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Let us suppose that $A=\{a,b,\ldots z\}$ is a minimal set of generators for a finite group $G$. Then, does the sequence $a,ab,abc,\ldots, abc\ldots z, abc\ldots za, abc\ldots zab,\ldots\}$ produce all the distinct elements of $G$. Note that the $z$ being in the end is just symbolic. It does not imply that there are just $26$ generators, rather it only implies that the group is finitely generated.

I think yes, but am not clear how to show that all the words in the sequence are actually distinct. What if two words get the same element of $G$? How about if $G$ is the symmetric group $S_n$ and $A$ is the set of transpositions $\{(12), (13),\ldots, (1 n)\}$. Can we ensure at least in this case? Thanks beforehand.

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    $\begingroup$ Also, that’s really bad notation. It would be better to use an indexed set of generators. $\endgroup$ – Arturo Magidin May 18 at 22:56
  • $\begingroup$ It's not clear if you're asking whether all elements of $G$ appear, or whether all the elements in the sequence are distinct. $\endgroup$ – verret May 18 at 23:03
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    $\begingroup$ For an example where not all elements of $G$ appear, consider $G=C_n\times C_n$ and $A=\{a,b\}$ the canonical generating set. Then the sequence you wrote produces only $2n$ elements, instead of $n^2$. $\endgroup$ – verret May 18 at 23:07
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If $n=|A|$ and $t$ is the order of the ordered product of the elements of $A$ (what you write $abc\cdots z$ but which, as noted in the comments, is bad notation, since it suggests there are $26$ elements), then clearly the sequence has at most $tn$ different elements. So in the case of $S_n$ at the end of your question, we will not get all elements for $n\geq 4$ at least.

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  • $\begingroup$ but sequences of longer length (with repeated elements ) are being taken as well $\endgroup$ – Yorch May 18 at 23:17
  • $\begingroup$ Let $x=abc\cdots z$ and let $W$ be the set $\{a,ab,abc,\ldots,x\}$. Note that this is a finite sequence, with $|W|\leq |A|$. Note that every element in the "long" sequence under consideration can be written in the form $x^iw$, for some $w\in W$. So the total number of elements in the sequence is bounded above by $|A|$ times the order of $x$. $\endgroup$ – verret May 18 at 23:21
  • $\begingroup$ Oh yeah my bad sorry. $\endgroup$ – Yorch May 18 at 23:25
  • $\begingroup$ yes, my question was was silly though! $\endgroup$ – vidyarthi May 18 at 23:40
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No, you don’t always get the full set of elements.

Take $G=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, written additively, and take as your minimal generating set $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. Your list of elements is: $$(1,0,0), (1,1,0), (1,1,1), (0,1,1), (0,0,1), (0,0,0)$$ after which it repeats. So you never get $(1,0,1)$ or $(0,1,0)$.

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  • $\begingroup$ oh makes sense,but then what could the intended thing we need to prove be. $\endgroup$ – Yorch May 18 at 23:25
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    $\begingroup$ @Onir: I don’t know what a true statement would be; what the OP wanted to be true just isn’t true, though. $\endgroup$ – Arturo Magidin May 18 at 23:26
  • $\begingroup$ Thanks, I at least thought it would be true for abelian groups, but your example shows otherwise. $\endgroup$ – vidyarthi May 18 at 23:39

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