Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

Is G (2) ∼ = G (4) ? We have just seen that |Z(G (4) )| ≥ 2. On the other hand an element
g = u k a ℓ is in the center of G (2) ⇔ e = xgx −1 for all x = u i a j ∈ G (2) . Applying the
multiplicative form of the group law in G (2) worked out in equation (30), we get
e = (u i a j )(u k a ℓ )(a −j u −i )
= u i (a j u k a −j )(a ℓ u −i a −ℓ )a ℓ
= u i a (−1)j k (a −(−1) ℓ i )a ℓ
= u (−1)j k+[1−(−1) ℓ ]i a ℓ
for all i, j. By unique decomposition we must have a ℓ = e; hence ℓ ≡ 0 (mod 4) and [1−(−1) ℓ ] =
0. We then get
u (−1)j k = e and hence k ≡ 0 (mod 7)
Therefore g = u k a ℓ = e is the only central element in G (2) and G (2) cannot be isomorphic to
G (4) .
Abelian groups and the Sylow theorems. If G is a finite abelian group the Sylow theorems
provide an explicit natural direct product decomposition. This is not the final answer if we want
to know the detailed structure of finite abelian groups, but it is a big step in that direction. The
proof uses the observation that in an abelian group all Sylow p-subgroups are automatically
normal subgroups, and hence by Theorem 6.3.1(b) there is just one Sylow p-subgroup for each
prime divisor of the order |G|.
6.3.6 Theorem. Any finite abelian group is isomorphic to a direct product Sp1 × . . . × Spr
where n = �r i=1 pni i is the prime decomposition of the order |G| = n and Spi is the unique Sylow
pi-subgroup in Gof order p ni
i . This direct product decomposition is canonical: the subgroups Spi
are uniquely determined, as are the primes pi and their exponents ni.
Note: The components Spi need not be cyclic groups. For instance, if p = 2 we might have
S2 = Z2 × Z4 which cannot be isomorphic to the cyclic group Z8 of the same size. Determining
the fine structure of the components Spi would require further effort, leading to the Fundamental
Structure Theorem for finitely generated abelian groups. The coarse decomposition 6.3.6 is the
first step in that direction.
Proof: Let’s write Si for the unique Sylow pi-subgroup. For each index index 1 ≤ i ≤ r the
product set Hi = �i j=1 Sj is a normal subgroup (G is abelian). We now apply Lagrange’s
theorem and the counting principle 3.4.7 to show that its order is |Hi| = �i j=1 pnj j . Obviously
|H1| = |S1| = p n1
1 . When i = 2, the order of the subgroup H1 ∩ S2 = S1 ∩ S2 must divide both
p n1
1 and pn2 2 , and hence the intersection H1 ∩S2 is trivial; it follows immediately from 3.4.7 that
H2 = H1S1 has cardinality |H1|·|S2|/|H1 ∩S2| = p n1
1 pn2 2 . At the next stage we have H3 = H2S3
and H2 ∩ S3 is again trivial because these subgroups have different prime divisors; applying
3.4.7 we get |H3| = |H2| · |S3| = p n1
1 pn2 . Continuing inductively we prove our claim.
2 pn3 3
This already implies that G is a direct product. In fact, since |G| = |Hr| we see that G
is equal to the product set S1S2 . . .Sr. It remains only to check that if a1a2 . . . ar = e with
ai ∈ Si, then each ai = e. Let q be the smallest index such that a non-trivial decomposition of
the identity occurs. Certainly q > 1 and aq �= e, and then a −1
q = a1 . . . aq−1. But on the left
we have an element of Sq and on the right an element of the subgroup Hq−1. These subgroups
have trivial intersection, which is impossible if aq �= e. Thus every element in G has a unique
decomposition of the form a1a2 . . . ar, and G is the direct product of its Sylow subgroups. �
Essentially, this result says that to understand the the internal structure of any finite abelian
group it suffices to analyze the abelian p-groups – those with just one prime divisor and order
p k for some k
In order for the Sylow subgroups to be useful indicators of the overall structure for noncommutative
G it is necessary to prove that they are pervasive in G. Here’s what that means.
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