Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

## § 3.4 Isomorphisms

Definition 3.4.1. Let G1 and G2 be groups, and let µ : G1 -> G2 be a function. Then µ is said to be a group isomorphism if

(i)   µ is one-to-one and onto and

(ii)   µ (ab) = µ (a) µ (b)   for all a,b in G1.
In this case, G1 is said to be isomorphic to G2, and this is denoted by G1 G2.
(a) The inverse of a group isomorphism is a group isomorphism.

(b) The composition of two group isomorphisms is a group isomorphism.

Proposition 3.4.3. Let µ : G1 -> G2 be an isomorphism of groups.

(a) If a has order n in G1, then µ (a) has order n in G2.

(b) If G1 is abelian, then so is G2.

(c) If G1 is cyclic, then so is G2.

Proposition 3.4.4. Let G1 and G2 be groups, and let µ : G1 -> G2 be a function such that

µ (ab) = µ (a) µ (b)

for all a,b in G1. Then µ is one-to-one if and only if µ (x) = e implies x = e, for all x in G1.

Proposition 3.4.5. If m,n are positive integers such that gcd(m,n)=1, then

Zm × Zn Zmn.

## § 3.4 Isomorphisms: Solved problems

A group isomorphism µ : G1 -> G2 can be thought of as simply renaming the elements of G1, since it is a one-to-one correspondence. The condition that

µ (ab) = µ (a) µ (b),     for all a,b in G1

makes certain that multiplication can be done in either group and then transferred to the other, since the inverse function µ-1 also respects the multiplication of the two groups.

In terms of the respective group multiplication tables for G1 and G2, the existence of an isomorphism guarantees that there is a way to set up a correspondence between the elements of the groups in such a way that the group multiplication tables will look exactly the same.

From an algebraic perspective, we should think of isomorphic groups as being essentially the same. The problem of finding all abelian groups of order 8 is impossible to solve, because there are infinitely many possibilities. But if we ask for a list of abelian groups of order 8 that comes with a guarantee that any possible abelian group of order 8 must be isomorphic to one of the groups on the list, then the question becomes manageable. In fact, we can show (in Section 7.5) that the answer to this particular question is the list

Z8,     Z4 × Z2,     Z2 × Z2 × Z2.

In this situation we would usually say that we have found all abelian groups of order 8, up to isomorphism.

To show that two groups G1 and G2 are isomorphic, you should actually produce an isomorphism µ : G1 -> G2. To decide on the function to use, you probably need to do some experimentation (compute some products) in order to understand why the group operations are similar.

In some ways it is harder to show that two groups are not isomorphic. If you can show that one group has a property that the other one does not have, then you can decide that two groups are not isomorphic (provided that the property would have been transferred by any isomorphism). Suppose that G1 and G2 are isomorphic groups. If G1 is abelian, then so is G2; if G1 is cyclic, then so is G2. Furthermore, for each positive integer n, the two groups must have exactly the same number of elements of order n. Each time you meet a new property of groups, you should ask whether it is preserved by any isomorphism.

21. Show that Z17× is isomorphic to Z16.     Solution

22. Let µ : R× -> R× be defined by µ (x) = x3, for all x in R. Show that µ is a group isomorphism.     Solution

23. Let G1, G2, H1, H2 be groups, and suppose that µ1 : G1 -> H1 and µ2 : G2 -> H2 are group isomorphisms. Define

µ : G1 × G2 -> H1 × H2

by setting

µ (x1,x2) = ( µ1 (x1), µ2 (x2)),

for all (x1, x2) in G1 × G2. Prove that µ is a group isomorphism.     Solution

24. Prove that the group Z7× × Z11× is isomorphic to the group Z6 × Z10.     Solution

25. Define µ : Z30 × Z2 -> Z10 × Z6 by

µ ( [n]30 , [m]2 ) = ( [n]10 , [4n+3m]6 ),

for all ( [n]30 , [m]2 ) in Z30 × Z2 . First prove that µ is a well-defined function, and then prove that µ is a group isomorphism.     Solution

26. Let G be a group, and let H be a subgroup of G. Prove that if a is any element of G, then the subset

aHa-1 = { g in G | g = aha-1 for some h in H }

is a subgroup of G that is isomorphic to H.     Solution

27. Let G, G1, G2 be groups. Prove that if G is isomorphic to G1 × G2, then there are subgroups H and K in G such that
(i)   H K = { e },
(ii)   HK = G, and
(iii)   hk=kh for all h in H and k in K.     Solution

28. Show that for any prime number p, the subgroup of diagonal matrices in GL2 (Zp ) is isomorphic to Zp× × Zp×.     Solution

29. (a) In the group G = GL2 (R) of invertible 2 × 2 matrices with real entries, show that

H =

is a subgroup of G.
(b) Show that H is isomorphic to the group R of all real numbers, under addition.
Solution

30. Let G be the subgroup of GL2 (R) defined by

G = .

Show that G is not isomorphic to the direct product R× × R.     Solution

31. Let H be the following subgroup of the group G = GL2 (Z3).

H =

Show that H is isomorphic to the symmetric group S3.     Solution

32. Let G be a group, and let S be any set for which there exists a one-to-one and onto function µ : G -> S. Define an operation on S by setting x1 · x2 = µ ( µ -1 (x1) µ -1 (x2) ), for all x1, x2 in S. Prove that S is a group under this operation, and that µ is actually a group isomorphism.     Solution

## § 3.4 Lab questions

To answer the experimental questions, use the Groups15 applet written by John Wavrik of UCSD.

Lab 1. The aim of this problem is to show that Table 3.3.2 in the text (see page 103 of the text or the online table) and the table shown by Groups15 for the cyclic group Z6 actually describe the same group. Rewrite Table 3.3.2, using A=e, B=a, C=a2, D=a3, E=a4, and F=a5. Check that the new table is the same as the one listed by Groups15 for Z6. Solution

Lab 2. The aim of this problem is to see whether Table 3.3.3 in the text (see page 104 of the text or the online table) and the table shown by Groups15 for the group S3 describe the same group. Rewrite Table 3.3.3, using A=e, B=a, C=a2, D=b, E=ab, and F=a2b. Check whether the new table is the same as the one listed by Groups15 for S3. If not, search for a different correspondence that will produce the same table. This will show that the two groups are isomorphic. Solution

The previous problem is really a specific application of Exercise 3.3.14 in the text, which asks you to prove that any group of order 6 is either cyclic or isomorphic to S3.

Corollary 3.2.12 shows that any group of prime order is cyclic. The analysis of groups of order 4 given on pages 103 and 104 of the text shows that if |G| = 4, then either G has at least one element of order 4, and is therefore cyclic, or G has three elements of order 2 (not counting the identity). In the latter case G must be isomorphic to Z2 × Z2. These comments give you enough information to do a deeper investigation of the lab questions in Section 3.3.

Lab 3. Show that the group Z3 Z4 (of order 12) in Groups15 is the internal semidirect product of a normal subgroup of order 3 and a cyclic subgroup of order 4. (See the lab problem 3.3.10.)
Note: The normal subgroup is isomorphic to Z3, and the second subgroup is isomorphic to Z4, so this is what justifies the use of the notation Z3 Z4.

Lab 4. Show that the group D6 (of order 12) in Groups15 is the internal semidirect product of a normal subgroup isomorphic to Z6 and and a subgroup isomorphic to Z2. Show that D6 can also be expressed as the internal semidirect product of a normal subgroup isomorphic to Z3 and a subgroup isomorphic to Z2 × Z2. (See the lab problem 3.3.8.)

Lab 5. Show that the group A4 (of order 12) in Groups15 is the internal semidirect product of a normal subgroup isomorphic to Z2 × Z2 and a subgroup isomorphic to Z3. (See the lab problem 3.3.9.)

Lab 6. Find an isomorphism between the group Z12, as described in Groups15, and the group Z3 × Z4.

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