MHB Abelian Groups of Order $2100$: Elements of Order $210$

[Megus1]

Find all the abelian groups of order $2100.$ For each group, give an example of an element of order $210.$

$2100 = 2^2 \cdot 3 \cdot 5^2 \cdot 7,$ then

$G_1 = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb Z_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7 \cong \mathbb Z_{10} \times \mathbb{Z}_{210}$

$G_2 = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{5^2} \times \mathbb{Z}_7 \cong \mathbb Z_{2} \times \mathbb{Z}_{1050}$

$G_3 = \mathbb{Z}_{2^2} \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7 \cong \mathbb Z_{5} \times \mathbb{Z}_{420}$

$G_4 = \mathbb{Z}_{2^2} \times \mathbb{Z}_3 \times \mathbb{Z}_{5^2} \times \mathbb{Z}_7 \cong \mathbb{Z}_{2100}$

How do I find the elements of order 210? I don't get it well, for example for $\mathbb Z_{1050},$ an element of order 210 is $\dfrac{1050}{210}=5,$ but why?

 

[Amer]

Find all the abelian groups of order $2100.$ For each group, give an example of an element of order $210.$

$2100 = 2^2 \cdot 3 \cdot 5^2 \cdot 7,$ then

$G_1 = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb Z_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7 \cong \mathbb Z_{10} \times \mathbb{Z}_{210}$

$G_2 = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{5^2} \times \mathbb{Z}_7 \cong \mathbb Z_{2} \times \mathbb{Z}_{1050}$

$G_3 = \mathbb{Z}_{2^2} \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7 \cong \mathbb Z_{5} \times \mathbb{Z}_{420}$

$G_4 = \mathbb{Z}_{2^2} \times \mathbb{Z}_3 \times \mathbb{Z}_{5^2} \times \mathbb{Z}_7 \cong \mathbb{Z}_{2100}$

How do I find the elements of order 210? I don't get it well, for example for $\mathbb Z_{1050},$ an element of order 210 is $\dfrac{1050}{210}=5,$ but why?

what is the element "e" order ? it is the least number "n" such theat ne = Identity or if the operation is product e^n = I
when we add 5 to itself 210 we will get 1050 which is equal 0 the identity of $\mathbb Z_{1050},$ . using that we can find an element in
$\mathbb Z_{2} \times \mathbb{Z}_{1050}$
with order 210 which is (0,5)