My Proof of Structure Theorem for Finite Abelian Groups

1. Mar 17, 2013

Site

Hello! If anybody has a minute, I'd appreciate a quick look-through of my proof that a finite abelian group can be decomposed into a direct product of cyclic subgroups. I'm new to formal writing (as well as Latex) and all feedback is greatly appreciated!

http://www.scribd.com/doc/130897466/Structure-of-Finite-Abelian-Groups-Brian-Blake

2. Mar 18, 2013

jbunniii

Your Lemma 1 is not valid. Consider for example a cyclic group $G = \langle x\rangle$ with $|G| = p^2$. The subgroup $H = \langle x^p \rangle$ has order $p$, so $G/H$ and $H$ are both cyclic with order $p$. But clearly $G/H \times H$ is not cyclic; it has order $p^2$ but any element has order $1$ or $p$.

The problem is your claim that $\phi(g_1 g_2) = (x^{k+j}H, h_1 h_2)$. This is not true in general. Let's take a concrete example with $G$ as above.

Suppose $p = 3$, so $G = \{e, x, x^2, \ldots x^8\}$, and $H = \{e, x^3, x^6\}$, and $G/H = \{H, xH, x^2 H\}$.

Take $g_1 = x^{4} = x^1 x^3 \in x H$ and $g_2 = x^{5} = x^2 x^3 \in x^2 H$. Then $g_1 g_2 = x^{9} = e = x^0 x^0 \in H$.

Then:
$$\phi(g_1) = (x H, x^{3})$$
$$\phi(g_2) = (x^2 H, x^{3})$$
but
$$\phi(g_1 g_2) = (H, x^{0})$$
whereas
$$\phi(g_1)\phi(g_2) = (x H \cdot x^2 H, x^{3}\cdot x^{3}) = (H, x^{6}) \neq \phi(g_1 g_2)$$

Last edited: Mar 18, 2013
3. Mar 18, 2013

jbunniii

4. Mar 18, 2013

Site

Thank you very much! I guess I was thinking incorrectly that the order of x was also the order of the quotient group. Back to the drawing board!

5. Mar 18, 2013

jbunniii

No problem, feel free to post an update when you have a revised proof. It was a good attempt - I knew the conclusion of Lemma 1 was wrong, but it took me a while to spot the problem with your proof.

Also, your writing style and Latex are good, so nothing to worry about there.