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 to is a homomorphism which is bijective. We call two groups and isomorphic if there exists an isomorphism from to and denote this by . Some immediate things about isomorphisms are:
Theorem: Let and be groups and an isomorphism. Then,
Note first that
so that, by an earlier theorem we have that . But, with equal validity
so that . The conclusion follows.
: Evidently since is an isomorphism. Also, by part of the previous theorem we know that if is an isomorphism then is an isomorphism, thus is symmetric. Finally, transitivity follows since if are isomorphisms then is bijective and by of the last problem we know that is a homomorphism.
: This follows by applying in the previous problem both ways.
: This comes from applying in the previous problem both ways.
It turns out though that in the above we’ve proven that the set of isomorphisms from is a group under composition. We call such an isomorphism an automorphism on and denote this group by .
Similarly to the definition of a linear transformation we may define the kernel of a homomorphism to be
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 (this follows from of the first theorem) and:
Theorem: Let . Then, is injective if and only if is trivial.
Proof: This is clear since if and only if . Thus, if is trivial than this would imply that or . The converse is clearly true since and so .
We finish this post with a common theorem, and nice example of isomorphisms:
Theorem: Let with . Then, if and if .
Proof: Assume first that . We then define
This is evidently injective since implies that or that . It’s also surjective since, as was proven in an earlier post the elements of are precisely . Thus, it suffices to prove that is a homomorphism. To do this we merely note that
from where there conclusion follows.
Now, assume that . Note firstly that is evidently a surjective homomorphism. Moreover, we note that if then which contradicts that . Thus, is, in fact, an isomorphism.
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