3.3: Isomorphisms
As we have been exploring various groups, I’m sure you’ve noticed that some groups seem to look and behave the same. For example, if we choose the same colors for our arrows and ignore the labels on the vertices, the Cayley diagram for \(D_3\) with generating set \(\{s,s'\}\) looks the same as the Cayley diagram for \(S_3\) with generating set \(\{s_1, s_2\}\) . That is, if we pick the appropriate colors and set the Cayley diagram for \(D_3\) (with generating set \(\{s,s'\}\) ) on top of the Cayley diagram for \(S_3\) (with generating set \(\{s_1, s_2\}\) ) such that the identities match up, then the two Cayley diagrams are identical up to relabeling the rest of the vertices. Figure \(\PageIndex{1}\) should make this clear. This act of matching up the Cayley diagrams establishes a correspondence between the elements of the two groups: \[\begin{aligned} e & \mapsto e\\ s & \mapsto s_1\\ s' &\mapsto s_2\\ ss' &\mapsto s_1s_2\\ s's & \mapsto s_2s_1\\ ss's & \mapsto s_1s_2s_1\end{aligned}\]
Notice that each correspondence is compatible with the correspondence of the generators, namely: \(s \mapsto s_1\) and \(s' \mapsto s_2\) . Given this correspondence, it should not be surprising that the subgroup lattices for \(D_3\) and \(S_3\) have the same structure.
The goal of this section is to formalize this phenomenon by introducing the notion of an isomorphism . First, let’s develop a little more intuition.
If two groups \(G_1\) and \(G_2\) have generating sets \(T_1\) and \(T_2\) such that we can color the edges of the corresponding Cayley diagrams so that the diagrams are identical up to relabeling of the vertices, then we say that there is a matching between \(G_1\) and \(G_2\) . Above, we showed that \(D_3\) and \(S_3\) have a matching. It’s important to emphasize that the existence of a matching between two groups depends on our choice of generating set. If two Cayley diagrams do not look alike, it does not immediately imply that there is not a matching between the groups since it might be the case that choosing different generating sets for the two groups leads to a matching.
Perhaps you’ve noticed that the Cayley diagram for \(R_4\) with generating set \(\{r\}\) looks like the Cayley diagram for the subgroup \(\langle j\rangle=\{\pm 1,\pm j\}\) with generating set \(\{j\}\) in \(Q_8\) . That is, there is a matching between \(R_4\) and \(\langle j\rangle\) , which we’ve depicted in Figure \(\PageIndex{2}\). Similarly, the Cayley diagram for \(S_2\) with generating set \(\{s\}\) looks like the Cayley diagram for the subgroup \(\langle -1\rangle=\{\pm 1\}\) with generating set \(\{-1\}\) in \(Q_8\) . The matching between \(S_2\) and \(\langle -1\rangle\) is depicted in Figure [fig:isoS2] . It’s fairly easy to see that there is also a matching between \(S_2\) and the subgroup \(\langle v\rangle =\{e,v\}\) of \(V_4\) . Since there is a matching between \(S_2\) and \(\langle -1\rangle\) and a matching between \(S_2\) and \(\langle v\rangle\) , there is a matching between \(\langle -1\rangle\) and \(\langle v\rangle\) .
We have seen two different Cayley diagrams for \(D_3\) , one with generating set \(\{s,r\}\) and one with generating set \(\{s,s'\}\) . As Figure \(\PageIndex{1}\) illustrates, there is a matching between \(D_3\) and \(S_3\) that relies on the generating sets \(\{s,s'\}\) and \(\{s_1,s_2\}\) , respectively. Find a different matching between \(D_3\) and \(S_3\) that utilizes the generating set \(\{r,s\}\) for \(D_3\) .
The next theorem follows immediately from the definition of matching.
If there is a matching between \(G_1\) and \(G_2\) using the generating sets \(T_1\) and \(T_2\) , respectively, then \(|G_1|=|G_2|\) and \(T_1\) and \(T_2\) have the same cardinality.
Unfortunately, the converse of the previous theorem is not true in general. That is, two groups that have the same order may or may not have a matching.
Loosely speaking, if two groups have a matching, then the two groups have the same structure and characteristics. In other words, the two groups essentially do the “same kind" of thing. In particular, the corresponding elements in each group have the same characteristics.
On the other hand, if one group has a property that the other does not have, then the two groups cannot have a matching. For example, if one group is abelian and the other is not, then the two groups cannot have a matching. Moreover, for each element \(g\) in one group with the property \(g^k=e\) for some \(k\in \mathbb{Z}\) , there must be a corresponding element in the other group with the same property. Otherwise, there cannot be a matching between the two groups.
Determine whether there is a matching between \(D_4\) and \(\text{Spin}_{1\times 2}\) .
Determine whether there is a matching between \(R_4\) and \(V_4\) .
Determine whether there is a matching between \(D_3\) and \(R_6\) .
Determine whether there is a matching between any pair of the following groups: \(R_8\) (i.e., the group of rotational symmetries of a regular octagon), \(D_4\) , \(Q_8\) .
Consider two light switches on a wall side by side. Consider the group of actions that consists of all possible actions that you can do to the two light switches. For example, one action is toggle the left light switch while leaving the right alone. Let’s call this group \(L_2\) .
- How many distinct actions does \(L_2\) have?
- Can you find a minimal generating set for \(L_2\) ? If so, give these actions names and then write all of the actions of \(L_2\) as words in your generator(s).
- Using your generating set from part (b), draw the corresponding Cayley diagram for \(L_2\) .
- Determine whether there is a matching between \(L_2\) and either of \(R_4\) or \(V_4\) .
Consider three light switches on a wall side by side. Consider the group of actions that consists of all possible actions that you can do to the three light switches. Let’s call this group \(L_3\) . It should be easy to see that \(L_3\) has 8 distinct actions.
- Can you find a minimal generating set for \(L_3\) ? If so, give these actions names and then write all of the actions of \(L_3\) as words in your generator(s).
- Using your generating set from part (a), draw the corresponding Cayley diagram for \(L_3\) .
- Is \(L_3\) cyclic? Briefly justify your answer.
- Is \(L_3\) abelian? Briefly justify your answer.
- Determine whether there is a matching between \(L_3\) and any of \(R_8\) , \(D_4\) , \(\text{Spin}_{1\times 2}\) , or \(Q_8\) .
Suppose \(G\) is a finite group and consider the group table for \(G\) . A coloring for the group table is an assignment of a unique color to each element of the group. For example, Figure \(\PageIndex{4}\) depicts a coloring for the group table of \(V_4\) .
We say that two finite groups have an identical table coloring , if we can arrange the rows and columns of each table and choose colorings for each table so that the pattern of colors is the same for both tables. Clearly, this is only possible if the two groups have the same order. In Problem 2.5.5 , we showed that \(R_4\) and \(V_4\) never have an identical table coloring.
Determine whether \(V_4\) and \(L_2\) have an identical table coloring.
Suppose there is a matching between finite groups \(G_1\) and \(G_2\) . Explain why \(G_1\) and \(G_2\) must have an identical table coloring.
Is the converse of the previous problem true? That is, if \(G_1\) and \(G_2\) are finite groups that have an identical table coloring, will there be a matching between \(G_1\) and \(G_2\) ?
Suppose there is a matching between \(G_1\) and \(G_2\) and suppose \(T_1\) is a generating set for \(G_1\) . Explain why there must be a generating set \(T_2\) for \(G_2\) and an appropriate choice of colors such that the Cayley diagrams for \(G_1\) and \(G_2\) using the generating sets \(T_1\) and \(T_2\) , respectively, are identical up to relabeling of the vertices.
The last few problems have led us to the following theorem.
If \(G_1\) and \(G_2\) are two finite groups, then there is a matching between \(G_1\) and \(G_2\) if and only if \(G_1\) and \(G_2\) have an identical table coloring.
As you’ve likely discovered, matchings and identical table coloring (or the lack thereof) are great for developing intuition about when two groups have identical structure, but the process of finding matchings and identical table colorings is cumbersome. Moreover, it turns out to not be a very useful approach for proving theorems. We need a different approach if we want to develop the general theory any further.
If two finite groups \(G_1\) and \(G_2\) have an identical table coloring, then
the product of corresponding elements yields the corresponding result.
This is the essence of what it means for two groups to have the same structure.
Let’s try to make this a little more precise. Suppose \((G_1,*)\) and \((G_2,\odot)\) are two finite groups that have an identical table coloring and let \(x_1,y_1\in G_1\) . Then these two elements have corresponding elements in the group table for \(G_2\) , say \(x_2\) and \(y_2\) , respectively. In other words, \(x_1\) and \(x_2\) have the same color while \(y_1\) and \(y_2\) have the same color. Since \(G_1\) is closed under its binary operation \(*\) , there exists \(z_1\in G_1\) such that \(z_1=x_1*y_1\) . But then there must exist a \(z_2\in G_2\) such that \(z_2\) has the same color as \(z_1\) . What must be true of \(x_2\odot y_2\) ? Since the two tables exhibit the same color pattern, it must be the case that \(z_2=x_2\odot y_2\) . This is what it means for the product of corresponding elements to yield the corresponding result. Figure \(\PageIndex{5}\) illustrates this phenomenon for group tables.
We can describe the identical table matching between \(G_1\) and \(G_2\) using a function. Let \(\phi:G_1\to G_2\) be the one-to-one and onto function that maps elements of \(G_1\) to their corresponding elements in \(G_2\) . Then \(\phi(x_1)=x_2\) , \(\phi(y_1)=y_2\) , and \(\phi(z_1)=z_2\) . Since \(z_2=x_2\odot y_2\) , we obtain \[\phi(x_1*y_1)=\phi(z_1)=z_2=x_2\odot y_2=\phi(x_1)\odot \phi(y_1).\] In summary, it must be the case that \[\phi(x_1*y_1)=\phi(x_1)\odot \phi(y_1).\] We are now prepared to state a formal definition of what it means for two groups to be isomorphic.
Let \((G_1,*)\) and \((G_2,\odot)\) be two groups. Then \(G_1\) is isomorphic to \(G_2\) , written \(G_1\cong G_2\) , if and only if there exists a one-to-one and onto function \(\phi:G_1\to G_2\) such that \[\label{hom_property} \phi(x*y)=\phi(x)\odot \phi(y).\] The function \(\phi\) is referred to as an isomorphism . Equation \(\PageIndex{3}\) is often referred to as the homomorphic property .
It should be clear from the development that two finite groups are isomorphic if and only if they have an identical table coloring. Moreover, since two finite groups have an identical table coloring if and only if there is a matching between the two groups, it must be the case that two groups are isomorphic if and only if there is a matching between the two groups. The upshot is that we have three different ways to think about what it means for two groups to be isomorphic:
- There exists generating sets for the two groups such that the respective Cayley diagrams are identical up to relabeling of the vertices.
- There exists a choice of colors and an arrangement of the rows and columns of the group tables such that the two tables exhibit the same pattern of colors.
- There exists a bijective function between the two groups that satisfies the homomorphic property.
Using the work that you did earlier in this section, determine which of the following groups are isomorphic to each other: \(S_2\) , \(\langle -1\rangle\) in \(Q_8\) , \(R_3\) , \(R_4\) , \(V_4\) , \(L_2\) , \(\langle i\rangle\) in \(Q_8\) , \(\langle sr, sr^3\rangle\) in \(D_4\) , \(R_5\) , \(R_6\) , \(D_3\) , \(S_3\) , \(R_7\) , \(R_8\) , \(D_4\) , \(\text{Spin}_{1\times 2}\) , \(Q_8\) , \(L_3\) .
Consider the groups \((\mathbb{R},+)\) and \((\mathbb{R}^+,\cdot)\) , where \(\mathbb{R}^+\) is the set of positive real numbers. It turns out that these two groups are isomorphic, but this would be difficult to discover using our previous techniques because the groups are infinite. Define \(\phi:\mathbb{R}\to \mathbb{R}^+\) via \(\phi(r)=e^r\) (where \(e\) is the natural base, not the identity). Prove that \(\phi\) is an isomorphism.
For each of the following pairs of groups, determine whether the given function is an isomorphism from the first group to the second group.
- \((\mathbb{Z},+)\) and \((\mathbb{Z},+)\) , \(\phi(n)=n+1\) .
- \((\mathbb{Z},+)\) and \((\mathbb{Z},+)\) , \(\phi(n)=-n\) .
- \((\mathbb{Q},+)\) and \((\mathbb{Q},+)\) , \(\phi(x)=x/2\) .
Show that the groups \((\mathbb{Z},+)\) and \((2\mathbb{Z},+)\) are isomorphic.
Perhaps one surprising consequence of the previous problem is that when dealing with infinite groups, a group can have a proper subgroup that it is isomorphic to. Of course, this never happens with finite groups.
Once we know that two groups are isomorphic, there are lots of interesting things we can say. The next theorem tells us that isomorphisms map the identity element of one group to the identity of the second group. This was already clear using Cayley diagrams and groups tables, but you should try to prove the theorem directly using Definition: Isomorphic.
Suppose \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) . If \(e_1\) and \(e_2\) are the identity elements of \(G_1\) and \(G_2\) , respectively, then \(\phi(e_1)=e_2\) .
The next theorem tells us that isomorphisms respect inverses.
If \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) , then \(\phi(g^{-1})=[\phi(g)]^{-1}\) .
It turns out that “isomorphic" ( \(\cong\) ) determines an equivalence relation on the class of all possible groups. The next two theorems justify that \(\cong\) is symmetric and transitive.
If \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) , then the function \(\phi^{-1}:G_2\to G_1\) is an isomorphism.
If \(\phi:G_1\to G_2\) and \(\psi:G_2\to G_3\) are isomorphisms from the groups \((G_1,*)\) to \((G_2,\odot)\) and \((G_2,\odot)\) to \((G_3,\star)\) , respectively, then the composite function \(\psi\circ\phi\) is an isomorphism of \(G_1\) and \(G_3\) .
The only thing left to do in order to justify the next theorem is prove that \(\cong\) is reflexive.
If \(\mathcal{G}\) is any nonempty collection of groups, then the relation \(\cong\) is an equivalence relation on \(\mathcal{G}\) .
Mathematicians love to classify things. In particular, mathematicians want to classify groups. One can think of this pursuit as a taxonomy of groups. In order to simplify the task, one can classify isomorphism classes (i.e., the equivalence classes determined by \(\cong\) ) instead of classifying groups. If two groups are isomorphic, then we say that the groups are the same up to isomorphism . If there are \(k\) isomorphism classes of order \(n\) , then we say that there are \(k\) groups of order \(n\) up to isomorphism .
Explain why all groups with a single element are isomorphic. Justify your answer using group tables.
In light of the previous problem, we say that there is one group of order one up to isomorphism.
Suppose that \((G,*)\) is a group of order 2 such that \(G=\{e,a\}\) . Complete the following group table for \(G\) .
\(\begin{array}{l|l|l}
* & e & a \\
\hline e & & \\
\hline a & &
\end{array}\)
Explain why every group of order 2 must be isomorphic to \(S_2\) .
The previous problem implies that up to isomorphism, there is only one group of order 2.
Suppose \((G,*)\) is a group of order 3 such that \(G=\{e,a,b\}\) . Complete the following group table for \(G\) .
\(\begin{array}{l|l|l|l}
* & e & a & b \\
\hline e & & & \\
\hline a & & & \\
\hline b & & &
\end{array}\)
Explain why every group of order 3 must be isomorphic to \(R_3\) .
Suppose \((G,*)\) is a group of order 4 such that \(G=\{e,a,b,c\}\) . Assuming that \(e\) is the identity, the first row and first column of the corresponding group table must be completed as follows.
\(\begin{array}{c|c|c|c|c}
* & e & a & b & c \\
\hline e & e & a & b & c \\
\hline a & a & ? & & \\
\hline b & b & & & \\
\hline c & c & & &
\end{array}\)
The cell with the question mark cannot be filled with an \(a\) . So, this entry must be either \(e\) , \(b\) , or \(c\) . However, it should be easy to see that the cases with \(b\) and \(c\) are symmetric. Thus, there are two cases: (i) the entry with the question mark is filled with \(e\) , or (ii) the entry with the question mark is without loss of generality filled with \(b\) . Complete the group table in each of these two cases. Are either of the resulting groups isomorphic to \(R_4\) or \(V_4\) . What conclusion can you make about groups of order 4?
So far we’ve seen that there are unique groups up to isomorphism of orders 1, 2, and 3, but that there are two groups up to isomorphism of order 4. A natural question to ask is: how many groups are there of order \(n\) ?
In a future chapter we will be able to prove that there is only one group up to isomorphism of order 5, namely those groups isomorphic to \(R_5\) .
We’ve seen three groups of order 6, namely \(R_6\) , \(D_3\) , and \(S_3\) . However, \(D_3\cong S_3\) while \(R_6\) is not isomorphic to either of these. So, we can conclude that there are at least two groups up to isomorphism of order 6. But are there others? It turns out that the answer is no, but why?
The group \(R_7\) is the group of rotational symmetries of a regular 7-sided polygon. This group has order 7. Are there other groups of order 7 that are not isomorphic to \(R_7\) ? It turns out that the answer is no, but why?
We’ve encountered several groups of order 8, namely \(D_4\) , \(\text{Spin}_{1\times 2}\) , \(Q_8\) , \(R_8\) , and \(L_3\) . Of these, only \(D_4\) and \(\text{Spin}_{1\times 2}\) are isomorphic. Thus, there are at least four groups up to isomorphism of order 8. Are these the only isomorphism types? It turns out that there are five groups of order 8 up to isomorphism.
Let’s return to proving some general statements about isomorphisms.
Suppose \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) . If \(G_1\) is cyclic, then \(G_2\) is cyclic.
Is the converse of Theorem \(\PageIndex{7}\) true? That is, if \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) and \(G_2\) is cyclic, is \(G_1\) necessarily cyclic? If the converse is true, then prove it. If the converse is false, provide a counterexample.
Suppose \(\phi:G_1\to G_2\) is an isomorphism from the group \((G_1,*)\) to the group \((G_2,\odot)\) . If \(G_1\) is abelian, then \(G_2\) is abelian.
If \(\phi:G_1\to G_2\) is a function, not necessarily an isomorphism, and \(X\subseteq G_1\) , then the set \[\phi(X):=\{y\in G_2\mid \text{there exists } x\in X\text{ such that }\phi(x)=y\}.\] is called the image of \(X\) . The next theorem tells us that the image of a subgroup under an isomorphism is also a subgroup.
If \(\phi:G_1\to G_2\) is an isomorphism and \(H\leq G_1\) , then \(\phi(H)\leq G_2\) .
Suppose \(G\) is a group and let \(g\in G\) . Define \(\phi_g:G\to G\) via \(\phi_g(x)=gxg^{-1}\) . The map \(\phi_g\) is called conjugation by \(g\) .
If \(G\) is a group and \(g\in G\) , then conjugation by \(g\) is an isomorphism from \(G\) to \(G\) .
Now that you’ve proved the above theorems, it’s a good idea to review the key themes. If you were really paying attention, you may have noticed that in a few of the proofs, we did not use the fact that the function was one-to-one and onto despite assuming that the function was an isomorphism.
For which of the recent theorems could we remove either the assumption that the function is one-to-one or the assumption that the function is onto?
A function that satisfies the homomorphic property and may or may not be one-to-one or onto is called a homomorphism and will be the subject of a future chapter.
What claims can be made about the subgroup lattices of two groups that are isomorphic? What claims can be made about the subgroup lattices of two groups that are not isomorphic? What claims can be made about two groups if their subgroup lattices look nothing alike?