The Fundamental Theorem of Finitely Generated Abelian Groups states

Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$\Bbb{Z_{(p_1)^{r_1}}}\times \Bbb{Z_{(p_2)^{r_2}}}\times\dots\times \Bbb{Z_{(p_n)^{r_n}}}\times\Bbb{Z}\times\Bbb{Z}\dots\times\Bbb{Z}$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^{r_i}$ are unique.

The book (A First Course in Abstract Algebra by Fraleigh) omits the proof. The way I interpret the theorem is given below:

Let $\{a_1,a_2,\dots a_m,b_1,b_2,\dots,b_n\}$ be the finite list of generators of the abelian group $G$. Also let the order of $a_i=s_i$ (so $(a_i)^{s_i}=e$), whilst $b_j$ does not have a finite order for any $j$. This is the most general case in my opinion. Every element of $G$ can now be represented as $(a_1)^{t_1}(a_2)^{t_2}\dots (b_n)^{t_{m+n}}$ (remember that $G$ is abelian). This is easily seen to be isomorphic to $$\Bbb{Z_{s_1}}\times \Bbb{Z_{s_2}}\times\dots\times \Bbb{Z_{s_m}}\times\underbrace{\Bbb{Z}\times\Bbb{Z}\dots\times\Bbb{Z}}_{\text{n times}}$$.

Now let $s_1=(p_{s_1})^{r_{s_1}}(p_{s_2})^{r_{s_2}}\dots(p_{s_n})^{r_{s_n}}$, where all $p_{s_i}$ are primes. Then $$\Bbb{Z_{s_1}}\cong \Bbb{Z_{(p_{s_1})^{r_{s_1}}}}\times \Bbb{Z_{(p_{s_2})^{r_{s_2}}}}\times \Bbb{Z_{(p_{s_n})^{r_{s_n}}}}$$ Doing the same for all $\Bbb{Z_{s_i}}$, we get an expansion of the form that is given in the definition.

If my understanding is correct, then I don't see why the prime powers $(p_i)^{r_i}$ have to be unique. Two generators $a$ and $b$ of the group can have the same order, and hence the same factorization. Hence, in a decomposition of the group $G$, two primes powers (belonging to the factorizations of the orders of $a$ and $b$) could be the same!

Any help would be much appreciated. Thanks in advance.

  • $\begingroup$ Yes...so?? Where do you see any contradiction to the uniqueness? For example,. $\,\Bbb Z_2\times \Bbb Z_2\times\Bbb Z^4\;$ is a fin. gen. abelian group, and the primes powers $\;2=2^1\;$ are unique...is this what you meant? $\endgroup$ – DonAntonio Jan 25 '14 at 8:30
  • $\begingroup$ The key phrase is "$p_i$ are primes, not necessarily distinct": you may be confusing this with uniqueness of the exponents $r_i$ for any given $p_i$ $\endgroup$ – zcn Jan 25 '14 at 8:33
  • $\begingroup$ Yes. According to the theorem, it should be impossible to factor an abelian group as $\Bbb{Z_2}\times\Bbb{Z_2}\times\Bbb{Z_4}$ as we can't have the same powers of the same primes. $\endgroup$ – algebraically_speaking Jan 25 '14 at 8:33
  • $\begingroup$ @user115654- My understanding is, if $p_1\neq p_2$, we're done. If $p_1=p_2$, then this implies $r_1\neq r_2$. Is my understanding flawed? $\endgroup$ – algebraically_speaking Jan 25 '14 at 8:36
  • $\begingroup$ If the primes $p_1$ and $p_2$ are not distinct, this does not imply that $r_1 = r_2$. The only thing that is unique is the exponent $r_i$ for a given $p_i$, regardless of the other primes. It may happen that another $p_j$ is the same prime as $p_i$, but $p_j$ will have its own exponent $r_j$ that is unique to $p_j$ (and it is possible that $r_i = r_j$) $\endgroup$ – zcn Jan 25 '14 at 8:39

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