How do you prove uniqueness for the fundamental theorem of finite abelian groups? The book I'm using has this not very well written proof that I can't follow.
So following this proof, I multiply by $p$
$$(1)p\mathbb{Z}_{p^{\alpha _1}} \oplus p\mathbb{Z}_{p^{\alpha _2}}\oplus.....\oplus p\mathbb{Z}_{p^{\alpha _n}}=p\mathbb{Z}_{p^{\beta _1}} \oplus p\mathbb{Z}_{p^{\beta _2}}\oplus.....\oplus p\mathbb{Z}_{p^{\beta _m}}$$
From the second isomorphism we have this relation $$(2)\frac{\mathbb{Z}_ p }{p \mathbb{Z}_{p^{\alpha_1}}} \cong \mathbb{Z}_{p^{\alpha_1}}$$
After this I get lost, I can apply the inductive hypothesis to (1) but I have a feeling that in (1), $m$ and $n$ are not the same as in before we multiply by $p$. And I also don't understand why $\alpha_1 -1 = \beta -1...$. What's the order of $p\mathbb{Z}_{p^{\alpha_1}}?$ And any other help.? Thanks,
Update: Am I allowed to do this $\frac{p\mathbb{Z}}{p^{\alpha_1}\mathbb{Z}}= \frac{\mathbb{Z}}{p^{\alpha_1-1}\mathbb{Z}} =\mathbb{Z}_{p^{\alpha_1-1}}$ Hence that's where the $\alpha_1 -1 = \beta_1 -1....>$ terms came from?
My favorite way to see the uniqueness is the following: given your $p$-group $G$, consider the groups $p^{i-1} G / p^i G$. Each of these is a module over $\mathbb{Z} / p$, the field with $p$ elements---that is, each is a vector space over the field with $p$ elements. Record their dimensions as follows:
$$\lambda_i=\mathrm{dim}_{\mathbb{Z}/p}(p^{i-1} G / p^i G).$$
Now consider any decomposition
$$G=\mathbb{Z} / p^{\alpha_1} \oplus \mathbb{Z} / p^{\alpha_2} \oplus \cdots . $$ You might as well arrange things so that the sequence $\alpha_1 \geq \alpha_2 \geq \cdots$ is non-increasing. Then the basic fact is that the sequence of $\lambda_i$'s defines the conjugate partition, that is $$\lambda_i=\#\{j \ | \ \alpha_j \geq i \}. $$ If you are familiar with Young diagrams of partitions, this just means that the Young diagram of $\lambda$ is the reflection about the diagonal of the Young diagram of $\alpha$. Evidently $\lambda$ is determined by the isoclass of $G$, hence so is $\alpha$.