Say we have a group G and we know some of the elements (but not all). How does one determine the order and list all the elements of the group in an intuitive way?
In this case G is the smallest subgroup of $S_4$, and it contains the permutations $\tau_1 = (2,3,4)$, $\tau_2 = (1,3,4)$, $\tau_3 = (1,2,4)$, $\tau_4 = (1,2,3)$, and $\mu_{12}=(1,2)(3,4)$, $\mu_{13}=(1,3)(2,4)$, $\mu_{14}=(1,4)(2,3)$.
What i want to do: find the order of G and list all the elements of G by using the symbols above and give the corresponding permutations.
My thought process so far: $S_4$ has $4! = 24$ which means that G's order must be a divisor (possibilities are 8, 12, 24 ignoring 1,2,3,4 and 6). Since G is a group, it must contain the identity element, the inverses for all elements mentioned above, and all possible products of the elements*.
- is where I get stuck, because one has many possibilities? Hence why i think i am missing something. How does one go about this?
In this particular case, you may note that all listed permutation are even, so you are within $A_{4}$, of order $12$. Your group has at least $8$ elements, but then it has elements of order $3$, so it cannot have $8$ elements.
You may finish noting that $$ V = \{ 1, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \} $$ is a (normal) subgroup of $A_{4}$, and the elements of $A_{4}$ outside $V$ have all order $3$.
To obtain all the elements of order $2$ from the given elements, try conjugating $(123)$ and $(234)$ by the elements of $V$. For instance $$ (123)^{(12)(34)} = (214). $$