Group Homomorphism is Continuous if the Preimage of Every Neighborhood of 0 is a Neighborhood of 0

Homomorphisms and Isomorphisms

We've looked at groups defined by generators and relations. We've also developed an intuitive notion of what it means for two groups to be the same. This sections will make this concept more precise, placing it in the more general setting of maps between groups.

Homomorphisms

A homomorphism is a map between two groups which respects the group structure. More formally, let G and H be two group, and f a map from G to H (for every g∈G, f(g)∈H). Then f is a homomorphism if for every g₁,g₂∈G, f(g₁g₂)=f(g₁)f(g₂). For example, if H<G, then the inclusion map i(h)=h∈G is a homomorphism. Another example is a homomorphism from Z to Z given by multiplication by 2, f(n)=2n. This map is a homomorphism since f(n+m)=2(n+m)=2n+2m=f(n)+f(m).

Activity 1: A treasure trove of maps

1. Suppose f:G→H is a homomorphism between two groups, with the identity of G denoted e_G and the identity of H denoted e_H. Show that f(e_G)=e_H, that is, identity is sent to identity by any homomorphism. (Hint: use the fact that e=ee and the defining property of homomorphisms.)
2. Consider the map f:Z₉→Z₃ given by f(R_m)=R_3m (recall that R_m is a counterclockwise rotation by m degrees). Is this a homomorphism?
3. Find a homomorphism from Z₆ to Z₃.
4. Is the map f:Z₆→Z₅ given by f(R_m)=R₀ (the identity) a homomorphism?
5. Find a homomorphism from F₂ to Z×Z. (Hint: why not try sending the generators of one to the generators of the other?)

The last part of the above activity hints at a key fact: a homomorphism is determined by what elements it sends the generators to. Suppose the two elements a,b∈G generate G. Then for any group H and homomorphism f:G→H, f(aba^-1bb)=f(a)f(b)f(a)^-1f(b)f(b). Thus we need only specify f(a) and f(b) to define the homomorphism for every element in G.

Activity 2: Extending homomorphisms

1. Prove the equality above by showing that for any homomorphism f and elements a,b,c, f(abc)=f(a)f(b)f(c) and f(a^-1)=f(a)^-1.
2. Find all homomorphisms from Z to Z and from F₂ to Z₃×Z₃. (Hint: use generators.)

The approach of picking where generators of a group go and then "extending" the homomorphism to the rest of the group very often comes in handy. However, this can only be done when the elements the generators are sent to satisfy all the relations between the generators themselves. This is a key point, which the following problems will, I hope, make clear.

3. Recall that Z₃ is generated by R₁₂₀. Suppose we try to define a homomorphism f:Z₃→Z by letting f(R₁₂₀)=1, sending a generator to a generator. Does this extend to a homomorphism? What relation does R₁₂₀ satisfy that 1∈Z does not?
4. There are many homomorphisms from F₂ to Z×Z. Take for instance f(a)=(1,0) and f(b)=(0,2). What are some homomorphisms from Z×Z to F₂? (Hint: suppose f is such a homomorphism and f((1,0))=w₁, f((0,1))=w₂, where w₁ and w₂ are words in F₂. What relation must w₁ and w₂ satisfy?)
5. If you're itching for a challenge, try to find all the homomorphisms from Z×Z to F₂. What do they have in common?

Putting the above idea into the language of Cayley graphs, we get that if f:G→H is a homomorphism, and elements a and b generate G, then any loops in the Cayley graph of G with respect to generators a and b must be sent to similarly oriented (possibly trivial) loops in the Cayley graph of H with respect to generators f(a) and f(b). Informally, the Cayley graph of H (with respect to f(a) and f(b)) may have "additional" loops to those in the Cayley graph of G, but it may not have fewer. Here's a picture explanation to make this more clear.

Give a homomorphism f:G→H, the set of all elements h∈H such that h=f(g) for some g∈G is called the image of G and is denoted f(G). It is the set of elements in H which f maps some element of G to. If f(G)=H, we say that f is surjective or onto. Similarly, we denote by f^-1(h) all the elements in G which f maps to h. For example, the homomorphism f:Z₆→Z₃ given by f(R_m)=R_2m is a surjective homomorphism and f^-1(R₁₂₀)={R₆₀,R₂₄₀}.

Activity 3: Two kernels of truth

1. Suppose f:G→H is a homomorphism, e_G and e_H the identity elements in G and H respectively. Show that the set f^-1(e_H) is a subgroup of G. This group is called the kernel of f. (Hint: you know that e_G∈f^-1(e_H) from before. Use the definition of a homomorphism and that of a group to check that all the other conditions are satisfied.)
2. If f^-1(e_H)={e_G}, only the identity of G, how many elements does the set f^-1(h) have for any other h∈H? (Hint: suppose v,w∈G and f(v)=f(w)=h. What is f(vw^-1)? Use the definition of a homomorphism.)

For f:G→H a homomorphism, if f^-1(identity) has only one element—and by problem 2 above we know that this means that f maps each element of G to a distinct element of H—then we say that f is injective or one-to-one. A homomorphism which is both injective and surjective is called an isomorphism, and in that case G and H are said to be isomorphic.

Activity 4: Isomorphisms and the normality of kernels

1. Find all subgroups of the group D₄. Which of them are isomorphic? Which are normal? (Hint: use the Cayley graph and/or product table to help you.)
2. Show that for any two groups G and H, the kernel of any homomorphism f:G→H is a normal subgroup of G. (Hint: Call the kernel K. Consider the image of g^-1Kg under f, i.e. the set f(g^-1Kg) for some g∈G. What does it equal?)

Activity 5: The final proof

Given a surjective homomorphism f:G→H, let K be it's kernel. Show that the quotient group G/K is isomorphic to H. (Hint: first construct a homomorphism q from G/K to H, and then show that it's surjective and injective. You have only the given homomorphism f to work with, so why not try q(gK)=f(g)? Is this a homomorphism? Is it injective and surjective?)

This theorem is called the first isomorphism theorem, and it will serve as a concluding note in this brief introduction to group theory.

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Source: https://pi.math.cornell.edu/~mec/2008-2009/Victor/part6.htm

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