Clarification of GENERATORS.

Question

Can someone help me understand the concept of generators. I believe that the motivation behind this concept is that subgroups often have a great number of elements due to the closure conditions under multiplication (assuming that the operation is multiplication). Furthermore, say I have two elements $x,y$. If I let $\langle H \rangle$ be the set generated by these two elements, it means that: $\langle H \rangle = {{x^i,y^j \vert i,j \in \mathbb{Z}}}$. So in any case, unless I find a nice condition, these elements will generate an infinite set comprising of the products of different combinations of these elements in relation to their powers. Obviously, this generalizes to many more elements Am I correct in my understanding?

With that said, can someone help me with this question:

Let $\langle H \rangle <S_5$ be the subgroup of the symmetric group generated by (23),(34). Describe H in such a way that it allows you to compute the order of H. Furthermore, what is it's order?

I think it would be silly to go by trying to use the definition of generators since there are an infinite number of elements...unless I can find a nice condition/order on the elements that keeps it finite. Any help would be greatly appreciated. Thank you !

Answer 1

Your initial generators only mention 2,3,4 so no matter how you multiply them, you won't get any elements that have anything other than 2,3,4 in their cycle structure. Therefore the number of permutations generated is finite. In fact, they can only permute three elements: 2,3,4 so there cannot be more than 3! = 6 of them. Now try multiplying them up to see that you get all 6 elements.

Answer 2

It is a useful general fact about groups that if $t$ and $u$ are elements of order $2$ in any group, they generate a dihedral group (including the possibility that they commute and generate a group of order $4$). It is quite a good illustration of why thinking in terms of generators can be useful. Consider $h = tu.$ Then $uhu = u(tu)u = (ut)uu = ut = u^{-1}t^{-1}= (tu)^{-1} = h^{-1}.$ I won't give all details, but this implies that the cyclic subgroup $\langle h \rangle$ is a normal subgroup of $\langle t, u \rangle$ of index $2$.