Does the generator always have to be contained within the group. If so, why? Just because those are the rules?
For example, if I have a cyclic group, (Z, +) I can write its presentation as $\langle1\rangle$ – just the generator and no relations. Then if I want to write the presentation for the group of even integers under addition id have $\langle2\rangle$. Will the math police come for me if I just wrote the group as
$\langle$1 : some relation saying that odds are a no no$\rangle$
I though the whole point of presentation notation was to build up a big mass of stuff with our generators, and shave what we want off using our relations. So why could I not say 1 is my generator for the set of even numbers, it just happens to not be in the group of evens, and when constructing my group using the generator 1 I have to use some relation to only include the even products of the generator.
The reason I ask the question:
I was asked to prove that every subset, H, of some Cyclic group Cn was also a cyclic group. Well, I knew that all elements of H must also be contained in Cn, all elements of Cn are generated by c$\in$ Cn, and so its pretty clear we can represent any h$\in$H as ck for some integer k. So yea, the proof is trivial because c$\in$Cn generates H. But then we can easily have subgroups where our generator on the group is not contained within the subgroup — but even still we can represent the new generator on the group as some power of the generator on our overgroup, so why not just call that our generator for the subgroup even if it isn't contained in the group. Could we not just use relations to make everything kosher? Can we not do this because I have a bad understanding of presentations/relations, or just because "them's the rules". Thanks.
Best Answer
The generators must be elements of the group, because the definition of a group presentation $\langle S \mid R\rangle$ implies that all elements of $S$ are elements of the group $\langle S \mid R\rangle$.
So your "trivial" proof that a subgroup $H$ of a cyclic group $C_n$ must be cyclic doesn't work. If $c$ generates $C_n$, then $c$ generates all of $C_n$. Thus it cannot generate a proper subgroup of $C_n$.