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This same problem had me find the orders of $S_4 \times A_5$ and $S_5 \times A_4$ and show that they are not isomorphic, so I'm not sure if these facts have anything to do with the largest cyclic subgroup. I can't find any similar questions with these types of groups, but based off this, would it just be $lcm(4,5)$?

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  • $\begingroup$ No there is no element of order $20$ in $A_4 \times S_5$. $\endgroup$ – Derek Holt Nov 7 '17 at 18:27
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In $S_5$, the order of an element $x$ can be $1, 2, 3, 4, 5,$ or $6$. (Think about the cycle decomposition of $x$.)

In $A_4$, the order of an element $y$ can be $1, 2,$ or $3$. (Again think about cycle decomposition, and we know that a $4$-cycle is not in $A_4$.)

So if you pick an element $(x,y) \in S_5\times A_4$, what is the order of that element? Well, $x$ has some order in $S_5$, and $y$ has some order in $A_4$, and we have to make both coordinates go to the identity so the order is the LCM: $$ \text{order of } (x,y) = \operatorname{lcm}(\text{order of } x, \text{order of } y). $$

So if order of $x$ is $1, 2, 3, 4, 5,$ or $6$, and order of $y$ is $1, 2,$ or $3$, what's the maximum number we can get for the LCM?

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