yield a contradiction.

Edit: Here is a another try: Notice that $\left\langle g\right\rangle\cap H=\left\{e\right\}$. Now consider $G/H$.For each $x\in G$ we denote by $\bar{x}$ the corresponding class in $G/H$. We claim that $|\bar{g}|=|g|=p^m$ (here $|x|$ is the order of an element). Suppose that $|\bar{g}|<p^m$. Then $\bar{g}^{p^{m-1}}=\bar{e}$, hence $g^{p^{m-1}}\in H$. But then $g^{p^{m-1}}\in \left\langle g\right\rangle\cap H=\left\{e\right\}$ contradicts the fact that $|g|=p^m$. Thus $|\bar{g}|=p^m$. It follows that the order of $\bar{g}$ is maximal in $G/H$.

By the induction hypothesis $G/H\cong \mathbb{Z}_{p^m}\times K$ where $K$ is a product of cyclic groups. Now consider the subgroup $L:=\left\{x\in G\mid \bar{x}\in K\right\}$. Then $|L|=p^i|K|$ for some $i$, in fact $|L|=|H||K|$ (Why?). Suppose that $x\in \left\langle g\right\rangle\cap L$, then $\bar{x}\in \left\langle \bar{g}\right\rangle\cap K=\left\{\bar{e}\right\}$ (Why?), and thus $x\in H$, hence $x\in \left\langle g\right\rangle\cap H=\left\{e\right\}$. We just proved that $\left\langle g\right\rangle\cap L=\left\{e\right\}$.

Thus $|\left\langle g\right\rangle L|=|\left\langle g\right\rangle||L|=p^i(|\left\langle \bar{g}\right\rangle||K|)=|H||G/H|=|G|$. By maximality of $H$ we definitely need that $|\left\langle g\right\rangle H|=|G|$ which then shows the claim.

• Well, if $x^{i}$ is an element of the subgroup then $x$ must be, because of closure? Apr 28 '17 at 15:35
• I quite like the first approach, but how would you explain that $x^i$ being in the group implies that so is $x$? Apr 29 '17 at 11:13
• I honestly don't know, I though the first approach was going to work, but I cannot conclude that $x\in \left\langle g \right\rangle H$. Maybe someone else will find an easy argument. Obviously the more advanced Sylow theory and structure theorems for modules over PID's easily solve the problem, but not very elementary. Apr 29 '17 at 17:17