I have to solve the following problem:
We've got a group: $G =\mathbb Z_{36} \times \mathbb Z_{60}$.
And a subgroup of $G$ called $H = \left< ( 5+36\mathbb Z , 10+60\mathbb Z ) \right>$.
The operation is : $+$
Which is the order of the element $\overline{(4,2)}$ in the group $G/H$ ?
If I'm not wrong, the order of the group G/H, which is written as |G/H|, is equal to |G|/|H| when G is finite (and in this case, I think that |G|=|G1 x G2|=|G1|x|G2|=36*60=2160). But I don't know if this is okay because I think this number is quite big.
Anyway, how can I calculate the order of an element in G/H; in this case the order of the element (4,2) in the group G/H? (*above the 4 and 2 there is a line)?
Best Answer
You can compute the order of an element $g$ in a multiplicative group as the smallest positive integer $k$ such that $g^k=1$.
If the group is a quotient group, say $G/H$, the order is computed the same, but, since $gH=1H$ if and only if $g\in H$, the problem becomes to find the smallest positive integer $k$ such that $g^k\in H$.
With additive notation it's the same, just substitute multiples. Therefore you need to find the smallest positive integer $k$ such that $$ k(4+36\mathbb{Z},2+60\mathbb{Z})\in H $$ Now use the definition of $H$.