# What does it mean “unique” for this author?

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 $a_{i_l}\in N_{i_l}$.

• 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 $e$. – Gregory Grant Mar 26 '15 at 14:36
• @GregoryGrant so for the author the element $g$ I mentioned is not unique? – user42912 Mar 26 '15 at 14:40
• 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 $N_{i_k}$. – Gregory Grant Mar 26 '15 at 14:49
• Sorry I guess I didn't answer your question very well. – Gregory Grant Mar 26 '15 at 14:49
• 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 $i_k$. – Gregory Grant Mar 26 '15 at 14:51

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 $ab = ba$.