Finding the order of an element in a group

Question

Let $N$ ≤ ($\mathbb Z^2$, +) be the subgroup given by $\;N =\bigl\{ (x, y) ∈ \mathbb Z^2\mid 3x+5y ≡ 0 \pmod{30}\bigr\}$ .

In the quotient $G = \mathbb Z^2/N$, find the order of the element $(1, 1) + N$.

I've came across the problem and I don't know where and how to start. Please any help?

Answer 1

The order of an element $E$ is the least positive integer $n$ such that $nE=0$. So we’re looking for the least $n$ such that $n(1,1)$ gets sent to zero, that is the least $n$ such that $(n,n)$ is in the kernel $N$. Thus we have the equation $3n+5n \equiv 0$ mod $30$, solving this we want the least $n$ such that $8n$ is a multiple of $30$. By unique factorization, we know that for $8n$ to be a multiple of $30$, n must have $15$ as a factor. Trying $n=15$ solves our equation so our least solution is $n=15$.

Answer 2

First off, note that $30 \cdot ((1,1) + N) = (30,30) + N$. Now, $3\cdot 30 + 5\cdot 30 = 8\cdot 30 \equiv 0 \text{ mod } 30$, so $(30,30) \in N$ which tells us $30 \cdot ((1,1) + N) = N$.

So $(1,1) + N$ has order at most $30$, but remember that if $k\cdot g = 0$ in any group, then this means that the order of the element $g$ divides $k$. Thus, the only possible orders for $(1,1) + N$ are the divisors of $30$.

Then you can start multiplying $(1,1) + N$ by the divisors of $30$. Which will give you zero?

Answer 3

The "n-th power" of that coset in the quotient is $(n,n)+N$. Now, $(n,n)+N=N$ iff $8n \equiv 0 \pmod {30}$. The least value of $n$ fulfilling this equation is $15$.