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 $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$?

  • $\begingroup$ You must assume that $G$ is abelian because there are non-abelain groups of order 8, such as the dihedral group of order 8 (the group of symmetries of the square). $\endgroup$ – lhf Sep 26 '13 at 2:09
  • $\begingroup$ @lhf Right. I put that in the title but forgot to put it in the question. Edited. $\endgroup$ – tylerc0816 Sep 26 '13 at 12:00

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

$$G\cong N\times K\cong (\mathbb{Z}/4\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$$

  • $\begingroup$ I'm not sure what $2G = 0$ means. Also, we haven't covered vector spaces yet. Could you elaborate on 'It then follows that $G$ is a $\mathbb{F}_2$ vector ...'? Thank you for your help! $\endgroup$ – tylerc0816 Sep 26 '13 at 11:59
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    $\begingroup$ $2G = 0$ means that, in multiplicative notation, $g\cdot g = 1$ for all $g \in G$. The notation comes from when you use additive notation in which case $2g = g+g = 0$. $\endgroup$ – RghtHndSd Sep 26 '13 at 13:10
  • $\begingroup$ Maybe you also know this: math.stackexchange.com/questions/759647/… $\endgroup$ – annie heart Apr 18 '14 at 20:27

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