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Given the following two groups $\mathbb{Z}_{10}\times \mathbb{Z}_{10}$, $\mathbb{Z}_{10}\times \mathbb{Z}_{15}$.

a) give the order of each group?

b) in each group, what is the largest order of an element?

c) for the second group, give an element of order 6.

So for a) the order of a group is the number of elements so 100 and 150 would be the order of the groups receptively right?

b) To find this you take the least common multiple(lcm) so lcm(10,10)=10? I get confused by the lcm and lcm(10,15)=30? I feel like this is wrong though so if someone could let me know if this is right i would appreciate it.

c) I don't know how to go about finding an element of order 6 .

So basically just looking for some help with c and confirmation that a and b are done correctly thank you for all your help.

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    $\begingroup$ For (a) you are correct and for (b) you are correct as well. Then this should follow for (c) let $z=(x,y) \in \mathbb{Z}_{10} \times \mathbb{Z}_{15}$ then $|z| = \textrm{lcm}(|x|,|y|)$. So you need an element of order two in the first slot and one of order 3 in the second. $\endgroup$ – Faraad Armwood Apr 11 '16 at 3:34
  • $\begingroup$ I hope this doesn't sound dumb but i don't know how i would go about finding an element of order 2 or order 3. Could you give me a starting point? $\endgroup$ – pkid58 Apr 11 '16 at 3:40
  • $\begingroup$ Doesn't sound dumb at all. Need to get $x$ such that $2x = 0 \ \textrm{mod}(10)$ and $y$ such that $3y = 0\ \textrm{mod}(15)$. $\endgroup$ – Faraad Armwood Apr 11 '16 at 3:49
  • $\begingroup$ so (0,0)? would be the element of order 6? $\endgroup$ – pkid58 Apr 11 '16 at 3:55
  • $\begingroup$ Remember that the order of an element is the minimal such positive integer. By definition order(0,0) = 1. $\endgroup$ – Faraad Armwood Apr 11 '16 at 3:59
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I posted a comment which answers everything, but here is proof for the main fact that you are using.

If $(x,y) \in \mathbb{Z}_m \oplus \mathbb{Z}_n$ and $k \cdot (x,y) = (0,0)$ for some $k \in \mathbb{Z}^+$ then $kx = ky = 0$ i.e $k$ divides $m,n$. The order$(x,y) := \textrm{min} \{j:j \cdot (x,y) = (0,0) \}$. The minimal $j$ which divides $m,n$ i.e the minimal $j$ which divides $mn$. However, this is just the lcm and so $\textrm{ord}(x,y) = \textrm{lcm}\left(\textrm{ord}(x), \textrm{ord}(y)\right)$.

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