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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.

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  • $\begingroup$ What does the notation $G = [a_{1}, \ldots, a_{n}]$ stand for? And what is $X$? $\endgroup$ Commented Apr 28, 2014 at 19:10
  • $\begingroup$ One idea I have is to use Sylow's Theorem to show if $|G| = p_{1}^{k_{1}}\cdots p_{n}^{k_{n}}$ is the prime factorization of the order of $G$, then there exist unique normal subgroups $G_{i}$ whose order is $p_{i}^{k_{i}}.$ $\endgroup$ Commented Apr 28, 2014 at 19:11

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

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  • $\begingroup$ I'd think a better (and more general) approach to prove the isomorphism $G\cong\langle a_{1}\rangle\times G'$ is to establish two necessary conditions that (1) there exists a bijection between $G$ and $\langle a_{1}\rangle\times G'$; and (2) there exists a homomorphism from $G$ to $\langle a_{1}\rangle\times G'$. $\endgroup$
    – hchar
    Commented Oct 14, 2019 at 0:40

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