Abstract Nonsense

Crushing one theorem at a time

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<k 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


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December 28, 2010 - Posted by | Algebra, Group Theory, Uncategorized | ,

3 Comments »

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  2. […] not difficult to see that is deserving of it’s name in that it really is an automorphism. Moreover, consider that for any fixed and variable we have […]

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