9.3 Group homomorphism and Isomorphism

After studying sets, we looked at how functions allows us to go from one set to another. A group is a set with some extra structure, and so to learn about groups we want to consider functions between groups which take account of this structure somehow.

A “group homomorphism” is a function between two groups that links the two group operations in the following way.

Definition 9.12:
Let \((G,\ast)\) and \((H,\bullet)\) be groups. A group homomorphism is a function \(\varphi:G\to H\) such that \[\varphi(x\ast y)=\varphi(x)\bullet \varphi(y)\] for all \(x,y\in G\).
Remark:
If the groups \(G\) and \(H\) are written multiplicatively, then \(\varphi:G\to H\) is a homomorphism if \[\varphi(xy)=\varphi(x)\varphi(y)\] for all \(x,y\in G\). But it should be noted that on the left hand side of that equation we multiply \(x\) and \(y\) in \(G\), but on the right hand side we multiply \(\varphi(x)\) and \(\varphi(y)\) in \(H\), so this still links the group operations of different groups.
Remark:
If you are doing Linear Algebra, then you may find it helpful to note a similarity between the definitions of a group homomorphism and a linear map. These are both functions that “commute with the operations” in the sense that the definition of a homomorphism says that multiplying two elements and then applying the homomorphism gives the same as applying the homomorphism to the two elements and then multiplying the resulting elements, and the definition of a linear map says the same for the operations of addition and multiplication by scalars in place of the group operation. In fact, many of the basic facts about homomorphisms are very similar to basic facts about linear maps, usually with pretty much the same proof.
Example:
If \(G\) and \(H\) are groups and \(e_H\) is the identity element of \(H\), the function \(\varphi:G\to H\) given by \[\varphi(x)=e_H\] for all \(x\in G\) is a homomorphism (the trivial homomorphism).
Example:
If \(H\leq G\) are groups, then the inclusion map \(i:H\to G\) given by \[i(x)=x\in G\] for all \(x\in H\) is a homomorphism. This is injective but not surjective (unless \(H=G\)).

If we have a group homomorphism, there are certain things we can deduce:

Lemma 9.13:

Let \(\varphi:G\to H\) be a homomorphism, let \(e_G\) and \(e_H\) be the identity elements of \(G\) and \(H\) respectively, and let \(x\in G\). Then

  1. \(\varphi(e_G)=e_H\),

  2. \(\varphi(x^{-1})=\varphi(x)^{-1}\),

  3. \(\varphi(x^i)=\varphi(x)^i\) for any \(i\in\mathbb{Z}\).

Proof.

We go through all three statements.

  1. We have \(\varphi(e_G)=\varphi(e_Ge_G)=\varphi(e_G)\varphi(e_G)\), so by uniqueness of the identity \(\varphi(e_G)=e_H\).

  2. We have \(e_H=\varphi(e_G)=\varphi(xx^{-1})=\varphi(x)\varphi(x^{-1})\), so by uniqueness of inverses \(\varphi(x^{-1})=\varphi(x)^{-1}\).

  3. This is true for positive \(i\) by a simple induction (it is true for \(i=1\), and if it is true for \(i=k\) then \[\varphi(x^{k+1})=\varphi(x^k)\varphi(x)=\varphi(x)^k\varphi(x)=\varphi(x)^{k+1},\] so it is true for \(i=k+1\)). Then by b.) it is true for negative \(i\), and by a.) it is true for \(i=0\).

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As we have seen, bijective functions allows us to treat two sets of the same cardinality as essentially the same sets, just changing the name of the elements. A similar story occurs with groups. Suppose \(G=\langle x\rangle=\{e,x,x^2\}\) and \(H=\langle y\rangle=\{e,y,y^2\}\) are two cyclic groups of order \(3\). Strictly speaking they are different groups, since (for example) \(x\) is an element of \(G\) but not of \(H\). But clearly they are “really the same” in some sense: the only differences are the names of the elements, and the “abstract structure” of the two groups is the same. To make this idea, of two groups being abstractly the same, precise, we introduce the idea of an isomorphism of groups.

Definition 9.14:
Let \((G,\ast)\) and \((H,\bullet)\) be groups. An isomorphism from \(G\) to \(H\) is a bijective homomorphism. That is, a bijective function \(\varphi:G\to H\) such that \[\varphi(x\ast y)=\varphi(x)\bullet\varphi(y)\] for all elements \(x,y\in G\).
Remark:
As when we defined homomorphisms, we used different symbols for the two group operations to point out that the isomorphism links the two different operations. As ever, we’ll usually write the groups multiplicatively, in which case the defining property of an isomorphism becomes \[\varphi(xy)=\varphi(x)\varphi(y),\] but again it should be stressed that this involves two different kinds of “multiplication”: on the left hand side of the equation we are multiplying in \(G\), but on the right hand side in \(H\).
Etymology:

Homomorphism comes from the Greek homo to mean “same” and morph to mean “shape”. Isomorphism comes from the Greek iso to mean “equal” (and morph to mean “shape”).

A homomorphism is a map between two groups that keeps the same shape, i.e.  preserves the group operation (but not necessarily anything else), while an isomorphism is a map between two groups that have equal shape (it preserves the group operation and many other useful properties as we will see below).

Example:
Let \(G=\langle x\rangle\) and \(H=\langle y\rangle\) be two cyclic groups of the same order. Then \(\varphi:G\to H\) defined by \(\varphi(x^i)=y^i\) for every \(i\in\mathbb{Z}\) is an isomorphism, since it is a bijection and \[\varphi(x^{i+j})=y^{i+j}=y^iy^j=\varphi(x^i)\varphi(x^j)\] for all \(i,j\).
Remark:
Since \(\varphi\) is a bijection, it pairs off elements of \(G\) with elements of \(H\), and then the defining property of an isomorphism says that we can use \(\varphi\) and its inverse \(\varphi^{-1}\) as a dictionary to translate between elements of \(G\) and elements of \(H\) without messing up the group operation. If we take the multiplication (Cayley) table of \(G\) and apply \(\varphi\) to all the entries, we get the multiplication table of \(H\): the groups \(G\) and \(H\) are “really the same” apart from the names of the individual elements.

Next we’ll prove some of the easy consequences of the definition.

Proposition 9.15:

Let \(\varphi:G\to H\) be an isomorphism between (multiplicatively-written) groups. Then \(\varphi^{-1}:H\to G\) is also an isomorphism.

Proof.

Since \(\varphi\) is a bijection, it has an inverse function \(\varphi^{-1}\) that is also a bijection.

Let \(u,v\in H\). Since \(\varphi\) is a bijection, there are unique elements \(x,y\in G\) with \(u=\varphi(x)\) and \(v=\varphi(y)\). Then \[\varphi^{-1}(uv)=\varphi^{-1}\left(\varphi(x)\varphi(y)\right) = \varphi^{-1}\left(\varphi(xy)\right)=xy=\varphi^{-1}(u)\varphi^{-1}(v),\] and so \(\varphi^{-1}\) is an isomorphism.

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Because of this Proposition, the following definition makes sense, since it tells us that there is an isomorphism from \(G\) to \(H\) if an only if there is one from \(H\) to \(G\).

Definition 9.16:
Two groups \(G\) and \(H\) are said to be isomorphic, or we say \(G\) is isomorphic to \(H\), if there is an isomorphism \(\varphi:G\to H\), and then we write \(G\cong H\), or \(\varphi:G\cong H\) if we want to specify an isomorphism.
Proposition 9.17:

Let \(G,H,K\) be three groups. If \(G\) is isomorphic to \(H\) and \(H\) is isomorphic to \(K\), then \(G\) is isomorphic to \(K\).

Proof.

Let \(\varphi:G\to H\) and \(\theta:H\to K\) be isomorphisms. Then the composition \(\theta\varphi:G\to K\) is a bijection and if \(x,y\in G\) then \[\theta\left(\varphi(xy)\right)=\theta\left(\varphi(x)\varphi(y)\right) =\theta\left(\varphi(x)\right)\theta\left(\varphi(y)\right),\] and so \(\theta\varphi\) is an isomorphism.

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Proposition 9.18:

Let \(\varphi:G\to H\) be an isomorphism between (multiplicatively-written) groups, let \(e_G\) and \(e_H\) be the identity elements of \(G\) and \(H\) respectively, and let \(x\in G\). Then

  1. \(\varphi(e_G)=e_H\),

  2. \(\varphi(x^{-1})=\varphi(x)^{-1}\).

  3. \(\varphi(x^i)=\varphi(x)^i\) for every \(i\in\mathbb{Z}\).

  4. \(\operatorname{ord}_G(x)=\operatorname{ord}_H\left(\varphi(x)\right)\).

Proof.

The first three statements follows directly from Proposition 9.13 since an isomorphism is an homomorphism.

For the last statement, by a.) and c.), we have \(x^i=e_G\Leftrightarrow\varphi(x)^i=e_H\), and so \(\operatorname{ord}_G(x)=\operatorname{ord}_H\left(\varphi(x)\right)\).

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To prove that two groups are isomorphic usually requires finding an explicit isomorphism. Proving that two groups are not isomorphic is often easier, as if we can find an “abstract property” that distinguishes them, then this is enough, since isomorphic groups have the same “abstract properties”. We’ll make this precise, and prove it, with some typical properties, after which you should be able to see how to give similar proofs for other properties, just using an isomorphism to translate between properties of \(G\) and of \(H\).

Proposition 9.19:

Let \(G\) and \(H\) be isomorphic groups. Then \(|G|=|H|\).

Proof.

This follows just from the fact that an isomorphism is a bijection \(\varphi:G\to H\).

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Proposition 9.20:

Let \(G\) and \(H\) be isomorphic groups. If \(H\) is abelian then so is \(G\).

Proof.

Suppose that \(\varphi:G\to H\) is an isomorphism and that \(H\) is abelian. Let \(x,y\in G\). Then \[\varphi(xy)=\varphi(x)\varphi(y)=\varphi(y)\varphi(x)=\varphi(yx),\] since \(\varphi(x),\varphi(y)\in H\), which is abelian. Since \(\varphi\) is a bijection (and in particular injective) it follows that \(xy=yx\). So \(G\) is abelian.

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Proposition 9.21:

Let \(G\) and \(H\) be isomorphic groups. If \(H\) is cyclic then so is \(G\).

Proof.

Suppose \(\varphi:G\to H\) is an isomorphism and \(H=\langle y\rangle\) is cyclic. So every element of \(H\) is a power of \(y\). So if \(g\in G\) then \(\varphi(g)=y^i\) for some \(i\in\mathbb{Z}\), and so \(g=\varphi^{-1}(y)^i\). So if \(x=\varphi^{-1}(y)\) then every element of \(G\) is a power of \(x\), and so \(G=\langle x\rangle\).

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Notation:

If \(n\in \mathbb{Z}_+\), we write \(C_n\) to mean a (multiplicatively-written) cyclic group of order \(n\).

This notation is sensible since a cyclic group can only be isomorphic to another cyclic group of the same order, and since two cyclic groups of the same order are isomorphic.

Remark:

We can use this proposition to show that there exists groups of the same order which are not isomorphic. Consider the groups \(C_8\) and \(D_8\), both have order \(8\), however we have seen \(D_8\) is not abelian, and hence not cyclic.

This again highlights that a group is a set with a binary operation: a bijection exists between the sets \(C_8\) and \(D_8\), but it does not preserve the group operations.
Proposition 9.22:

Let \(G\) and \(H\) be isomorphic groups and \(n\in \mathbb{Z}\cup \{\infty\}\). Then \(G\) and \(H\) have the same number of elements of order \(n\).

Proof.

By Proposition 9.18, an isomorphism \(\varphi:G\to H\) induces a bijection between the set of elements of \(G\) with order \(n\) and the set of elements of \(H\) with order \(n\).

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The idea of isomorphism gives an important tool for understanding unfamiliar or difficult groups. If we can prove that such a group is isomorphic to a group that we understand well, then this is a huge step forward.

The following example of a group isomorphism was used for very practical purposes in the past.

Example:

The logarithm function \[\log_{10}:(\mathbb{R}_+,\cdot)\to(\mathbb{R},+)\] is an isomorphism from the group of positive real numbers under multiplication to the group of real numbers under addition (of course we could use any base for the logarithms). It is a bijection since the function \(y\mapsto 10^y\) is the inverse function, and the group isomorphism property is the familiar property of logarithms that \[\log_{10}(ab)=\log_{10}(a)+\log_{10}(b).\] Now, if you don’t have a calculator, then addition is much easier to do by hand than multiplication, and people used to use “log tables” to make multiplication easier. If they wanted to multiply two numbers, they would look up the logarithms, add them and then look up the “antilogarithm”.

In group theoretic language they were exploiting the fact that there is an isomorphism between the “difficult” group \((\mathbb{R}_+,\cdot)\) and the “easy” group \((\mathbb{R},+)\).