The Basis Theorem for Finite Abelian Groups I am using Pinter's Abstract algebra book to prove the basis theorem for finite abelian groups (Every finite abelian group is a direct product of cyclic groups of prime power order.) $G$ is an abelian group of order $p^km$, $p^k$ and $m$ are relatively prime. $G=[a_1,a_2,...a_n]$
There are a series of problems you do to eventually prove it. I'm stuck at this one
$G\simeq\langle a_1 \rangle×G'$, where G' is a subgroup of G and $G'=[a_2,...,a_n]$. It also says to conclude that $G\simeq \langle a_1\rangle \times \langle a_2\rangle \times \cdots \times \langle a_n\rangle$.
My ideas are to show for any $x$ in $G$, $x=a_1^{k_1}\times a_2^{k_2} \times \cdots \times a_n^{k_n}$. I know that $a_1^{k_1}$ is in $\langle a_1\rangle$ and the rest is G', $x=\langle a_1\rangle$ so $G\simeq\langle a_1\rangle \times G'$. After that I'm lost.
 A: For the particular part of the question where it is required to show that $G\cong\langle a_{1}\rangle\times G'$, you need to show that $G$ is an inner direct product of the subgroups $\langle a_{1}\rangle$ and $G'$. 
A group $G$ is an inner direct product of subgroups $H$ and $K$ if:
(1) $H$ and $K$ are normal in $G$;
(2) Every $g\in G$ can be written in the form $hk$ for some $h\in H$ and $k\in 
K$, and;
(3) $H\cap K=\{e\}$ where e is the identity in $G$. 
The first condition is satisfied since every subgroup of an abelian group is necessarily normal. The second condition is satisfied since any $x\in G$ can be written in the form $a_{1}^{k_{1}}a_{2}^{k_{2}}\cdots a_{n}^{k_{n}}$ for some $k_{1},\dots,k_{n}\in\mathbb{Z}$ (by $(D1)$ from the definition of "decomposable" given in the question) and $a_{1}^{k_{1}}\in\langle a_{1}\rangle$, $a_{2}^{k_{2}}\cdots a_{n}^{k_{n}}\in G'$. The third condition can be checked by supposing that $\langle a_{1}\rangle \cap G'$ contains an element $x$ other than the identity; using $(D2)$ from the "decomposable" definition supplied in the question will lead to a contradiction. 
The conclusion that $G\cong\langle a_{1}\rangle\times\cdots\times\langle a_{n}\rangle$ can be made using induction. If we define $$G^{(m)}=\{a_{m+1}^{l_{m+1}}a_{m+2}^{l_{m+2}}\cdots a_{n}^{l_{n}}\mid l_{m+1},\dots,l_{n}\in\mathbb{Z}\}\text{ for $m\in\{0,1,\dots,n-1\}$}$$ so that $G=G^{(0)}$, $G'=G^{(1)}$ etc. then we know $G\cong\langle a_{1}\rangle\times G^{(1)}$. Assuming $G\cong\langle a_{1}\rangle\times\cdots\times\langle a_{m-1}\rangle\times G^{(m-1)}$ as the inductive hypothesis, if it can be shown that $G^{(m-1)}\cong\langle a_{m}\rangle\times G^{(m)}$, then $G\cong\langle a_{1}\rangle\times\cdots\times\langle a_{m}\rangle\times G^{(m)}$ is implied. As this holds for all $m\in\{0,1,\dots,n-1\}$ the result $$G\cong\langle a_{1}\rangle\times\cdots\times\langle a_{n-1}\rangle\times G^{(n-1)}$$ is obtained, where $G^{(n-1)}=\{a_{n}^{l_{n}}\vert l_{n}\in\mathbb{Z}\}=\langle a_{n}\rangle$.
