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Does every Abelian finite group have a linearly independent generator?

Here is my solution:

Let G be a Abelian finite group. Assume $\mbox{ord}(G)=n$, then exists $x_1, x_2, \ldots, x_n\in G$ such that $$ G = \langle x_1\rangle \oplus \langle x_2\rangle \oplus \cdots \oplus \langle x_n\rangle$$

So $S = \{x_1, \ldots, x_n\}$ is a linearly independent generator of $G.$

However, $\mathbb{Z}/n$ group is Abelian finite but it has not linearly independent generator. Because $n\neq 0$ and $n\overline{a} = \overline{na} = \overline{n}\overline{a} = \overline{0}, \forall \overline{a}\in \mathbb{Z}/n.$

I have dont nothing wrong? Can you expalin for me? Thank you very much!

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  • $\begingroup$ If by "$x_1, \dots, x_n$ are linearly independant" you mean $\forall (a_1, \dots, a_n) \in \mathbb Z^n, \sum a_ix_i = 0 \Rightarrow a_i = 0$ you see that this is wrong by considering $\mathbb Z/n \mathbb Z$ as you did. $\endgroup$ – user171326 Dec 14 '16 at 13:54
  • $\begingroup$ In fact, you can notice that $nx = 0$ for any $x \in G$, where $n$ is the order of $G$. Thus there is never linearly independant element in a finite abelian group $G$. $\endgroup$ – user171326 Dec 14 '16 at 13:55
  • $\begingroup$ @ N.H. We know that $S = \{x_1, \ldots, x_n\}\subseteq G$ is linearly independent generator if and only if $$S = \langle x_1\rangle \oplus \cdots \oplus \langle x_n\rangle$$ and if $G$ is a Abelian group finite then $G$ has a decomposition direct sum of every cyclics subgroup. Hence, set $S = \{x_1, \ldots, x_n\}$ is a linearly independent generator of $G$ I have dont nothing wrong? $\endgroup$ – lony Dec 14 '16 at 14:15
  • $\begingroup$ Well, your "if and only if" characterization of linearly independence is just incorrect, then. $\endgroup$ – Dustan Levenstein Dec 14 '16 at 14:22
  • $\begingroup$ @ Dustan Levenstein. Yes! $\langle S\rangle = \langle x_1\rangle \oplus \cdots \oplus \langle x_n\rangle$ $\endgroup$ – lony Dec 14 '16 at 14:29
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As already explained, 'linearly independent' doesn't make sense in this context. However, there is still a sense of mutual independence, in that generators $x_1$, ..., $x_n$ can be chosen, with G isomorphic to the direct product of the cyclic groups generated by them, such that the intersection of any two such cyclic groups contains only the identity element. Then it is possible to prove that the vector ($m_1$, $m_2$, ... $m_n$) represents the identity of G exactly when $m_1$ is a multiple of the order of $x_1$, $m_2$ of $x_2$, ..., $m_n$ of $x_n$ - i.e if you limit the possible values of $m_1$ to { 0, 1, ..., ord($x_1$) - 1 }, etc, then $m_1$ must be 0, etc.

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  • $\begingroup$ @ PMar. We know that $S = \{x_1, \ldots, x_n\}\subseteq G$ is linearly independent generator if and only if $$\langle S\rangle = \langle x_1\rangle \oplus \cdots \oplus \langle x_n\rangle$$ and if $G$ is a Abelian group finite then $G$ has a decomposition direct sum of every cyclics subgroup. Hence, set $S = \{x_1, \ldots, x_n\}$ is a linearly independent generator of $G$ I have dont nothing wrong? $\endgroup$ – lony Dec 14 '16 at 14:23
  • $\begingroup$ @Iony Your algebra is rough but OK. It is just that the phrase 'linearly independent', as used for vector spaces, doesn't carry over to this context, at least not without some adjustment. No group-theory text or article which I have read has ever used that phrase when discussing this sort of group representation. $\endgroup$ – PMar Dec 14 '16 at 15:15
  • $\begingroup$ @Iony I would object to the phrase 'G has a decomposition direct sum of every cyclics subgroup', in that the word every is not correct. Consider the group $Z_p$ x $Z_p$ (p prime) - this group has (p+1) cyclic subgroups of order p, but representing it as a product requires only two of those subgroups. $\endgroup$ – PMar Dec 14 '16 at 15:18

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