# 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$$?$$

• As $5$ and $3$ are in different factors, you're perfectly right. – Bernard Feb 7 at 18:15
• What does it mean to be in different factors? – Mathsaddict Feb 7 at 18:16
• 5 is in $\mathbf Z_{30}$ and $3$ in $\mathbf Z_{12}$. – Bernard Feb 7 at 18:19
• @Mathsaddict, I edited your post to fix the angled braces. Can you clarify if you are considering direct product or free product? – Cheerful Parsnip Feb 7 at 18:44
• @Mathsaddict, okay I edited to put in the direct product symbol. – Cheerful Parsnip Feb 7 at 22:53

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.