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As homework the teacher asks us to determine how many elements are there in $\langle (123) , (234) \rangle \subset S_4$ .

I've started doing all the multiplications between the elements, and I've counted so far $9$ different elements. But I think there is a easier way to determine the order of this subgroup of $S_4$. My argument is:

$(123)$, $(234)$ are even permutations, so they can generate only even permutations. So our subgroup has at most $12$ elements. I've counted so far $9$ distinct elements, so using the Lagrange Theorem it must have 12 elements.

Is it correct this argument?

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Yes, your argument is correct. If you know orbit-stabilizer, then here is a quicker method: These two permutations can move 1 to 2 to 3 to 4, so the subgroup of elements that move 1 to 1 has index 4 and 4 divides the order of the group. Also there is an element of order 3, so Lagrange says that 3 divides the order of the group, and thus 12 divides the order of the group. Since all the permutations are even, it is a subgroup of the alternating group, which has order 12. The only possibility is that it is the alternating group.

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