Forward to §7.6 problems | Back to §7.4 problems | Up | Table of Contents | About this document


§ 7.5 Finite Abelian Groups: solved problems

7. Find all abelian groups of order 108 (up to isomorphism).

Solution: The prime factorization is 108 = 22 · 33. There are two possible groups of order 4: Z4 and Z2 × Z2 . There are three possible groups of order 27: Z27 , Z9 × Z3 , and Z3 × Z3 × Z3 . This gives us the following possible groups:

Z4 × Z27
Z2 × Z2 × Z27
Z4 × Z9 × Z3
Z2 × Z2 × Z9 × Z3
Z4 × Z3 × Z3 × Z3
Z2 × Z2 × Z3 × Z3 × Z3 .

Next problem | Next solution | Table of Contents







































































8. Let G and H be finite abelian groups, and assume that G × G is isomorphic to H × H. Prove that G is isomorphic to H.

Solution: Let p be a prime divisor of |G|, and let q = pm be the order of a cyclic component of G. If G has k such components, then G × G has 2k components of order q. An isomorphism between G × G and H × H must preserve these components, so it follows that H also has k cyclic components of order q. Since this is true for every such q, it follows that G H.

Next problem | Next solution | Table of Contents







































































9. Let G be an abelian group which has 8 elements of order 3, 18 elements of order 9, and no other elements besides the identity. Find (with proof) the decomposition of G as a direct product of cyclic groups.

Solution: We have |G| = 27. First, G is not cyclic since there is no element of order 27. Since there are elements of order 9, G must have Z9 as a factor. To give a total of 27 elements, the only possibility is G Z9 × Z3.
Check: The elements 3 and 6 have order 3 in Z9, while 1 and 2 have order 3 in Z3. Thus the following 8 elements have order 3 in the direct product: (3,0), (6,0), (3,1), (6,1), (3,2), (6,2), (0,1), and (0,2).

Next problem | Next solution | Table of Contents







































































10. Let G be a finite abelian group such that |G| = 216. If | 6 G | = 6, determine G up to isomorphism.

Solution: We have 216 = 23 · 33, and 6G Z2 × Z3 since it has order 6.. Let H be the Sylow 2-subgroup of G, which must have 8 elements. Then multiplication by 3 defines an automorphism of H, so we only need to consider 2H. Since 2H Z2, we know that there are elements not of order 2, and that H is not cyclic, since 2 Z8 Z4. We conclude that H Z4 × Z2.
A similar argument shows that the Sylow 3-subgroup K of G, which has 27 elements, must be isomorphic to Z9 × Z3.
Using the decomposition in Theorem 7.5.4, we see that
G     Z4 × Z2 × Z9 × Z3.
(If you prefer the form of the decomposition in Proposition 7.5.5, you can also give the answer in the form
G     Z36 × Z6.)

Next problem | Next solution | Table of Contents







































































11. Apply both structure theorems to give the two decompositions of the abelian group Z216×

Solution: Z216×     Z8× × Z27×     Z2 × Z2 × Z27×

Since 27 is a power of an odd prime, it follows from Corollary 7.5.11 that Z27× is cyclic.
This can also be shown directly by guessing that 2 is a generator.
Since Z27× has order 33 - 32 = 18, an element can only have order 1, 2, 3, 6, 9 or 18. We have
22 = 4,
23 = 8,
26 82 10, and
29 23 · 26 8 · 10 -1,
so it follows that 2 must be a generator.

We conclude that Z216×     Z2 × Z2 × Z18.
To give the first decomposition, recall that Theorem 7.5.4 states that any finite abelian group is isomorphic to a direct product of cyclic groups of prime power order. In this decomposition we need to split Z18 up into cyclic subgrops of prime power order, so we finally get the decomposition
Z216×     Z2 × Z2 × Z2 × Z9.

On the other hand, Proposition 7.5.5 describes the second decomposition, where any finite abelian group is written as a direct product of cyclic groups in which the orders any component is a divisor of the previous one. To do this we need to group together the largest prime powers that we can. In the first decomposition, we can combine Z2 and Z9 to get Z18 as the first component. We end up with
Z216×     Z18 × Z2 × Z2
as the second way of breaking Z216× up into a direct product of cyclic subgroups.

Next problem | Next solution | Table of Contents







































































12. Let G and H be finite abelian groups, and assume that they have the following property. For each positive integer m, G and H have the same number of elements of order m. Prove that G and H are isomorphic.

Solution: We give a proof by induction on the order of |G|. The statement is clearly true for groups of order 2 and 3, so suppose that G and H are given, and the statement holds for all groups of lower order. Let p be a prime divisor of |G|, and let Gp and Hp be the Sylow p-subgroups of G and H, respectively. Since the Sylow subgroups contain all elements of order a power of p, the induction hypothesis applies to Gp and Hp. If we can show that Gp Hp for all p, then it will follow that G H, since G and H are direct products of their Sylow subgroups.

Let x be an element of Gp with maximal order q = pm. Then < x > is a direct factor of Gp by Lemma~7.5.3, so there is a subgroup G' with Gp = < x > × G'. By the same argument we can write Hp = < y > × H', where y has the same order as x.

Now consider < xp > × G' and < yp > × H'. To construct each of these subgroups we have removed elements of the form (xk,g'), where xk has order q and g' is any element of G'. Because x has maximal order in a p-group, in each case the order of g' is a divisor of q, and so (xk,g') has order q since the order of an element in a direct product is the least common multiple of the orders of the components. Thus to construct each of these subgroups we have removed (pm - pm-1) · |G'| elements, each having order q. It follows from the hypothesis that we are left with the same number of elements of each order, and so the induction hypothesis implies that < xp > × G' and < yp > × H' are isomorphic. But then G' H', and so Gp Hp, completing the proof.


Forward to §7.6 | Back to §7.4 | Up | Table of Contents