4.0: Homomorphisms
 Page ID
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When we think of the integers, it is useful not just to think of individual numbers, but relations between numbers; people do this automatically, comparing two numbers to see which is bigger. And if neither number is bigger, we see that the two numbers are equal. In fact, this is a very important way to approach mathematics: We consider not just objects but also how objects are related to one another. So far, we have thought extensively about group as objects. But are there interesting relationships between groups? If so, can we come up with a notion of when two groups are the same?
We can relate groups using special functions between different groups called homomorphisms. There's a way to think of relationships between numbers as functions, too: if you have \(k\) cows and \(p\) chickens, we can try to make a onetoone function from cows to chickens. If it's impossible, we know the number of cows is larger. If there are chickens left over (ie, the function is not onto) then the number of chickens is larger. But if the function is onetoone and onto, then the two numbers \(p\) and \(k\) are equal.
Symmetry and LengthPreserving Functions
Think for a moment of the symmetries of the equilateral triangle. These were given by functions from the triangle back to itself. But not just any functions: the symmetries don't distort the triangle in any way: For example, The center of the triangle never gets moved closer to one of the vertices. In particular, the symmetries are lengthpreserving functions, are known as isometries. We can be quite precise about what a lengthpreserving function is.
Definition 4.0.0: Length Preserving Functions
Let \(d(x,y)\) denote the distance between two points \(x\) and \(y\). Then a function \(f\) is lengthpreserving if \(d(x,y)=d(f(x),f(y))\) for every pair of points \(x, y\). In other words, distances before we apply the function are the same as distances after we apply the operation.
We should look at lots of examples to build up some intuition! Take the integers with the operation of addition. Define \(phi\):
Exercise 4.0.1
Find a function on the interval \([0,1]\) that changes the endpoints but is not a symmetry. In other words, find some \(f: [0,1]\rightarrow [0,1]\) such that \(f(0)=1\), \(f(1)=0\), but \(f\) is not lengthpreserving.
So symmetries are a special kind of function which preserve distances. When we try to relate groups to one another, we use special kinds of functions between the groups.
Homomorphisms
A group is a set with an operation which obeys certain rules. So we'll consider functions that preserve the operation. That is, functions for which it doesn't matter whether we perform our group operation before or after applying the function. More precisely:
Definition 4.0.2: Homomorphism
Let \(G\) and \(H\) be groups, and \(\phi:G \rightarrow H\). Then \(\phi\) is a homomorphism if \(\phi(gh)=\phi(g)\phi(h)\). If a homomorphism is also a bijection, then it is called an isomorphism.
We should look at lots of examples to build up some intuition! Take the integers with the operation of addition. Define \(\phi:\mathbb{Z}\rightarrow\mathbb{Z}\) by \(\phi(n)=2n\). Note that the definition of homomorphism works regardless of the symbol we're using for the group operation, and for \(\mathbb{Z}\) we use addition. Then to show that \(\phi\) is a homomorphism, we need to check that \(\phi(n+m)=\phi(n)+\phi(m)\); the operation before applying \(\phi\) is the same as the operation after applying \(\phi\). So we check! \(\phi(n+m)=2(n+m)\), while \(\phi(n)+\phi(m)=2m+2n\). Since \(2(n+m)=2n+2m\), we see that \(\phi\) is a homomorphism.
We can also have homomorphisms between groups where the operations are written differently! For example, there is a homomorphism between the integers modulo \(n\) (\(mathbb{Z}_n\)) and the \(n\)th roots of unity. Remember that \(mathbb{Z}_n\) is written with an addition operation, while the \(n\)th roots of unity are written with multiplication. We define \(\rho\) by \(\rho(k)=e^{\frac{ik}{2\pi}}\). Then we check that \(\rho\) is a homomorphism! Since the operations are written differently (addition and multiplication), we need to check whether \(\rho(k+j)=\rho(k)\rho(j)\). This isn't so bad: \(\rho(k+j)=e^{\frac{i(k+j)}{2\pi}}\). On the other hand, \(\rho(k)\rho(j)=e^{\frac{i(k)}{2\pi}}e^{\frac{i(j)}{2\pi}}=e^{\frac{i(k+j)}{2\pi}}\). So this is a homomorphism; in fact, it is an isomorphism, since the \(n\)th roots of unity and \(\mathbb{Z}_n\) have the same number of elements.
Isomorphisms are very special homomorphisms. If two groups are isomorphic, it is impossible to tell them apart using just the tools of group theory. True, the two groups may look very different, but they are structurally identical. When we saw the integers modulo \(n\), we saw four different realizations of 'the same' group; they were all isomorphic.
Exercise 4.0.3
Define \(\rho: \mathbb{Z}_n\rightarrow \mathbb{Z}_{2n}\) by \(\rho(x)=2x\) for each \(x\) in \(\mathbb{Z}_n\). Show that \(rho\) is a isomorphism. (Hint: Show that the map is a homomorphism, and argue that the two sets have the same cardinality.)
Exercise 4.0.4
Let \(H\) be a subgroup of \(G\). Define the inclusion \(\iota: H\rightarrow G\) by \(\iota(x)=x\) for each \(x\in H\). Show that \(\iota\) is a homomorphism.
There are many things we can say about homomorphisms with just a little work. We'll prove two basic statements right away.
Proposition 4.0.5 

Let \(\phi: G\rightarrow H\) be a homomorphism. Then:

Proof 4.0.6 


Exercise 4.0.7
 Show that for any homomorphism \(\phi\), we have \(\phi(x^n)=\phi(x)^n\).
 Show that if two finite cyclic groups have the same order, then they are isomorphic.
This tells us that group homomorphisms, in addition to preserving the group operation, also preserve inverses and exponents. Thus, group homorphisms also preserve inverses and exponents!
Contributors
 Tom Denton (Fields Institute/York University in Toronto)