Page 122 - DMTH403_ABSTRACT_ALGEBRA

Unit 10: Finite Abelian Groups

Solution: The prime factorization is 108 = 2 · 3 . There are two possible groups of order 4: Z and Notes
3
2
4
Z × Z . There are three possible groups of order 27: Z , Z × Z , and Z × Z × Z . This gives us
3
3
3
3
2
2
9
27
the following possible groups:
Z × Z 27
4
Z × Z × Z 27
2
2
Z × Z × Z 3
4
9
Z × Z × Z × Z 3
2
2
9
Z × Z × Z × Z 3
4
3
3
Z × Z × Z × Z × Z .
3
3
3
2
2
Example: 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 = p be the order of a cyclic component of G.
m
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
Example: Let G be a finite 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 Z as a factor. To give a total of 27 elements, the only
9
possibility is G  Z × Z .
9
3
Check: The elements 3 and 6 have order 3 in Z , while 1 and 2 have order 3 in Z . Thus, the
3
9
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).
Example: Let G be a finite abelian group such that |G| = 216. If | 6 G | = 6, determine G up
to isomorphism.
Solution: We have 216 = 2 · 3 , and 6G  Z × Z since it has order 6. Let H be the Sylow
3
3
3
2
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  Z , we know that there are elements not of order
2
2, and that H is not cyclic, since 2 Z  Z . We conclude that H  Z × Z .
4
8
4
2
A similar argument shows that the Sylow 3-subgroup K of G, which has 27 elements, must be
isomorphic to Z × Z .
3
9
Using the decomposition, we see that
G  Z × Z × Z × Z . 3
4
9
2
(If you prefer the form of the decomposition, you can also give the answer in the form G  Z × Z .)
36
6
Example: Apply both structure theorems to give the two decompositions of the finite
abelian group Z  216
Solution: Z   Z × Z  Z × Z × Z 


216 8 27 2 2 27
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