Hello,

I'm am stuck on this problem (or so I think) XD

Source:Let $\displaystyle n$ be a positive integer. Let $\displaystyle H=\{0,\pm n,\pm2n,\pm3n,\dots\}$. Find all left cosets of $\displaystyle H$ in $\displaystyle \mathbb{Z}$. How many are there?Contemporary Abstract Algebra, 6e, by J. Gallian.

Since $\displaystyle n\in\mathbb{Z}^+$, I thought that it would be safe to say that the left cosets would be

$\displaystyle \begin{aligned}

0+H & = \{0,\pm n,\pm2n,\pm 3n,\dots\}= nk+H,\,k\in\mathbb{Z}^+\\

1+H & = \{\dots,1-2n,1-n,1,1+n,1+2n,\dots\}=nk+1+H,k\in\mathbb{Z}^+\\

2+H & = \{\dots,2-2n,2-n,2,2+n,2+2n,\dots\}=nk+2+H,k\in\mathbb{Z}^+\\

& \,\vdots\\

\left(n-1\right)+H & = \{\dots,-2n-1,-n-1,-1,n-1,2n-1,\dots\}=2nk-1+H,k\in\mathbb{Z}^+

\end{aligned}$

From this, it would appear that there are $\displaystyle n$ distinct left cosets. But is there another way to verify that this is the case? Lagrange's Theorem wouldn't apply to this, since $\displaystyle \left|n\mathbb{Z}\right|$ and $\displaystyle \left|\mathbb{Z}\right|$ are infinite.

Does this look right, or am I way over my head? XD

I would appreciate any suggestions!