Cosets and Lagrange's Theorem

• May 15th 2009, 06:37 PM
Chris L T521
Cosets and Lagrange's Theorem
Hello,

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

Quote:

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

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

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

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

I would appreciate any suggestions! :D
• May 15th 2009, 07:33 PM
PaulRS

Okay, let us see: $
aH = bH \Leftrightarrow H = \left( {a^{ - 1} b} \right)H \Leftrightarrow a^{ - 1} b \in H
$
or equivalently: $
n\left| {\left( {b - a} \right)} \right.
$
( written in another way: $
b \equiv a\left( {\bmod .n} \right)
$
)

... ;)

There are $n$ distinct left-cosets indeed.
• May 15th 2009, 07:36 PM
Gamma
Yeah, that is exactly right. $H=n\mathbb{Z}$. It is a normal subgroup as $\mathbb{Z}$ is abelian. You will notice that

$\mathbb{Z}/H = \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$

This is precisely how you define the modular numbers. So you can say $[G:H]=n=|\mathbb{Z}_n|$