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I need to know the order of the subgroup of the symmetric group $S_5$ that is generated by the two elements $(123)$ and $(345)$.

I do know that the order of $S_5$ itself is 120 and thus it must be a factor of that. I also know that through the Orbit-Stabilizer theorem, it must be a multiple of 5 (due to the orbit of any element in {1,2,3,4,5} being order 5). It must also be a multiple of 3 because each element given is of order 3. So this leaves possible subgroup sizes of 30, 60, and 120.

Is there anything else I can do here to solve this? And is this process work for all similar problems to this?

Thanks

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It is $A_5$ of order $60$. Note that $(123)(345)=(12345)$ and since $(12345)$ and $(123)$ generates $A_5$, it must be $A_5$. This is because they are even permutations and so no odd permutations are generated.

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