In $S_4$, what does the expression for the cyclic group $\langle (13),(1234)\rangle$ mean? In $S_4$, what does the expression for the cyclic group $\langle (13),(1234)\rangle$ mean?
I apologize if this is too basic, but I haven't come across such an expression anywhere in my book. Also, not much success on the web regarding what does $\langle (13),(1234)\rangle$ actually mean? How is it evaluated?
Thank you for your help.
 A: This is not "cyclic", that would be the case if it is only one element of $S_4$. But this is a generalization - the group generated by two elements namely $(13)$ and $(1234)$. So write down all elements that can be made with powers and products of these elements, for example $(13)\cdot(1234)\cdot(1234)\cdot (13)\cdot (1234)$ is an element. Can you work out which subgroup of $S_4$ the group $\langle(13),(1234)\rangle$ represents?
A: $\langle(13),(1234)\rangle$ is NOT a cyclic group! A cyclic group is a group generated by a single element, the one you wrote is a subgroup of $S_4$ generated by TWO elements, $(13)$ and $(1234)$.
In general given a group $G$, taken $g_1,\dots,g_n\in G$, the expression $\langle g_1,\dots,g_n\rangle$ denotes the smallest subgroup of $G$ which contains the $g_i$'s which is the intersection of all the subgroups of $G$ containing all the $g_i$'s.
In your case, you can write down explicitly the elements of the subgroup $\langle(13),(1234)\rangle$: it's formed by all the product of integers power of the elements $(13)$ and $(1234)$. Hence a typical element of $\langle(13),(1234)\rangle$ is $(13)^{k_1}(1234)^{k_2}(13)^{k_3}\dots$ with $k_i\in\mathbb Z$. Being $(13)$ an element of order $2$ you can wlog think the element of your group as $(13)(1234)^{h_1}(13)(1234)^{h_2}\dots$ and $(1234)^{h_1}(13)(1234)^{h_2}(13)\dots$ with $h_i\in\mathbb Z$.
