3.3: Isomorphic Groups

One of the key ideas we've discussed in determining whether binary structures are essentially “the same” or “different.” We approach this rigorously using the concept of isomorphic groups.

Definition: Isomorphic and Nonisomorphic

We say that two groups \(G\) and \(G'\) (or binary structures \(S\) and \(S'\)) are isomorphic (or \(G\) is isomorphic to \(G'\)), and write \(G\simeq G'\text{,}\) if there exists an isomorphism from \(G\) to \(G'\text{.}\) We say that \(G\) is isomorphic to \(G'\) via \(\phi\) if \(\phi\) is an isomorphism from \(G\) to \(G'\text{.}\) If there exists no isomorphism from \(G\) and \(G'\text{,}\) then we say that \(G\) and \(G'\) are nonisomorphic (or \(G\) is not isomorphic to \(G'\)), and write \(G\not\simeq G'\text{.}\)

Note

Just because a particular map (even an “obvious” one) from group \(G\) to group \(G'\) is not an isomorphism, we do not know that \(G\) and \(G'\) are not isomorphic! For instance, the map \(\phi: \mathbb{Z}\to \mathbb{Z}\) defined by \(\phi(x)=2x\) for all \(x\) is not an isomorphism (since it's not onto), but \(\mathbb{Z}\) is isomorphic to itself, as we will see in Part 1 of Theorem \(3.3.1\).

Isomorphic groups have the same structure as far as algebraists are concerned. Again, picture two houses that are identical except for the colors they are painted. Though they differ in some ways (one house is red while the other is green), they are structurally identical. Isomorphic groups have identical structures, though the elements of one group may differ greatly from those of the other. Returning to the house analogy: if two houses are structurally identical, we can learn many things about one house by looking at the other (e.g., how many bathrooms it has, whether it has a basement, etc.). Similarly, suppose we know a great deal about group \(G\) and are given a new group, \(G'\text{.}\) If we prove that \(G'\) is isomorphic to \(G\text{,}\) then we can likely deduce information about \(G'\) from the information we know about \(G\text{.}\)

Theorem \(\PageIndex{1}\)

Let \(\langle G,\cdot\rangle\text{,}\) \(\langle G',\cdot'\rangle\text{,}\) and \(\langle G'',\cdot''\rangle\) be groups.

1. Group \(G\) is isomorphic to itself.
2. If \(\phi\) is an isomorphism from \(G\) to group \(G'\text{,}\) then there exists an isomorphism from \(G'\) to \(G\text{.}\) Hence, \(G\simeq G'\) if and only if \(G'\simeq G\text{.}\)
3. If \(G\simeq G'\) and \(G'\simeq G''\text{,}\) then \(G\simeq G''\text{.}\)

Proof

For Part 1: The identity map \(1_G:G→G\) defined by \(1_G(a)=a\) for all \(a∈G\) is clearly an isomorphism.

For Part 2: Since \(ϕ\) is an isomorphism, it's a bijection, hence has inverse \(ϕ^{−1}\). From Theorem \(1.2.1\), we know that \(ϕ^{−1}\) must also be a bijection (in this case, from \(G′\) to \(G\)). So it suffices to show that \(ϕ^{−1}\) is a homomorphism. Let \(a,b∈G′\). We want to show that \(ϕ^{−1}(a \cdot′b)=ϕ^{−1}(a) \cdot ϕ^{−1}(b)\). Since \(ϕ\) is 1-1, it suffices to show that

\(ϕ(ϕ^{−1}(a \cdot ′b))=ϕ(ϕ^{−1}(a) \cdot ϕ^{−1}(b))\).

Notice, we have \(ϕ(ϕ^{−1}(a \cdot ′b))=a \cdot ′b\); further, we have

\(ϕ(ϕ^{−1}(a) \cdot ϕ^{−1}(b))=ϕ(ϕ^{−1}(a)) \cdot ′ϕ(ϕ^{−1}(b))=a \cdot ′b\).

 

This shows that \(ϕ^{−1}(a \cdot ′b)=ϕ^{−1}(a) \cdot ϕ^{−1}(b)\), as desired.

For Part 3: Since \(G \simeq G′\) and \(G′ \simeq G′′\), there exist isomorphisms \(ϕ:G→G′\) and \(ψ:G′→G′′\). Define \(θ:G→G″\) by \(θ=ψ \circ ϕ\). Since \(ϕ\) and \(ψ\) are both bijections, \(θ\) is a bijection (Theorem \(1.2.3\)). Next, let \(a,b∈G\). The

\(\begin{array} &θ(a⋅b)&=ψ(ϕ(a⋅b)) &(\text{by definition of \(θ\)}) \\ &=ψ(ϕ(a) \cdot ′ϕ(b)) &(\text{since \(ϕ\) is a homomorphism}) \\ &=ψ(ϕ(a)) \cdot ′′ψ(ϕ(b)) &(\text{since \(ψ\) is a homomorphism})\\ &=θ(a) \cdot ′′θ(b) (\text{by definition of } θ). \end{array}\)

Thus, \(θ\) is a homomorphism, and hence, since it is also a bijection, an isomorphism. 

 

 

Remark

To show that given groups \(G\) and \(G'\) are isomorphic, we must do three things:

1. Define a function \(\phi\) from \(G\) to \(G'\) (or from \(G'\) to \(G\text{,}\) as we have Theorem \(3.3.1\)).
2. Prove that \(\phi\) is a homomorphism.
3. Prove that \(\phi\) is a bijection.

For Part 1: The identity map \(1_G:G\to G\) defined by \(1_G(a)=a\) for all \(a\in G\) is clearly an isomorphism.

For Part 2: Since \(\phi\) is an isomorphism, it's a bijection, hence has inverse \(\phi^{-1}\text{.}\) From Theorem 1.2.9, we know that \(\phi^{-1}\) must also be a bijection (in this case, from \(G'\) to \(G\)). So it suffices to show that \(\phi^{-1}\) is a homomorphism. Let \(a,b\in G'\text{.}\) We want to show that \(\phi^{-1}(a\cdot'b)=\phi^{-1}(a)\cdot\phi^{-1}(b)\text{.}\) Since \(\phi\) is 1-1, it suffices to show that

\begin{equation*} \phi(\phi^{-1}(a\cdot'b))=\phi(\phi^{-1}(a)\cdot\phi^{-1}(b)). \end{equation*}

Notice, we have \(\phi(\phi^{-1}(a\cdot'b))=a\cdot'b\text{;}\) further, we have

\begin{equation*} \phi(\phi^{-1}(a)\cdot\phi^{-1}(b))=\phi(\phi^{-1}(a))\cdot'\phi(\phi^{-1}(b))=a\cdot'b. \end{equation*}

This shows that \(\phi^{-1}(a\cdot'b)=\phi^{-1}(a)\cdot\phi^{-1}(b)\text{,}\) as desired.

For Part 3: Since \(G\simeq G'\) and \(G'\simeq G''\text{,}\) there exist isomorphisms \(\phi:G\to G'\) and \(\psi:G'\to G''\text{.}\) Define \(\theta:G\to G''\) by \(\theta=\psi \circ \phi\text{.}\) Since \(\phi\) and \(\psi\) are both bijections, \(\theta\) is a bijection (Theorem 1.2.11). Next, let \(a,b\in G\text{.}\) The

\begin{align*} \theta(a\cdot b)\amp =\psi(\phi(a\cdot b))\amp \amp \text{ (by definition of \(\theta\)) }\\ \amp =\psi(\phi(a)\cdot'\phi(b))\amp \amp \text{ (since \(\phi\) is a homomorphism) }\\ \amp =\psi(\phi(a))\cdot''\psi(\phi(b))\amp \amp \text{ (since \(\psi\) is a homomorphism) }\\ \amp =\theta(a)\cdot''\theta(b)\amp \amp \text{ (by definition of \(\theta\)) } . \end{align*}

Thus, \(\theta\) is a homomorphism, and hence, since it is also a bijection, an isomorphism.

Note

Remember, you can show that \(\phi\) is a bijection by proving that it's one-to-one and onto, or by showing that it has an inverse.

Note

Do NOT try to prove that a function \(\phi\) is an isomorphism WITHOUT DEFINING \(\phi\text{!}\)

We know provide some terminology that will be helpful for our study of the structures of groups.

Definition: Unique Group

Given a certain property (or properties), we say there is a unique group with that property (or properties) up to isomorphism if any two groups sharing that property (or properties) are isomorphic to one another.

This may seem a little abstruse at the moment, but seeing examples will help illuminate the concept.

Example \(\PageIndex{1}\)

1. If \(G\) and \(G'\)are groups with \(|G|=|G'|=1\text{,}\) then \(G\simeq G'\text{,}\) since the map from \(G\) to \(G'\) sending \(G\)'s identity (and sole) element to \(G'\)'s identity (and sole) element is clearly an isomorphism. This is why we can mildly abuse terminology and call any group of order 1 the trivial group instead of a trivial group: \(G\) and \(G'\) may technically be different groups, but structurally they are identical, so we can consider them to be more or less “the same.” Thus, there is a unique group of order 1, up to isomorphism.
2. Let \(G\) be a group with \(|G|=2\text{.}\) Then \(G\) must consist of an identity element \(e\) and a nonidentity element \(a\text{,}\) and have the following group table. Compare the respective group tables for \(G\) and for the specific two-element group \(\mathbb{Z}_2\text{.}\)

\(*\)	\(e\)	\(a\)
\(e\)	\(e\)	\(a\)
\(a\)	\(a\)	\(e\)

 

\(+\)	\(0\)	\(1\)
\(0\)	\(0\)	\(1\)
\(1\)	\(1\)	\(0\)

Note that the first table looks exactly like the second table if we replace \(*\) with \(+\text{,}\) each \(e\) with \(0\text{,}\) and each \(a\) with \(1\text{.}\) This shows that groups \(G\) and \(\mathbb{Z}_2\) have identical structures; more precisely, it shows that the function \(\phi\) from \(G\) to \(\mathbb{Z}_2\) defined by \(\phi(e)=0\) and \(\phi(a)=1\) is an isomorphism. Since any group of order \(2\) is isomorphic to \(\mathbb{Z}_2\text{,}\) using Theorem \(3.3.1\) we see that there is a unique group of order \(2\), up to isomorphism.

3. A similar argument shows that there is a unique group of order \(3\) up to isomorphism: specifically, any group of order \(3\) is isomorphic to \(\mathbb{Z}_3\text{.}\)

4. We will see later, in Example \(3.3.6\), that there is not a unique group of order \(4\) up to isomorphism: that is, there are two nonisomorphic groups of order \(4\).

Theorem \(\PageIndex{2}\)

The groups \(\langle \mathbb{R},+\rangle\) and \(\langle \mathbb{R}^+, \cdot\rangle\) are isomorphic.

Proof

Define \(ϕ:\mathbb{R}→\mathbb{R}^+\) by \(ϕ(x)=e^x\). Our map \(ϕ\) is a homomorphism since for every \(x,y∈\mathbb{R}\), we have

\(ϕ(x+y)=e^{x+y}=e^xe^y=ϕ(x)ϕ(y)\).

Moreover, \(ϕ\) is a bijection, since it has inverse function \(\ln x: \mathbb{R}^+→\mathbb{R}\). Hence, \(\mathbb{R}≃\mathbb{R}^+\) via isomorphism \(ϕ\).

Example \(\PageIndex{2}\)

Let \(n\in \mathbb{Z}^+\text{.}\) Then the groups \(\langle n\mathbb{Z},+\rangle\) and \(\langle \mathbb{Z},+\rangle\) are isomorphic. The proof of this is left as an exercise for the reader.

We have now seen examples in which we have proved that two groups are isomorphic. How, though, do we prove that two groups are not isomorphic? It is usually wildly impractical, if not impossible, to check that no function from one group to the other is an isomorphism. For instance, it turns out that \(\mathbb{R}^*\) is not isomorphic to \(GL(2,\mathbb{R})\) (see Example \(3.3.3\)), but there are infinitely many bijections from \(\mathbb{R}^*\) to \(GL(2,\mathbb{R})\)—it is impossible to check that each one is not an isomorphism. Instead, we use group invariants.

Definition: Group Invariant

A group property is called a group invariant if it is preserved under isomorphism.

Group invariants are structural properties. Some examples of group invariants are:

1. Cardinality (since any isomorphism between groups is a bijection);
2. Abelianness (the proof that this is a group invariant is left as an exercise for the reader);
3. Number of elements which are their own inverses (proven by an argument similar to that in Example \(3.3.6\)).

A nonexample of a group invariant is the property of being a subset of \(\mathbb{R}\text{.}\)

Example \(\PageIndex{3}\)

The group \(\mathbb{R}\) cannot be isomorphic to the group \(GL(2,\mathbb{R})\) since the former group is abelian and the latter nonabelian.

Example \(\PageIndex{4}\)

The groups \(\mathbb{R}\) and \(\mathbb{Q}\) cannot be isomorphic since the former group is uncountable and the latter countable.

 Sometimes we must resort to trickier methods in order to decide whether or not two groups are isomorphic.

Example \(\PageIndex{5}\)

The groups \(\mathbb{Z}\) and \(\mathbb{Q}\) are not isomorphic. We use contradiction to prove this. Suppose that \(\mathbb{Z}\) and \(\mathbb{Q}\) are isomorphic via isomorphism \(\phi :\mathbb{Q} \to \mathbb{Z}\text{.}\) Let \(a\in \mathbb{Q}\text{.}\) Then \(\dfrac{a}{2} \in \mathbb{Q}\) with \(\dfrac{a}{2} + \dfrac{a}{2} =a\text{.}\) Then

\begin{equation*} \phi \left(\dfrac{a}{2} \right)+\phi \left(\dfrac{a}{2} \right)=\phi \left(\dfrac{a}{2}+\dfrac{a}{2}\right)=\phi(a); \end{equation*}

since \(\phi \left( \dfrac{a}{2} \right)\) is in \(\mathbb{Z}\text{,}\) \(\phi(a)\) must be evenly divisible by \(2\). But \(a\) was arbitrary in \(\mathbb{Q}\) and \(\phi\) is onto, so this means every element of \(\mathbb{Z}\) must be evenly divisible by \(2\), which is clearly false. Thus, \(\mathbb{Z}\not\simeq \mathbb{Q}\text{.}\)

Example \(\PageIndex{6}\)

The groups \(\mathbb{Z}_4\) and \(V=\mathbb{Z}_2^2\) are not isomorphic. Why? Well, they are both abelian of order \(4\), so we cannot use cardinality or commutativity to prove they are nonisomorphic. The gist of our argument will be to note that every element in \(V\) is its own inverse; so if \(V\) and \(\mathbb{Z}_4\) are isomorphic (hence structurally identical) we must have that every element of \(\mathbb{Z}_4\) is also its own inverse, which doesn't hold (e.g., in \(\mathbb{Z}_4\text{,}\) \(3+3=2\text{,}\) not \(0\)).

A rigorous proof of the fact that \(V\) and \(\mathbb{Z}_4\) are not isomorphic is as follows: For now, denote the usual operation on \(V\) by \(*\text{.}\) Suppose that \(V\) and \(\mathbb{Z}_4\) are isomorphic, via isomorphism \(\phi\) from \(V\) to \(\mathbb{Z}_4\text{.}\) Then since \(\phi\) is onto, there exists an element \(a\in V\) such that \(\phi(a)=3\text{.}\) Then

\(\begin{array} &3+3& =\phi(a)+\phi(a)& \text{ (by definition of \(a\)) }\\ & =\phi(a*a) & \text{ (since \(\phi\) is a homomorphism) }\\ & =\phi((0,0)) & \text{ (since every element of \(V\) is its own inverse) }\\ & =0, & \end{array}\)

since \(\phi\) is a homomorphism, so sends identity element to identity element. But this is a contradiction, since \(3+3=2\neq 0\) in \(\mathbb{Z}_4\text{.}\) Thus, \(V\not\simeq \mathbb{Z}_4\text{.}\)