# Abstract Nonsense

## Review of Group Theory: Homomorphisms (Pt. II)

Point of post: This is a continuation of this post.

Just like with linear transformations we wish to consider not only the maps which preserve the binary operation (and scalar multiplication for linear transformations) but we also would like to consider the bijective ones…the isomorphisms. In particular an isomorphism from $G$ to $G'$ is a homomorphism $\theta:G\to G'$ which is bijective. We call two groups $G$ and $G'$ isomorphic if there exists an isomorphism from $G$ to $G'$ and denote this by $G\cong G'$. Some immediate things about isomorphisms are:

Theorem: Let $G$ and $G'$ be groups and $\theta:G\to G'$ an isomorphism. Then,

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \left|g\right|=\left|\theta(g)\right|\\ &\mathbf{(2)}\quad \cong\textit{ is an equivalence relation}\\ &\mathbf{(3)}\quad G\textit{ is abelian if and only if }G'\textit{ is}\\ &\mathbf{(4)}\quad G\textit{ is cyclic if and only if }G'\textit{ is}\end{aligned}

Proof:

$\mathbf{(1)}$ Note first that

$\left(\theta(g)\right)^{|g|}=\theta\left(g^{|g|}\right)=\theta(e_G)=e_{G'}$

so that, by an earlier theorem we have that $|\theta(g)|\mid |g|$. But, with equal validity

$g^{|\theta(g)|}=\left(\theta^{-1}\left(\theta(g)\right)\right)^{|\theta(g)|}=\theta^{-1}\left(\theta(g)^{|\theta(g)|}\right)=\theta^{-1}\left(e_{G'}\right)=e_G$

so that $|g|\mid |\theta(g)|$. The conclusion follows.

$\mathbf{(2)}$: Evidently $G\cong G$ since $\text{id}_G:G\to G$ is an isomorphism. Also, by part $\mathbf{(4)}$ of the previous theorem we know that if $\theta:G\to G'$ is an isomorphism then $\theta^{-1}:G'\to G$ is an isomorphism, thus $\cong$ is symmetric. Finally, transitivity follows since if $\theta:G\to G',\phi:G'\to G''$ are isomorphisms then $\phi\circ\theta:G\to G''$ is bijective and by $\mathbf{(9)}$ of the last problem we know that $\phi\circ\theta$ is a homomorphism.

$\mathbf{(3)}$: This follows by applying $\mathbf{(7)}$ in the previous problem both ways.

$\mathbf{(4)}$: This comes from applying $\mathbf{(8)}$ in the previous problem both ways.

$\blacksquare$

It turns out though that in the above we’ve proven that the set of isomorphisms from $G\to G$ is a group under composition. We call such an isomorphism an automorphism on $G$ and denote this group by $\text{Aut}(G)$.

Similarly to the definition of a linear transformation we may define the kernel of a homomorphism $\theta\in\text{Hom}\left(G,G'\right)$ to be

$\ker\theta=\left\{g\in G:\theta(g)=e_{G'}\right\}$

we’ll see (in the definition of normal subgroups and the first isomorphism theorem) that the kernel of homomorphism plays a big rule in the study of group theory. But, for now, we’ll have to suffice knowing that it’s always a subgroup of $G$ (this follows from $\mathbf{(4)}$ of the first theorem) and:

Theorem: Let $\theta\in\text{Hom}\left(G,G'\right)$. Then, $\theta$ is injective if and only if $\ker\theta$ is trivial.

Proof: This is clear since $\theta(g)=\theta(h)$ if and only if $\theta(gh^{-1})\theta(g)\theta(h)^{-1}=e_{G'}$. Thus, if $\ker\theta$ is trivial than this would imply that $gh^{-1}=e_G$ or $g=h$. The converse is clearly true since $\theta(e_G)=e_{G'}$ and so $\ker\theta=\theta^{-1}(\{e_{G'}\})=\{e_G\}$. $\blacksquare$

We finish this post with a common theorem, and nice example of isomorphisms:

Theorem: Let $G=\left\langle g\right\rangle$ with $|G|=\kappa$. Then, $G\cong \mathbb{Z}_n$ if $\kappa=n\in\mathbb{N}$ and $G\cong\mathbb{Z}$ if $\kappa=\aleph_0$.

Proof: Assume first that $|G|=n<\infty$. We then define

$\theta:G\to\mathbb{Z}_n:g^k\mapsto k$

This is evidently injective since $g^k\in\ker\theta$ implies that $k=0$ or that $g^k=e$. It’s also surjective since, as was proven in an earlier post the elements of $G$ are precisely $\{g^0,\cdots,g^{n-1}\}$. Thus, it suffices to prove that $\theta$ is a homomorphism. To do this we merely note that

$\theta\left(g^kg^\ell\right)=\theta\left(g^{\ell+k}\right)=\theta\left(g^{\ell+k\text{ mod }n}\right)=\ell+k\text{ mod }n=\theta(\ell)+\theta(k)$

from where there conclusion follows.

Now, assume that $|G|=\aleph_0$. Note firstly that $\theta:\mathbb{Z}\to G:k\mapsto g^k$ is evidently a surjective homomorphism. Moreover, we note that if $\theta(k)=\theta(\ell),\text{ }\ell then $g^{k-\ell}=e$ which contradicts that $|g|=|G|=\aleph_0$. Thus, $\theta$ is, in fact, an isomorphism. $\blacksquare$

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print

December 28, 2010 -