What is the order of the subgroup $\langle 5\rangle \times \langle 3\rangle$ in $Z_{30} \times Z_{12} ?$ What is the order of the subgroup $\langle 5\rangle \times \langle 3\rangle$ in $Z_{30} \times Z_{12} ?$
I think it should be $12$, since $O(5) = 6$ in $Z_{30}$ and $O(3) = 4$ in $Z_{12}$. The order of the group generated by both should be l.c.m of the orders of both elements.
Is this right$?$
 A: If we read strictly by text, we will find, that your subgroup is countably infinite as a non-cyclic subgroup of a free product of cyclic groups (Here you can read about what is free product: https://en.wikipedia.org/wiki/Free_product)
However, from the context you provided I can conclude, that the "free product" here was just a typo, and you actually intended to write $Z_{30} \times Z_{12}$, instead of $Z_{30} \ast Z_{12}$. In that case, lets write your group as $\langle a \rangle_{30} \times \langle b \rangle_{12}$, where your subgroup is generated by elements $a^5$ and $b^3$. All elements of this subgroup can be uniquely represented as $a^nb^m$, where $0 \leq n < 30$, $0 \leq m < 12$, $n$ divides $5$ and $m$ divides $3$. You can see, that there are only $6$ possible $n$-s, only $4$ possible $m$-s and any combination of them is valid. So by Fundamental Combinatoric Theorem we can conclude, that there are $24$ elements in your subgroup. So, in that case, its order is $24$. 
However, if you are looking for the exponent of the group, and not order, then it really is $LCM(6, 4) = 12$, in case, when we deal with direct product. In case, when we deal with free product, the exponent is not defined, as there are elements of infinite order.
