# Understanding the proof of finite abelian group G as the internal direct product of p-groups.

There are some questions that I have regarding the following Theorem (and its proof):

Theorem $\mathbf{13.2}$ Every finite abelian group $G$ is the internal direct product of $p$-groups.

$\text{P}\scriptstyle{\rm ROOF}$. If $|G|=1$, then the theorem is trivial. Suppose that the order of $G$ is greater than $1$, say $$|G|=p_1^{\alpha_1}\cdots p_n^{\alpha_n},$$ where $p_1,\ldots,p_n$ are all prime, and define $G_i$ to be the set of elements in $G$ order $p_i^k$ for some integer $k$. Since $G$ is an abelian group, we are guaranteed that $G_i$ is a subgroup of $G$ for $i=1,\ldots,n$. We must show that $$G=G_1G_2\cdots G_n.$$ That is, we must be able to write every $g\in G$ as a unique product $g_{p_1}\cdots g_{p_n}$ where $g_{p_i}$ is of the order of some power of $p_i$. Since the order of $g$ divides the order of $G$, we know that $$|g|=p_1^{\beta_1}p_2^{\beta_2}\cdots p_n^{\beta_n}$$ for integers $\beta_1,\ldots,\beta_n$. Letting $a_i=|g|/p_i^{\beta_i}$, the $a_i$'s are relatively prime; hence, there exist integers $b_1,\ldots,b_n$ such that $a_1b_1+\cdots+a_nb_n=1$. Consequently, $$g=g^{a_1b_1+\cdots+a_nb_n}=g^{a_1b_1}\cdots g^{a_nb_n}.$$ Since $g^{(a_ib_i)p_i^{\beta_i}}=g^{b_i|g|}=e,$ it follows that $g^{a_ib_i}$ must be in $G_i$. Let $g_i=g^{a_ib_i}$. Then $g=g_1\cdots g_n$ and $G_i\cap G_j=\{e\}$ for $i\neq j$.
$\quad$ To show uniqueness, suppose that $g=g_1\cdots g_n=h_1\cdots h_n$, with $h_i\in G_i$. Then $$e=(g_1\cdots g_n)(h_1\cdots h_n)^{-1}=g_1h_1^{-1}\cdots g_nh_n^{-1}.$$ The order of $g_ih_i^{-1}$ is a power of $p_i$; hence, the order of $g_1h_1^{-1}\cdots g_nh_n^{-1}$ is the least common multiple of the orders of the $g_ih_i^{-1}$. This must be $1$, since the order of the identity is $1$. Therefore, $|g_ih_i^{-1}|=1$ or $g_i=h_i$ for $i=1,\ldots,n$. $\tag*{$\square$}$

$1-$ Why $a_i$'s are relatively prime? They must not since for example $a_1=p_2^{\beta_2}p_3^{\beta_3} \dots p_n^{\beta_n}$ and $a_2=p_1^{\beta_1}p_3^{\beta_3} \dots p_n^{\beta_n}$ have $p_3^{\beta_3} \dots p_n^{\beta_n}$ in common.

$2-$ If $a$ and $b$ are relatively prime, I know a proof for 'there exist integers $c$ and $d$ such that $ac+bd=1$. I don't know how to generalize it even by induction method. How to prove that if $a_i$'s are relatively prime there exist integers $b_i$'s such that $a_1b_1+ \dots =1$?

$3-$ Is it necessary to show the uniqueness or it is just a 'supplementary' fact? I mean that I think that if the mentioned proof had finished at the point of $g=g_1\dots g_n$ (i.e. without whole of the last paragraph), the proof is still complete. If not, why?

I really appreciate any clear simple explanations.

PS - Source: Abstract Algebra, Theory and Applications by Thomas W. Judson

• For question $2$, I think it would help to look at the fact that $$\gcd(a_1, a_2, \ldots, a_n) = \gcd(a_1, \gcd(a_2, \gcd(\cdots, \gcd(a_{n-1}, a_n)\cdots)))$$ You can construct $\gcd(a_{n-1}, a_n)$ from a linear combination of $a_{n-1}$ and $a_n$. Then you can construct $\gcd(a_{n-2}, a_{n-1}, a_n) = \gcd(a_{n-2}, \gcd(a_{n-1}, a_n))$ with a linear combination of $a_{n-2}$ and $\gcd(a_{n-1}, a_n)$ (which is already a linear combination of $a_{n-1}$ and $a_n$). And so on. – Arthur Sep 11 '15 at 11:41

1. They are relatively prime "on the whole", there exists no $d$ (non-trivial) such that $d$ divides all $a_i$'s.
2. Take $a_1,...,a_s$ then in $\mathbb{Z}$ the ideal $(a_1,...,a_s)$ must be generated by one element (the ring $\mathbb{Z}$ is principal) let $d$ such that $(a_1,...,a_s)=(d)$, hence $d$ divides all $a_i$'s hence it must be $1$ (since they are relatively prime), in particular $1\in (a_1,...,a_s)$.