202306041006

Type :Note Tags : Algebra

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Free Abelian Group

Note

A free abelian group of rank $n$ is any group which is the direct sum of $n$ subgroups, each of which is isomorphic to $\mathbb{Z}$, equivalently, it is isomorphic to the additive group ${\mathbb{Z}}^n$.

The rank of such a group is well defined because the ${\mathbb{Z}}^n$‘s are pairwise non isomorphic.

Proposition 1:

Note

${\mathbb{Z}}^n\simeq {\mathbb{Z}}^m$ iff $m=n$.

Proof:

Let $m<n$ and let there be an isomorphism $\varphi :{\mathbb{Z}}^m\to {\mathbb{Z}}^n$, let $A$ be the corresponding matrix for this map. Let $\tilde{\varphi }:{\mathbb{Q}}^m\to {\mathbb{Q}}^n$ be the map corresponding to $A$ but on ${\mathbb{Q}}^m$. Then since $m<n$, this map is not injective, thus there is a vector $v\in {\mathbb{Q}}^n$ which has no preimage. Scaling the vector $v$, we get that there is a vector in ${\mathbb{Z}}^n$ which has no preimage, hence the map $\varphi$ is not injective, that’s a contradiction.

Proposition 2:

Note

Any subgroup of a free abelian group of rank $n$ is free of rank $\le n$.

Proof:

WLOG, let $G=\mathbb{Z}\oplus \mathbb{Z}\oplus \cdots \oplus \mathbb{Z}$ (n times). We will show by induction that $H$ is free of rank $\le n$. For $n=1$, we know any subgroup of $\mathbb{Z}$ is just $m\mathbb{Z}$ for some $m$, in this case we are done. Let it hold for $n-1$. Then let $\pi :G\to \mathbb{Z}$ be the projection onto the first coordinate, let $K$ be the kernel of this map. $K\simeq {\mathbb{Z}}^{n-1}$. Thus $H\cap {\mathbb{Z}}^{n-1}$ is free of rank $\le n-1$. Now $\varphi (H)\subset \mathbb{Z}$ is either $\{0\}$ or infinite cyclic. If it is $\{0\}$, then $H=H\cap K$ and we are done. Otherwise, $\varphi (H)=(\pi (h))$ for some $h\in H$, we will show that $H=\mathbb{Z}h\oplus (H\cap K)$.

Let $x\in H$ be any element, then $x=\frac{\pi (x)}{\pi (h)}h+(x-\frac{\pi (x)}{\pi (h)}h)$ The first term is in $\mathbb{Z}h$, and the second one is in $H\cap K$, thus $H=\mathbb{Z}h\oplus (H\cap K)$ (since it is easy to see that $\mathbb{Z}h\cap H\cap K=\{0\}$).

Proposition 3:

Note

Let $G,H$ be two free abelian groups of rank $n$ with $H\subset G$, then $G/H$ is a finite group.

Proof:

Take a basis for $G$, label it $\mathcal{G}=\{g_1,g_2,\dots g_n\}$. Take a basis for $H$, label it $\mathcal{H}=\{h_1,\dots ,h_n\}$. Now there is an integer matrix $M$ such that $\mathcal{H}=M\mathcal{G}$. Now we can take the vector space $G^{\prime }$ over $\mathbb{Q}$ whose basis is $\mathcal{G}$. In that vector space, $\mathcal{G}$ and $\mathcal{H}$ are both bases, hence $M$ is a base change matrix, hence is invertible.

Now $\det (M)M^{-1}$ is an integer matrix. This means $\det (M)g_i=$ some linear combination of $h_j^{\prime }s$ with integer coefficients. This means $\det (M)G\subset H$. This means $G/H$ is a finite group with size at most $|\det (M){|}^n$.

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