Without using the fundamental theorem of finite abelian groups, show that, if $G$ is a finite abelian group of order 8, then $G$ is isomorphic to one of $\mathbb{Z} / 8\mathbb{Z}$, $\mathbb{Z} / 4\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$, or $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$
I started by choosing some element $e \neq g \in G$. By Lagrange's theorem, the order of $g$ is either $2$, $4$, or $8$. If the order of $g$ is $8$, then $\langle g \rangle = G$, and so $G \cong \mathbb{Z} / 8 \mathbb{Z}$. I'm not totally sure what to do next. If the order of $g$ is $2$, do I consider $G / \langle g \rangle$ and show that this quotient is isomorphic to $\mathbb{Z} / 4 \mathbb{Z}$ or $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$?
Best Answer
Yes, you are on the right track, kind of! :)
If $G$ has an element of order $8$, then clearly $G$ is cyclic, so assume $G$ has no element of order $8$.
If $G$ also has no element of order $4$, then by Lagrange's theorem you have that every non-unit element of $G$ is of order $2$, and so $2G=0$. It then follows that $G$ is a $\mathbb{F}_2$ vector space, and thus, as a group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^3$.
So, we may assume that $g\in G$ has order $4$. Now, let $N=\langle g\rangle$. Choose $x\in G-N$ with $|x|=2$. Note that this is indeed possible. For, if not, then $G$ would have $6$ elements of order $4$, and $1$ element of order $2$. But, by a counting argument this is impossible. Let $K=\langle x\rangle$. Note then that $K\cap N=\{0\}$ (why?), $K,N\unlhd G$, and $KN=G$, since
$$\#(KN)=\frac{|K||N|}{|K\cap N|}=\frac{4\cdot 2}{1}=8$$
Thus, we have that
$$G\cong N\times K\cong (\mathbb{Z}/4\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$$