7: Isomorphism of Groups

Two groups may look different yet be essentially the same. This concept of sameness is formalized in mathematics by the concept of isomorphism (from the Greek: isos meaning the same and morphe meaning form). Before we give a precise definition of isomorphism, let’s look at some small groups and see if we can see whether or not they meet our intuitive notion of sameness.

Problem 7.1 Go through the examples of groups we have covered so far and make a list of all those with order \(\le 12\). List them according to their orders. For example, the following groups have order 6: \[GL(2,\mathbb{Z}_2),\quad \mathbb{Z}_6, \quad S_3, \quad U_7, \quad U_9, \quad \mathbb{Z}_2 \times \mathbb{Z}_3.\] Make a similar list for all integers from 1 to 12.

Definition 7.1:

Let \(G = \{ g_1, g_2, \dots, g_n \}\). Let \(o(g_i) = k_i\) for \(i = 1, 2, \dots, n\). We say that the sequence \((k_1, k_2, \dots, k_m)\) is the order sequence of the group \(G\). To make the sequence unique we assume that the elements are ordered so that \(k_1 \le k_2 \le \dots \le k_n\).

For example, the order sequence of \(S_3\) is \((1,2,2,2,3,3)\).

Problem 7.2 Consider the following list of properties that may be used to distinguish groups.

1. The order of the group.
2. The order sequence of the group.
3. Whether the group is abelian or not.

Look carefully at the groups in the list you made for the previous problem and see which may be distinguished by one or more of the three listed properties.

Definition 7.2:

Let \((G,*)\) and \((H,\bullet)\) be groups. A function \(f:G \to H\) is said to be a homomorphism from \(G\) to \(H\) if \[f(a*b) =f(a) \bullet f(b)\] for all \(a,b \in G\). If in addition \(f\) is one-to-one and onto, \(f\) is said to be an isomorphism from \(G\) to \(H\).

We say that \(G\) and \(H\) are isomorphic if and only if there is an isomorphism from \(G\) to \(H\). We write \(G \cong H\) to indicate that \(G\) is isomorphic to \(H\).

Examples 7.1 Some familiar examples of homomorphisms and isomorphisms are:

1. \(\det: GL(2,\mathbb{R}) \to \mathbb{R}-\{0 \}\) is a homomorphism since \[\det(AB) = \det(A)\det(B)\] for all \(A,B \in GL(2,\mathbb{R})\).
2. \(\textrm{sign}: S_n \to \{1,-1 \}\) is a homomorphism since \[\textrm{sign}(\sigma\tau) = \textrm{sign}(\sigma)\textrm{sign}(\tau)\] for all \(\sigma, \tau \in S_n\).
3. \(\log: \mathbb{R}^+ \to \mathbb{R}\), where \(\mathbb{R}^+\) denotes the positive real numbers and the operations are respectively multiplication and addition, is an isomorphism since from calculus we know that \(\log\) is one-to-one and onto and \[\log(xy) = \log(x) + \log(y)\] for all positive real numbers \(x\) and \(y\). [Here \(\log(x) = \ln(x)\).]
4. \(\exp: \mathbb{R}\to \mathbb{R}^+\), where \(\mathbb{R}^+\) denotes the positive real numbers and the operations are respectively addition and multiplication, is an isomorphism since from calculus we know that \(\exp\) is one-to-one and onto and \[\exp(x+y) = \exp(x)\exp(y)\] for all real numbers \(x\) and \(y\). Note that \(\exp(x)\) is an alternative notation for \(e^x\).

The last two examples show that the group of positive real numbers under multiplication is isomorphic to the group of all real numbers under addition.

Theorem \(\PageIndex{1}\) (Isomorphism is an Equavalence Relation)

If \(G\), \(H\) and \(K\) are groups then

1. \(G \cong G\),
2. If \(G \cong H\) then \(H \cong G\), and
3. If \(G \cong H\) and \(H \cong K\), then \(G \cong K\).

Problem 7.3 Prove Theorem 7.1.

Problem 7.4 Prove that every group of order 1 is isomorphic to the group \(U_2\).

Problem 7.5 Prove that every group of order 2 is isomorphic to the group \(\mathbb{Z}_2\).

Problem 7.6 Prove that every group of order 3 is isomorphic to the group \(\mathbb{Z}_3\).

Problem 7.7 Prove that if \(G\) and \(H\) are isomorphic groups then \(|G| = |H|\).

Problem 7.8 Prove that if \(G\) and \(H\) are groups then \(G \times H \cong H \times G\).

Theorem \(\PageIndex{2}\)

Let \((G,*)\) and \((H,\bullet)\) be groups and let \(f:G \to H\) be a homomorphism. Let \(e_G\) denote the identity of \(G\) and, \(e_H\) denote the identity of \(H\). Then

1. \(f(e_G) = e_H\),
2. \(f(a^{-1}) = f(a)^{-1}\), and
3. \(f(a^n) = f(a)^n\) for all \(n \in \mathbb{Z}\).

Problem 7.9 Prove parts 1 and 2 of Theorem 7.2.

Problem 7.10 Prove part 3 of Theorem 7.2 for \(n=2,-2,3,-3\).

The general case of Theorem 7.2 can be proved by induction on \(n\). We will come back to this later.

Problem 7.11 Restate Theorem 7.2 (a) in the case that both \(G\) and \(H\) use additive notation, (b) in the case where \(G\) uses additive notation and \(H\) uses multiplicative notation, and (c) in the case where \(G\) uses multiplicative notation and \(H\) uses additive notation.

Theorem \(\PageIndex{3}\)

Let \((G,*)\) and \((H,\bullet)\) be groups and let \(f:G \to H\) be an isomorphism. Then \(\mathrm{o}(a) = \mathrm{o}(f(a))\) for all \(a \in G\). It follows that \(G\) and \(H\) have the same number of elements of each possible order.

Problem 7.12 Prove Theorem 7.3. Hint: Use the Theorem 2.

Theorem \(\PageIndex{4}\)

If \(G\) and \(H\) are isomorphic groups, and \(G\) is abelian, then so is \(H\).

Problem 7.13 Prove Theorem 7.4.

Definition 7.3:

A group \(G\) is cyclic if there is an element \(a \in G\) such that \(\langle a \rangle = G\). If \(\langle a \rangle = G\) then we say that \(a\) is a generator for \(G\).

Problem 7.14 Find an example of a cyclic group that has more than one generator.

Theorem \(\PageIndex{5}\)

If \(G\) and \(H\) are isomorphic groups and \(G\) is cyclic then \(H\) is cyclic.

Problem 7.15 Prove Theorem 7.5.

Problem 7.16 Determine which of the following groups are cyclic and which are not cyclic.

1. \(\mathbb{Z}\) under ordinary addition.
2. \(\mathbb{Z}_n\) under addition modulo \(n\).
3. \(U_n\) for \(n \le 12\).
4. \(S_3\).
5. \(\mathbb{Z}_2 \times \mathbb{Z}_3\).
6. \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
7. \(\mathbb{Z}_3 \times \mathbb{Z}_5\).
8. \(A_3\).
9. \(S_4\).
10. \(GL(2,\mathbb{Z}_2)\).

Problem 7.17 [Challenge Problem] Prove that if \(G\) is a finite cyclic group of order \(n\) then \(G\) has \(\phi(n)\) generators. Hint: Let \(G = \langle a \rangle\). Show than an element \(b = a^i \in G\) is a generator of \(G\) if and only if \(\gcd(i,n) = 1\).

Theorem \(\PageIndex{6}\)

Let \(a\) be an element of a group \(G\).

1. If \(o(a) = \infty\) then \(\langle a \rangle \cong \mathbb{Z}\).
2. If \(o(a) = n \in \mathbb{N}\) then \(\langle a \rangle \cong \mathbb{Z}_n\).

Proof of 1: Assume that \(o(a) = \infty\). Define the function \(\varphi : \mathbb{Z} \to \langle a \rangle\) by the rule \(\varphi(n) = a^n\) for \(n \in \mathbb{Z}\). \(\varphi\) is onto by definition of \(\langle a \rangle\). To prove that \(\varphi\) is one-to-one let \(\varphi(n) = \varphi(m)\) for some \(n,m \in \mathbb{Z}\). Then \(a^n = a^m\). If \(n \neq m\) by symmetry we can assume \(n < m\). Then \[e = a^0 = a^{n - n} = a^na^{-n} = a^ma^{-n} = a^{m-n}.\] But \(n < m\) so \(m-n \in \mathbb{N}\). This contradicts the assumption that \(o(a) = \infty\). So we must have \(n = m\). This shows that \(\varphi\) is one-to-one. Since \[\varphi(n+m) = a^{n+m} = a^na^m = \varphi(n) \varphi(m)\] \(\varphi\) is a homomorphism and it follows that \(\varphi\) is an isomorphism. Hence \(\mathbb{Z}\cong \langle a \rangle\). By Theorem 7.1 \(\langle a \rangle \cong \mathbb{Z}\).

Proof of 2: Assume that \(o(a) = n \in \mathbb{N}\). For our proof we need to introduce the following notation from Appendix C.

Definition 7.4:

Let \(n \in \mathbb{N}\). For each \(a \in \mathbb{Z}\) by the Division Algorithm there are unique integers \(q\) and \(r\) satisfying \[a = nq + r \mbox{ \qquad and \qquad } 0 \le r < n.\] We denote \(r\) by \(a \bmod n\). That is, \(a \bmod n\) is the remainder when \(a\) is divided by \(n\).

Now using this we can define precisely addition modulo \(n\) by the rule: \[a \oplus b = (a+b) \bmod n.\] Note that here we write \(a + b\) for addition in \(\mathbb{Z}\) and \(a \oplus b\) for addition in \(\mathbb{Z}_n\). So to be precise, by \(\mathbb{Z}_n\) we mean the group \((\mathbb{Z}_n,\oplus)\).

Recall that \(\mathbb{Z}_n = \{0,1,2,\dots, n-1\}\). On the other hand by Theorem 4.2 since \(o(a) = n\) we have \[\langle a \rangle = \{ a^0, a^1, \dots , a^{n-1} \}.\] So the mapping \(\varphi:\mathbb{Z}_n \to \langle a \rangle\) defined by the rule \(\varphi(i) = a^i\) for \(i = 0, 1, 2, \dots , n-1\), is clearly one-to-one and onto. It remains to show that \(\varphi\) is a homomorphism. To prove this note first that \(i \oplus j= r\) means that \(i+j = qn + r\) where \(0 \le r \le n-1.\) Now we have \[\begin{aligned} \varphi(i\oplus j) &=& \varphi(r) = a^r = a^{i+j-qn} = a^ia^ja^{-qn} = a^ia^j(a^n)^{-q} \\ &=& a^ia^je^{-q} = a^ia^je = a^ia^j = \varphi(i)\varphi(j).\end{aligned}\] Hence \(\varphi(i \oplus j) = \varphi(i)\varphi(j)\). That is, \(\varphi\) is a homomorphism. Since it is also one-to-one and onto it is an isomorphism. Hence \(\mathbb{Z}_n \cong \langle a \rangle\) and by Theorem 7.1 \(\langle a \rangle \cong \mathbb{Z}_n\). \(\blacksquare\)

Problem 7.18 Prove that if \(G\) is a cyclic group then \(G\) is isomorphic to \(\mathbb{Z}\) or \(\mathbb{Z}_n\).

The above shows that a group generated by one element is easily describable. What about groups that are not generated by one element but are “generated” by two (or more elements)? The following exercise introduces a notation to make precise such matters.

Problem 7.19 [Challenge Problem] Let \(G\) be a group and let \(S \subset G\). Define \(\langle S \rangle\) to be the subset of \(G\) whose elements have the form \(s_1^{\epsilon_1}s_2^{\epsilon_2} \cdots s_n^{\epsilon_n}\) where \(n \in \mathbb{N}\), \(s_i \in S\) and \(\epsilon_i = \pm1\) for \(i = 1, 2, \dots , n\). Prove

1. \(\langle S \rangle\) is a subgroup of \(G\).
2. \(\langle S \rangle\) is the smallest subgroup of \(G\) that contains \(S\), that is, if \(K\) is a subgroup of \(G\) and \(S \subset K\) then \(\langle S \rangle \subset K\).
3. Show that for \(n \ge 3\) the group \(S_n\) is not cyclic, but \(S_n = \langle \{\tau,\sigma\} \rangle\) where \(\tau = ( 1 \; 2 )\) and \(\sigma = ( 1 \, 2 \, \cdots \, n)\).

Note that the above problem shows that although \(S_n\), \(n \ge 3\), is not cyclic, it is generated by two elements. However, unlike the cyclic groups one can say very little about groups generated by two elements.

You may be interested in the curious fact (first discovered by Philip Hall) that \((A_5)^{19}\) (i.e., the direct product of 19 copies of the alternating group of degree 5) can be generated by two elements, but \((A_5)^{20}\) cannot. On the other hand, the group \((\mathbb{Z}_2)^n\), that is, the direct product of \(n\) copies of \(\mathbb{Z}_2\), requires a minimum of \(n\) generators for each positive integer \(n\).

We state without proof the following theorem. A proof may be found, in any of the references .

Theorem \(\PageIndex{7}\)

If \(G\) is a finite group of order \(n\), then there is a subgroup \(H\) of \(S_n\) such that \(G \cong H\). \(\blacksquare\)

This makes precise the idea that every finite group is “contained” in \(S_n\) for some positive integer \(n\). For example, the group \(U_8=\{ 1,3,5,7 \}\) is isomorphic to the subgroup \[H = \{ \iota, (1 \ 2), (3 \ 4), (1 \ 2)(3 \ 4) \}\] of \(S_4\). Note that a group of order \(k\) may well be isomorphic to a subgroup of \(S_n\) where \(n < k\).

Problem 7.20 Find a group of order 120 which is ismorphic to a subgroup of \(S_n\) where \(n < 120\).