# how many element of order 2 and 5 are there

$G=Z_{10}\times Z_{15}$ Then $G$ contains exactly

1. one element of order $2$

2. five element of order $3$

3. 24 elements of order $5$

4. 24 elements of order $10$

what I think is 1 is correct, not able to guess others.

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Then don't guess. :) How would you find an element of order 3? What properties must one have? How might one describe a particular one? Could you then count descriptions? – Hurkyl Jun 13 '12 at 19:12

Let's go back to basics.

Suppose $a$ is a generator of $Z_{10}$ and $b$ is a generator of $Z_{15}$, then every element of your group is of the form $a^rb^s$ with $0 \leq r \leq 9$ and $0 \leq s \leq 14.$

That makes 150 elements.

Suppose $a^r$ has order $p$ (which must be 1, 2, 5 or 10) and $b^s$ has order $q$ (which must be 1, 3, 5 or 15) [these orders by Lagrange]. Then, since your group is abelian, it is trivial that the order of $a^rb^s$ is lcm($p,q$).

That should give you enough to determine the possible orders of elements in the product, to do some computations for each order, and to check (by counting elements) that your computations are correct.

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The Chinese remainder theorem can help here: $$G \cong Z_2 \times Z_5 \times Z_3 \times Z_5$$

A tuple $(\bar x_1,\bar x_2,\bar x_3, \bar x_4)$ of order 3, for example, must have $x_1 = 0$, $x_2 = 0$, $x_4 = 0$, since $3$ is invertible mod $2$ and $5$.

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g:=DirectProduct( CyclicGroup(10), CyclicGroup(15) );;

I:=[2,3,5,10];

for i in I do

S:=Size(Filtered(g,x->Order(x)=i));

Print("There are "); Print(S); Print(" element(s) of order "); Print(i); Print("\n");

od;

Result

There are 1 element(s) of order 2
There are 2 element(s) of order 3
There are 24 element(s) of order 5
There are 24 element(s) of order 10

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@Mex: there are just 2 elements of order 3. – Babak S. Jun 13 '12 at 19:54
Well if you are insistent at using a bulldozer to crack a peanut why not do it properly? I'm editing this... feel free to revert. – Nate Iverson Jun 13 '12 at 20:05
@NateIverson: Thanks. I am new in GAP so I did it basically. :) – Babak S. Jun 14 '12 at 4:46
Your original code was not bad, I think you probably already know enough to be dangerous ;) – Nate Iverson Jun 14 '12 at 15:05
Nice, I am relatively new to GAP...used it about a year ago for a bit, but I need to get it up and running again! +1 – amWhy Mar 6 '13 at 0:52