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I'm studying Hungerford's Abstract Algebra book. I would like to know what the author means by "unique" in this theorem:

The orders count? I mean the element $g\in G$ such that $g=a_{i_1}a_{i_2}=a_{i_2}a_{i_1}$ is considered unique?

I'm asking that because we know that if $G$ is an internal weak direct product of the family $\{N_i\mid i\in I\}$, we have $a_{i_k}a_{i_l}=a_{i_l}a_{i_k}$ for every $a_{i_k}\in N_{i_k}$ and $a_{i_l}\in N_{i_l}$.

Thanks in advance

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  • $\begingroup$ Unique as in literally one and only one way to write it with those conditions, including the order. That's why he doesn't allow any of them to be $e$. $\endgroup$ – Gregory Grant Mar 26 '15 at 14:36
  • $\begingroup$ @GregoryGrant so for the author the element $g$ I mentioned is not unique? $\endgroup$ – user42912 Mar 26 '15 at 14:40
  • $\begingroup$ I see what you mean if $G$ is abelian. But by order I really meant the correspondence between the elements and the $N_{i_k}$'s to which they belong. In other words what is unique is that each $a_{i_k}$ is in $N_{i_k}$. $\endgroup$ – Gregory Grant Mar 26 '15 at 14:49
  • $\begingroup$ Sorry I guess I didn't answer your question very well. $\endgroup$ – Gregory Grant Mar 26 '15 at 14:49
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    $\begingroup$ What he means is if $g$ can be written as a product of $a_{i_k}$'s which $a_{i_k}\in N_{i_k}$ and $g'$ can be written as a product of $a'_{i_k}$'s which $a'_{i_k}\in N_{i_k}$, then $a'_{i_k}=a_{i_k}$ for all $i_k$. $\endgroup$ – Gregory Grant Mar 26 '15 at 14:51
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It's not written very clearly, but actually the order is intended not to matter because pairs of elements of two normal subgroups which intersect only at the identity in fact must commute, as you say:

Suppose $N_1$ and $N_2$ are normal subgroups of $G$ and $N_1 \cap N_2 = \{e\}$. Then let $a \in N_1$ and $b \in N_2$. Then $aba^{-1}b^{-1} = (aba^{-1})b^{-1} \in N_2$ but also $aba^{-1}b^{-1} = a(ba^{-1}b^{-1}) \in N_1$, so $aba^{-1}b^{-1} = e$, or in other words $ab = ba$.

Here's a nice write-up I found on internal weak direct products with a more clearly stated characterization which you may find helpful.

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  • $\begingroup$ Thank you!!! I'm going to read this article. Thanks again. $\endgroup$ – user42912 Mar 26 '15 at 16:03

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