So let me tell you, as you prepare to go off into the world, remember those six rules: Trust yourself, Break some rules, Don’t be afraid to fail, Ignore the naysayers, Work like hell, and Give something back, Arnold Schwarzenegger.
Recall. The Fundamental Theorem of Finite Abelian Groups. Every finite Abelian group G is the direct sum of cyclic groups, each of prime power order. |G| = n = p_{1}^{r1}p_{2}^{r2}···p_{k}^{rk}. Then,
The number of terms in the direct product, as well as the order of the cyclic groups, are uniquely determined by the group.
How many elements of order 6 are there in ℤ_{6} ⊕ ℤ_{9}? The order of (a, b) is the least common multiple of the order of a and that of b, |(a, b)| = lcm(|a|, |b|). |(a, b)| = 6 ↭ the order of a is 6 (1, 5) and that of b is 1 (0) or 3 (3, 6), or the order of a is 2 (3) and that of b is 3 (3, 6). The desired elements of order 6 are: (1, 0), (5, 0), (1, 3), (1, 6), (5, 3), (5, 6), (3, 3), (3, 6).
How many Abelian groups of order 125? (up to isomorphism) Let |G| = 125 = 5^{3}, so there are three possibilities: ℤ_{53}, ℤ_{52} ⊕ ℤ_{5} or ⊕ ℤ_{5}⊕ ℤ_{5}⊕ ℤ_{5}. 💡Notice there is one Abelian group of order p^{k} for each partition of k, i.e., each way of writing k as a sum of positive integers. These integers become the exponents in the power of p.
Let |G| = 12 = 2^{2}·3, there are two possibilities, namely ℤ_{4} ⊕ ℤ_{3} and ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3}.
Let |G| = 540 = 2^{2}·3^{3}·5, there are six Abelian groups of order 540 up to isomorphism: ℤ_{4}⊕ℤ_{27}⊕ℤ_{5}, ℤ_{2}⊕ℤ_{2}⊕ℤ_{27}⊕ℤ_{5}, ℤ_{4}⊕ℤ_{9}⊕ℤ_{3}⊕ℤ_{5}, ℤ_{2}⊕ℤ_{2}⊕ℤ_{9}⊕ℤ_{3}⊕ℤ_{5}, ℤ_{4}⊕ℤ_{3}⊕ℤ_{3}⊕ℤ_{3}⊕ℤ_{5}, ℤ_{2}⊕ℤ_{2}⊕ℤ_{3}⊕ℤ_{3}⊕ℤ_{3}⊕ℤ_{5}.
Let |G| = 1176 = 2^{3}·3·7^{2}, then the list of the distinct isomorphism classes of Abelian groups is {ℤ_{8}⊕ℤ_{3}⊕ℤ_{49}, ℤ_{4}⊕ℤ_{2}⊕ℤ_{3}⊕ℤ_{49}, ℤ_{2}⊕ℤ_{2}⊕ℤ_{2}⊕ℤ_{3}⊕ℤ_{49}, ℤ_{8}⊕ℤ_{3}⊕ℤ_{7}⊕ℤ_{7}, ℤ_{4}⊕ℤ_{2}⊕ℤ_{3}⊕ℤ_{7}⊕ℤ_{7}, ℤ_{2}⊕ℤ_{2}⊕ℤ_{2}⊕ℤ_{3}⊕ℤ_{7}⊕ℤ_{7}}
How many Abelian groups are there of order 144? There is one Abelian group of order p^{k} for each partition of k, i.e., each way of writing k as a sum of positive integers. Since 144 = 2^{4}·3^{2}, there are five ways of partitioning 4 (4 -ℤ_{24}-, 3+1 -ℤ_{23}⊕ℤ_{2}–, 2+2, 2+1+1, 1+1+1+1+1), and two ways of partitioning 2 (2 -ℤ_{32}-, 1 +1 -ℤ_{32}⊕ℤ_{3}-), hence there are ten such groups. Similarly, there are 28 = 2·2·7 non-isomorphic Abelian groups of order 3781575 = 3^{2}·5^{2}·7^{5} since the number of partitions of 2 and 5 are 2 (2 and 1+1) and 7 (5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1+ 1, 1+1+1+1+1) respectively.
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Which one? The elements in ℤ_{4} ⊕ ℤ_{3} can have order 1, 2, 3, 4, 6, 12 and the elements in ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3} can have order 1, 2, 3, 6, 12. Thus, if U(13) has an element of order 4 then it must be isomorphic to ℤ_{4} ⊕ ℤ_{3} and we are done. 5 ∈ U(13), |5| = 4 (5^{0} = 1, 5^{1} = 5, 5^{2} = 25 =_{13} 12, 5^{3} = 125=_{13} 8, 5^{4} = 625 =_{13} 1) and so U(13) ≋ Z4 ⊕ Z3. Futhermore, U(13) ≋ ℤ_{4} ⊕ ℤ_{3} ≋ ℤ_{12} = ⟨1⟩, this also tells us that U(13) is cyclic.
|G| = 16 = 2^{4} ⇒ Partitions of 4 = 1 + 1 + 1 + 1, 2 + 1 + 1, 2 + 2, 3 + 1, 4, so we have 5 classes. G is isomorphic to one of these classes: ℤ_{2}⊕ℤ_{2}⊕ℤ_{2}⊕ℤ_{2}, ℤ_{4}⊕ℤ_{2}⊕ℤ_{2}, ℤ_{4}⊕ℤ_{4}, ℤ_{8}⊕ℤ_{2}, ℤ_{16}… but which one?
We can rule out…
Therefore, G ≋ ℤ_{4}⊕ℤ_{4}.
Which elements have order of elements of order 24 = lcm{8, 3}. Elements of order 8 could come from ℤ_{16} and ℤ_{8}. Elements of order 3 could come from ℤ_{9} and ℤ_{3}. If G has no elements of order 24, it narrows the possibilities to the following groups, ℤ_{4} ⊕ ℤ_{4} ⊕ ℤ_{9} ⊕ ℤ_{7}, ℤ_{4} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{9} ⊕ ℤ_{7}, ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{9} ⊕ ℤ_{7}, ℤ_{4} ⊕ ℤ_{4} ⊕ ℤ_{3} ⊕ ℤ_{3} ⊕ ℤ_{7}, ℤ_{4} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3} ⊕ ℤ_{3} ⊕ ℤ_{7}, ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3} ⊕ ℤ_{3} ⊕ ℤ_{7}.
Example. Let’s G be an Abelian group, |G| = 72 = 2^{3}·3^{2}. Construct a subgroup of order 12. By the Fundamental Theorem of Finite Abelian Groups, we have the following options: