Download Section III.15. Factor-Group Computations and Simple

III.15 Factor-Group Computations
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Section III.15. Factor-Group Computations
and Simple Groups
Note. In this section, we try to extract information about a group G by considering
properties of the factor group (or “quotient group”) G/N . We also introduce a new
class of groups, which have received much attention.
Example. Let G be a group with identity e. Consider the normal subgroup
N = {e}. Since N has one element, then all cosets of N have one element—in fact,
the coset gN = {ge} = {g}. So the factor group G/N is isomorphic to G.
Example. For any group G, the factor group G/G is isomorphic to {e}. This is
because there is only one coset of G (treating G as a normal subgroup of G) and
so G/G is a group with one element.
Note. The above two examples are extreme cases of “collapse” of the cosets of G
down to elements of G/N . If G is a finite group and N 6= {e} is a normal subgroup,
then G/N is a smaller group than G and so “may have a more simple structure
than G” (using quotation marks when referring to the text’s wording). There is a
correspondence between the multiplication of cosets in G/N and the multiplication
of their representatives in G (as given in Corollary 14.5).
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III.15 Factor-Group Computations
Lemma. If G is a finite group and N is a subgroup of G where |N | = |G|/2, then
N is a normal subgroup of G.
Proof. Since all cosets of N must be the same size and the cosets partition G,
then there are only two cosets of N , namely N and aN where a ∈ G \ N . Now for
any g ∈ G, (1) if g ∈ N then gN = N g = N , and (2) if g ∈ G \ N then gN = N g
since this is the only coset of N other than N itself. So gN = N g for all g ∈ G,
and N is a normal subgroup.
Example 15.4. Consider Sn /An. Recall that the symmetric group Sn has order
n! and the alternating group An has order n!/2. So, by Lemma, An is a normal
subgroup and hence Sn /An is defined. So there are two cosets in Sn /An . As in the
proof of Lemma, for any σ ∈ Sn , (1) if σ ∈ An (i.e., σ is an even permutation) then
σAn = An , and (2) if σ ∈ Sn \An (i.e., σ is an odd permutation) then σAn = Sn \An .
So the two cosets correspond to the even permutations and the odd permutations.
If we denote cosets as En and On, we get the multiplication table for Sn /An as:
En On
En En On
On On En
As the text comments, the factor group Sn /An does not give us details about
products of specific elements of Sn , but it does give us information about products
of types of elements of Sn (even and odd, in this case).
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Example 15.6. Falsity of the Converse of the Theorem of Lagrange.
We have claimed in the past that the alternating group A4 (of order 4!/2 = 12)
does not have a subgroup of order 6. Recall that Lagrange’s Theorem states that
the order of a subgroup divides the order of its group. This example shows that
there may be divisors of the order of a group which may not be the order of a
subgroup.
Suppose, to the contrary, that H is a subgroup of A4 of order 6. By Lemma,
H must be a normal subgroup of G. Then A4 /H has only two elements, H and
σH where σ ∈ An \ H. Since A4/H is a group of order 2, then it is isomorphic to
Z2 and the square of each element (coset) is the identity (H). So H · H = H and
(σH) · (σH) = σ2 H = H. So if α ∈ H then α2 ∈ H and if β ∈
/ H (then β ∈ σH)
then β 2 ∈ H. So, the square of every element of A4 is in H. But in A4 we have
(1, 2, 3) = (1, 3, 2)2 and (1, 3, 2) = (1, 2, 3)2
(1, 2, 4) = (1, 4, 2)2 and (1, 4, 2) = (1, 2, 4)2
(1, 3, 4) = (1, 4, 3)2 and (1, 4, 3) = (1, 3, 4)2
(2, 3, 4) = (2, 4, 3)2 and (2, 4, 3) = (2, 3, 4)2 .
So all 8 of the above (distinct) permutations are in H. This is a contradiction,
since we assumed |H| = 6. Therefore no such H exists.
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Exercise 15.6. Classify Z × Z/h(0, 1)i according to the Fundamental Theorem of
Finitely Generated Abelian Groups.
Solution. Notice h(0, 1)i = {0} × Z = H and the cosets are (x, y) + {0} × Z =
{x} × (Z + y) = {x} × Z for all (x, y) ∈ Z × Z. Now {x} × Z is isomorphic to Z
(define φ({x} × Z) = x and φ is an isomorphism). So Z × Z/h(0, 1)i is isomorphic
to Z.
Note. The previous exercise foreshadows the following result.
Theorem 15.8. Let G = H × K be the direct product of groups G and H. Then
H = {(h, e) | h ∈ H} is a normal subgroup of G. Also, G/H is isomorphic to K.
similarly G/K is isomorphic to H where K = {(e, k) | k ∈ K}.
Note.
The humble Theorem 15.8 reveals why we have been studying factor
groups—because they allow us to FACTOR GROUPS (sometimes, at least)! It
follows that if H is a normal subgroup of G = H × K then G ∼
= (G/H) × H since
H ∼
= H. So the reason we have addressed factor groups is so that we can take
a given group and “factor” (or better yet, “decompose”) it into “smaller” groups.
We know from the statements at the beginning of this section that H1 = {e} and
H2 = G are normal subgroups of G, but they do not yield interesting factor groups
because G/H1 ∼
= G and G/H2 ∼
= {e}. So very soon our attention will turn to
normal subgroups which are neither trivial nor improper.
Theorem 15.9. A factor group of a cyclic group is cyclic.
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Exercise 15.4. Classify (Z4 × Z8 )/h(1, 2)i.
Solution. Notice h(1, 2)i = {(1, 2), (2, 4), (3, 6), (0, 0)} = H and so |Z4×Z8 /h(1, 2)i| =
4 × 8/4 = 8. All groups are abelian, including the factor group. The abelian groups
of order 8 are Z2 × Z2 × Z2 , Z2 × Z4, and Z8 (we don’t distinguish between Z2 × Z4
and Z4 × Z2). We try to determine which of these three choices is correct by considering orders of elements of the factor group. Consider the coset (0, 1) + H. The
order of this coset is 8 since:
((0, 1) + H) + ((0, 1) + H) + · · · + ((0, 1) + H) = (0, k) + H
|
{z
}
k times
and the smallest value of k for which this yields the identity is k = 8. Since the
only choice from the three groups which has elements of order 8 is Z8, then the
factor group must be isomorphic to Z8.
Note. We now introduce a new class of groups, which have a rich history. Details
are given in the supplement to this section.
Definition 15.14. A group is simple if it is nontrivial and has no proper nontrivial
normal subgroups.
Example. For any prime p, the group Zp is simple because it has no proper nontrivial subgroups by Lagrange’s Theorem (and therefore has no proper nontrivial
normal subgroups).
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Note. We now see that simple groups play a role similar to that which prime
numbers play in number theory! That is, if G is simple then the only factor groups
of G are G/G ∼
= {e} and G/{e} ∼
= G. Notice that we cannot say that simple
groups are equivalent to indecomposable groups. For example, Zpn , where p is
prime, is indecomposable but it is not simple when n > 1. We also cannot say
that if H is a normal subgroup of G, then G ∼
= (G/H) × H (as illustrated by
considering indecomposable Z4 and normal subgroup Z2 ∼
= {0, 2}). The nature by
which group G relates to its normal subgroups will be spelled out in more detail in
the Jordan-Hölder Theorem (Theorem 35.15).
Theorem 15.15. The alternating group An is simple for n ≥ 5.
Note. The proof we give of Theorem 15.15 is not elegant! It requires a clunky
argument based on several cases. We defer it until the end of this section and give
it as a supplement to these notes.
Note. We are interested in when G/N is simple for normal subgroup N of G.
When this occurs, we have sort of found a maximal “factor” of G.
Note. Theorem 13.12 showed us that a group homomorphism sort of maps subgroups “back and forth.” The following result shows that the subgroup property
of normality is also mapped back and forth by homomorphisms.
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Theorem 15.16. Let φ : G → G0 be a group homomorphism. If N is a normal
subgroup of G, then φ[N ] is a normal subgroup of φ[G]. Also, if N 0 is a normal
subgroup of φ[G], then φ−1 [N 0 ] is a normal subgroup of G.
Note. The proofs of the two parts of Theorem 15.16 are Exercises 15.35 and 15.36.
Definition 15.17. A maximal normal subgroup of a group G is a normal subgroup
M not equal to G such that there is no proper normal subgroup N of G properly
containing M .
Theorem 15.18. M is a maximal normal subgroup of G if and only if G/M is
simple.
Note. Personally, I find the proof in the text hard to follow (it may just be
me, since the previous editions of Fraleigh present the same proof as given in the
7th edition). The proof depends on the use of the homomorphism of Theorem
14.9 which maps G to G/M . In addressing both “M is maximal” and “M/G
is simple” we must be concerned with nontrivial and proper subgroups. Since a
homomorphism is not one to one, this requires special attention to the details.
We now present a proof with many more details than the proof given in Fraleigh
(though still based on Fraleigh’s general argument’s). Notice that both claims in
the theorem are established by contradiction.
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Proof of Theorem 15.18. Let M be a maximal normal subgroup of G. Define
γ : G → G/M as γ(g) = gM . Then by Theorem 14.9, γ is a homomorphism
with Ker(γ) = M . Suppose G/M is not simple and that N 0 is a proper nontrivial
normal subgroup of G/M . Then by Theorem 15.16, γ −1 [N 0 ] is a normal subgroup
of G. Since N 0 is nontrivial then N 0 6= {e} = {M } (remember M is the identity
in G/M ). Also, we have Ker(γ) = M , so Ker(γ) 6= N 0 and φ−1 [N 0 ] 6= M . But
{e} = {M } ( N 0 and so M ( φ−1 [N 0 ]. So φ−1 [N 0 ] properly contains M . Since
N 0 is a proper subgroup of G/M , then it contains some but not all of the cosets of
M . Since the cosets of M partition G (Section II.10), then φ−1[N 0 ] contains some
but not all elements of G. That is, φ−1 [N 0 ] is a proper subgroup of G. Therefore,
φ−1 [N 0 ] is a nontrivial proper normal subgroup of G which properly contains M .
But this contradicts the maximality M . So no such N 0 exists and G/M is simple.
Now let G/M be simple. Suppose that M is not a maximal normal subgroup
and that there is a proper normal subgroup N of G properly containing M (i.e.,
M ( N ( G). Then by Theorem 15.16, γ[N ] is a normal subgroup of γ[G] = G/M
where γ is as defined above. Now Ker(γ) = {M } (remember that M is the identity
in G/M ). So γ[N ] is a nontrivial normal subgroup of G/M . Now we show that
γ[N ] is a proper subgroup of G/M . Recall that γ maps elements of G to cosets of
M . So the only way that γ[N ] = G/M is if N contains an element of each coset
of M . But we assumed M ( N , and so if N contains an element of each coset of
M , then (since N is a group and so is closed under the binary operation) N must
contain all cosets of M —that is, M = G. But this contradicts the choice of N as a
proper subgroup of G. Hence N does not contain an element from each coset of M
and γ[N ] 6= G/M . That is, γ[N ] ( G/M . Therefore, γ[N ] is a proper nontrivial
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subgroup of G/M . But this contradicts the fact that G/M is simple. Therefore,
no such N exists and M is a maximal normal subgroup of G.
Definition. For group G, define the center of G as
Z(G) = {z ∈ g | zg = gz for all g ∈ G}.
Note. The letter Z in the notation above is from the German zentrum for “center.”
In Exercise 5.52, it is shown that Z(G) is an abelian subgroup of G. Of course,
if G itself is abelian, then Z(G) = G. Since Z(G) is abelian, then it is a normal
subgroup of G.
Definition. For group G, consider the set
C = {aba−1 b−1 | a, b ∈ G}.
C is the commutator subgroup of G. (C is shown to in fact be a group below.)
Theorem 15.20. Let G be a group. Then the set C = {aba−1b−1 | a, b ∈ G} is a
subgroup of G. Additionally, C is a normal subgroup of G. Furthermore, if N is a
normal subgroup of G then G/N is abelian if and only if C ≤ N .
Revised: 10/25/2014