The Basis Theorem for Finite Abelian Groups

Solution 1:

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 $G^{(n-1)}=\{a_{n}^{l_{n}}\vert l_{n}\in\mathbb{Z}\}=\langle a_{n}\rangle$.