Words in the alphabet of minimal generators of group

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.

• Also, that’s really bad notation. It would be better to use an indexed set of generators. May 18 at 22:56
• It's not clear if you're asking whether all elements of $G$ appear, or whether all the elements in the sequence are distinct. May 18 at 23:03
• 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 $n^2$. May 18 at 23:07

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.

• but sequences of longer length (with repeated elements ) are being taken as well May 18 at 23:17
• 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 $x$. May 18 at 23:21
• Oh yeah my bad sorry. May 18 at 23:25
• yes, my question was was silly though! May 18 at 23:40

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 $$(0,1,0)$$.

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