Cosets, Lagrange's Theorem and Factor Groups

6 Cosets & Factor Groups

The course becomes markedly more abstract at this point. Our primary goal is to break apart a group into subsets such that the set of subsets inherits a natural group structure. This sounds strange, but it is precisely what you’ve already encountered when constructing a natural addition on the set of remainders modulo n: in Z6, when we write 3 + 5 = 2, we really mean [3] + [5] = [2], where

[3] = {3 + 6n : n ∈ Z} is the set of integers whose remainder is 3 when divided by 6. We shall call this set a coset to indicate that it is related to a subgroup of the integers, namely 3Z. That the above addition of sets makes sense and deﬁnes a group structure is merely an example of a more general construction.

6.1 Lagrange’s Theorem Before properly deﬁning cosets, we state what you should already have hypothesized, given the huge number of examples you’ve now seen. Cosets will turn out to provide a simple proof. Review any of the subgroup relations we’ve already seen and observe that the order of a subgroup is a divisor of the order of its parent group. This is a general result.

Theorem 6.1 (Lagrange). Suppose that G is a ﬁnite group. The order of any subgroup H ≤ G divides the order of G. Otherwise said

H ≤ G =⇒ |H| |G|

We have already proved the special case for subgroups of cyclic groups:1

If G is a cyclic group of order n, then, for every divisor d of n, G has exactly one subgroup of order d. More precisely, if G = hgi has order n, then k ∼ = n • g = Cd where d gcd(n,k)

• gk = gl ⇐⇒ gcd(n, k) = gcd(n, l)

We shall prove the full theorem shortly. For such a simple result, Lagrange’s Theorem is extremely powerful. For instance, you may have observed that there is only one group structure, up to isomorphism, of prime orders 2, 3, 5 and 7. This is also a general result.

Corollary 6.2. There is only one group (up to isomorphism) of each prime order p, namely Cp. Proof. Let p be prime and suppose that G is a group with |G| = p. Since p ≥ 2, we may choose some element x ∈ G \{e}. Consider the cyclic subgroup generated by x, namely hxi ≤ G. Lagrange =⇒ |hxi| divides p =⇒ hxi = 1 or p. Since hxi has at least two elements, its order must therefore be p, from which G = hxi is cyclic. Finally, recall that there is only one cyclic group of each order up to isomorphism, so G = Cp.

1Lagrange’s Theorem is sometimes misremembered as ‘the order of an element divides the order of a group.’ This is really only a statement about the cyclic subgroups of a group G and is not as general as Lagrange-proper.

1 6.2 Cosets and Normal Subgroups The idea for the proof of Lagrange’s Theorem is very simple. Given a subgroup H ≤ G, partition G into several subsets, each with the same cardinality as H. We call these subsets the cosets of H.

Deﬁnition 6.3. Let H ≤ G and choose g ∈ G. The subsets of G deﬁned by

gH := {gh : h ∈ H} and Hg := {hg : h ∈ H} are (respectively) the left and right cosets of H containing g. If the left- and right- cosets of H containing g are always equal (∀g ∈ G, gH = Hg) we say that H is a normal subgroup of G, and write H / G. If G is written additively, the left and right cosets of H containing g are

g + H := {g + h : h ∈ H} and H + g := {h + g : h ∈ H}

Even though our ultimate purpose is to prove Lagrange’s Theorem (a statement about ﬁnite groups), the concept of cosets is perfectly well-deﬁned for any group. Before seeing some examples, we state a straightforward lemma giving some conditions for identifying normal subgroups.

Lemma 6.4. 1. Every subgroup of an abelian group G is normal.

2. A subgroup H is normal in G if and only if ∀g ∈ G, ∀h ∈ H, ghg−1 ∈ H.

For non-abelian groups, most subgroups are typically not normal (although see example 3 below).

Proof. Part 1. should be obvious. For part 2., note ﬁrst that

H / G =⇒ ∀g ∈ G, h ∈ H, gh ∈ Hg =⇒ ∀g ∈ G, h ∈ H, ghg−1 ∈ H

Conversely, ghg−1 ∈ H =⇒ gh ∈ Hg =⇒ gH ⊆ Hg. If this holds for all g ∈ G and h ∈ H, then it certainly applies for g−1, whence

g−1hg ∈ H =⇒ hg ∈ gH =⇒ Hg ⊆ gH

We conclude that gH = Hg for all g ∈ G and so H is normal in G.

Examples

1. Consider the subgroup of Z12 generated by 4. This subgroup is cyclic of order 3: ∼ H = h4i = {0, 4, 8} = C3

Since the group operation is addition, we write cosets additively: for example, the left coset of h4i containing x ∈ Z12 is the subset

x + h4i = {x + n : n ∈ h4i} = {x, x + 4, x + 8}

2 where we might have to reduce x + 4 and x + 8 modulo 12. The distinct left cosets of h4i are as follows: 0 + h4i = {0, 4, 8} = 4 + h4i = 8 + h4i (usually written h4i, dropping the zero) 1 + h4i = {1, 5, 9} = 5 + h4i = 9 + h4i 2 + h4i = {2, 6, 10} = 6 + h4i = 10 + h4i 3 + h4i = {3, 7, 11} = 7 + h4i = 11 + h4i

There are only four distinct cosets: notice that each has the same number of elements (three) |Z12| = 12 = Z as the subgroup h4i, and that |h4i| 3 4. The cosets have partitioned 12 into equal-sized subsets. This is crucial for the proof of Lagrange’s Theorem. Observe also that the right cosets of h4i are the same as the left cosets, in accordance with Lemma 6.4). Observe ﬁnally that only one of the cosets is a subgroup, the others are merely subsets of Z12.

2. Recall the multiplication table for D3. With a little work, you should be able to compute that the left and right cosets of the subgroup H = {e, µ1} are as follows:

Left cosets Right cosets

eH = µ1 H = H = {e, µ1} He = Hµ1 = H = {e, µ1}

ρ1 H = µ3 H = {ρ1, µ3} Hρ1 = Hµ2 = {ρ1, µ2}

ρ2 H = µ2 H = {ρ2, µ2} Hρ2 = Hµ3 = {ρ1, µ3}

This time the left and right cosets of H are different (H is not a normal subgroup of D3), but all cosets still have the same cardinality.

1 3. Recall how we proved that the alternating subgroup An ≤ Sn has cardinality |An| = 2 |Sn|. Generalizing this approach, it should be clear that, for any α ∈ An and σ ∈ Sn, we have ασ even ⇐⇒ σ even ⇐⇒ σα even

Otherwise said, for any σ ∈ Sn, the cosets of An contiaining σ are ( An if σ even σAn = Anσ = Bn if σ odd

where Bn is the set of odd permutations in Sn. Note that An is a normal subgroup of Sn.

4. The vector space R2 is an Abelian group under addition. The real line R identiﬁed with the x-axis, is a subgroup. The cosets of R in R2 are all the sets v + R which are W horizontal lines. More generally, if W is a subspace of a vector space V, then the cosets v + W form sets parallel to W: only the zero coset W = 0 + W is a subspace. In the picture, W is line through the origin in R2; its cosets comprise all the lines in R2 parallel to W (including W itself!).

3 The examples should have suggested the following Theorem.

Theorem 6.5. The left cosets of H partition G.

Before reading the proof, look at each of the above examples and convince yourself that the result is satisﬁed in each case. The strategy is to deﬁne an equivalence relation, the equivalence classes of which are precisely the left cosets of H.

Proof. Deﬁne a relation ∼ on G by x ∼ y ⇐⇒ x−1y ∈ H. We claim that ∼ is an equivalence relation.

− Reﬂexivityx ∼ x since x 1x = e ∈ H. X Symmetryx ∼ y =⇒ x−1y ∈ H =⇒ (x−1y)−1 ∈ H, since H is a subgroup. But then

−1 y x ∈ H =⇒ y ∼ x. X

Transitivity If x ∼ y and y ∼ z then x−1y ∈ H and y−1z ∈ H. But H is closed, whence

−1 −1 −1 x z = (x y)(y z) ∈ H =⇒ x ∼ z. X

The equivalence classes of ∼ therefore partition G. We claim that these are precisely the left cosets of H. Speciﬁcally, we claim that

x ∼ y ⇐⇒ x and y lie in the same left coset of H ⇐⇒ xH = yH

However, note that x ∼ y ⇐⇒ x−1y ∈ H ⇐⇒ y ∈ xH. It follows that yH ⊆ xH. By symmetry, x ∼ y ⇐⇒ y−1x ∈ H ⇐⇒ xH ⊆ yH. We conclude that

x ∼ y ⇐⇒ xH = yH as required.

Aside: Partitions and Subgroups Each of the three conditions that ∼ be an equivalence relation corresponds to one of the properties characterizing H as a subgroup of G:

ReﬂexivityH contains the identity. SymmetryH is closed under inverses. TransitivityH is closed under the group operation.

It is precisely the fact that H is a subgroup which guarantees a partition. For example, if H is merely the subset H = {0, 1} of |Z3 = {0, 1, 2}, then its left ‘cosets’ are

0 + {0, 1} = {0, 1}, 1 + {0, 1} = {1, 2}, 2 + {0, 1} = {2, 1}, which do not partition Z3.

4 Proof of Lagrange’s Theorem. Suppose that H ≤ G and g ∈ G is a ﬁxed element. Then the function

φ : H → gH : h 7→ gh is a bijection (its inverse is φ−1 : gh 7→ g). It follows that every left coset of H has the same cardinality as H. Therefore

|G| = (number of left cosets of H) · |H| whence |H| divides |G|.

1 You should now revisit the proof that |An| = 2 |Sn|: it is nothing but a special case of Theorem 6.5 and the proof of Lagrange.

The right cosets of a subgroup H also partition G although, in general, they do this in a different way to the left cosets: observe the earlier example of the cosets of the subgroup {e, µ1} ≤ D3. The proof of Lagrange’s Theorem could also be argued with right cosets.

6.3 Indices The proof of Lagrange’s Theorem in fact tells us that the number of left and right cosets of H in G is |G| identical: both are equal to the quotient |H| . This leads us to the following deﬁnition. Deﬁnition 6.6. If H ≤ G then the index of H in G, written (G : H), is the number of left (or right) cosets of H in G. Strictly, the index is the cardinality of the set of left cosets:

(G : H) = |{gH : g ∈ G}|

( ) = |G| If G is finite G : H |H| and the index is simple to calculate. It is more difficult to consider the situation when G is infinite, although the concept is well-defined (the cardinalities of the sets of left and right cosets are identical).

Theorem 6.7. If K ≤ H ≤ G is a sequence of subgroups then

(G : K) = (G : H)(H : K)

If G is a ﬁnite group then the result is essentially trivial:

|G| |G| |H| (G : K) = = · = (G : H)(H : K) |K| |H| |K|

However, the Theorem is more general, for it applies also to inﬁnite groups.2

2Even when dealing with infinite groups, we will only think about examples where indices are finite so that their multiplication makes sense. If you are comfortable multiplying infinite cardinals then there is no problem removing this restriction and our proof works even in the general case.

5 Proof. Given K ≤ H ≤ G choose a single element gi from each left coset of H in G and a single element kj from each left coset of K in H. Clearly

(G : H) = |{gi}| and (H : K) = {hj}

We claim that the left cosets of K in G are precisely the sets (gihj)K. Certainly each set (gihj)K is a coset of K in G: our goal is to show that these sets partition G, whence the collection {(gihj)K} comprises all the left cosets of K in G. The partition conditions are conﬁrmed below: −1 • Every g ∈ G lies in some left coset of H, so ∃gi ∈ G such that g ∈ gi H. But then gi g ∈ H. −1 Every h ∈ H lies in some left coset of K in H, so ∃hj ∈ H such that gi g ∈ hjK. But then g ∈ (gihj)K so that every g ∈ G lies in at least one set (gihj)K.

• Suppose y ∈ gihjK ∩ gαhβK. Since the left cosets of H partition G, we have

y ∈ gi H ∩ gα H =⇒ gα = gi

−1 But then gi y ∈ hjK ∩ hβK =⇒ hβ = hj similarly, since the left cosets of K in H partition H. It follows that the sets (gihj)K are disjoint. 3 Since the left cosets of K in G are given by {(gihj)K}, it is immediate that

(G : K) = {gihj} = {gi} × {hj} = (G : H)(H : K)

Examples

1. Consider G = Z20 with its subgroups H = h2i and K = {10}. Certainly

K = {0, 10} ≤ H = {0, 2, 4, . . . , 18} ≤ G = {0, 1, 2, 3, . . .}

There are two (left)4 cosets of H in G, namely

H = {0, 2, 4, . . . , 18} and 1 + H = {1, 3, 5, . . . , 19}

whence the index is (G : H) = 2. Indeed, in the language of the proof, we could choose representatives g1 = 0 and g2 = 1. Meanwhile, K has ﬁve cosets in H:

K = {0, 10}, 2 + K = {2, 12}, 4 + K = {4, 14}, 6 + K = {6, 16}, 8 + K = {8, 18}

so that (H : K) = 5. We could choose representatives h1 = 0, h2 = 2, h3 = 4, h4 = 6, h5 = 8. There are ten cosets of K in G, in accordance with (G : K) = (G : H)(H : K):

K = {0, 10}, 1 + K = {1, 11}, 2 + K = {2, 12}, . . . , 9 + K = {9, 19}

In the language of the Theorem, these can be written

K = (g1 + h1) + K, 1 + K = (g2 + h1) + K, 2 + K = (g1 + h2) + K,...

. . . , 8 + K = (g1 + h4) + K, 9 + K = (g2 + h4) + K

3 The fact that {gihj} = {gi} × {hj} follows from the second bullet-point. 4The example is Abelian so left and right cosets are identical.

6 2. We consider the cosets for the sequence of subgroups

C3 ≤ S3 ≤ S4

where C3 = {e, (123), (132)} and S3 = {σ ∈ S4 : σ(4) = 4}. The left cosets for C3 in S3 are

eC3 = C3 = {e, (123), (132)} and (12)C3 = {(12), (23), (13)}

reﬂecting the fact that (S3 : C3) = 2. In the language of the Theorem, g0 = e and g1 = (12). Similarly, the left cosets for S3 in S4 are

eS3 = S3 = {e, (123), (132), (12), (23), (13)}

(14)S3 = S3 = {(14), (1234), (1324), (124), (14)(23), (134)}

(24)S3 = S3 = {(24), (1423), (1342), (142), (234), (13)(24)}

(34)S3 = S3 = {(34), (1243), (1432), (12)(34), (243), (143)}

with (S4 : S3) = 4 and h0 = e, h1 = (14), h2 = (24), h3 = (34). Finally, the left cosets of C3 in S4 are

eC3 = C3 = {e, (123), (132)} (12)eC3 = (12)C3 = {(12), (23), (13)}

(14)C3 = {(14), (1234), (1324)} (12)(14)C3 = (124)C3 = {(124), (14)(23), (134)}

(24)C3 = {(24), (1423), (1342)} (12)(24)C3 = (142)C3 = {(142), (234), (13)(24)}

(34)C3 = {(34), (1243), (1432)} (12)(34)C3 = {(12)(34), (243), (143)}

in accordance with (S4 : C3) = 8 = (S4 : S3) = (S3 : C3). There are patterns everywhere!

Aside: Indices in Inﬁnite Groups It is more challenging to think about indices for inﬁnite groups. Here are two classic examples:

1.O 2(R) is the set of rotations and reﬂections of the plane (or of the unit circle if you prefer). Its subgroup SO2(R) is the set of rotations. Both groups are inﬁnite (indeed uncountable), but we can compute the index:

(On(R) : SOn(R)) = 2

Intuitively this says that the rotations comprise half of group O2(R). To prove this, one per- forms what should be a now-familiar trick: the function 1 0 φ : SO (R) → O (R) \ SO (R) : A 7→ A 2 2 2 0 −1 is a bijection. . . 2. The sets Q and Z are both groups under addition. Their index is countably inﬁnite!

(Q : Z) = ℵ0 To see this, note that p + Z = q + Z ⇐⇒ p − q ∈ Z so that there is precisely one coset of Z in Q for every rational number in the interval [0, 1), a denumerable set.

7 6.4 Factor Groups The idea of factor groups is very simple, if abstract. Given a subgroup H ≤ G we may consider the set of left cosets . G = {gH : g ∈ G} H The question for this section is whether the set of left cosets has any interesting structure: that is, can we . view G as a group in a natural way?5 To see how this might work, let us recall the ﬁrst two examples H of cosets starting on page 2.

Turning the set of left cosets into a group Consider H ≤ Z12 where H = h4i = {0, 4, 8}, with (left) cosets

H, 1 + H, 2 + H, 3 + H

The set of left cosets is then Z . n o 12 = {H, 1 + H, 2 + H, 3 + H} = {0, 4, 8}, {1, 5, 9}, {2, 6, 10}, {3, 7, 11} H There is a nice pattern here: x and y are members of the same coset if and only if x ≡ y (mod 4). Z . Moreover, there are four cosets. Can we view 12 as merely the cyclic group Z4 in disguise? The H answer, of course, is yes. . To do this, we need to deﬁne a binary operation on Z12 : we need to see how to combine cosets H to create new ones. The obvious thing to do is to use the addition in Z12 that we already have: thus deﬁne

(a + H) ⊕ (b + H) := (a + b) + H

Z . The binary operation ⊕ on the set 12 comes naturally from the existing addition on Z12: we H haven’t created anything new, just using what we’ve been given.

There is a potential problem however: the freedom of choice built into the deﬁnition. Computing (a + H) ⊕ (b + H) requires three steps:

Choose representatives Make a choice of elements a and b in the respective cosets.

Add in the original group Compute a + b ∈ Z12. Take the coset Return the left coset (a + b) + H.

5There are typically many operations one could deﬁne on a set which would deﬁne a group structure. We want the structure to be natural: this means that it should appear without choice or imposition, and be inherited from the pre- existing structure on G. In this way the structure will encode some information about G itself.

8 If ⊕ is to make any sense at all, then the result (a + b) + H must be independent of the choices we made in the ﬁrst step.6 In this case there is no problem, as you can tediously check for yourself. For example, to compute (1 + H) ⊕ (2 + H) there are nine possibilities: 1 + 2 = 3 =⇒ (1 + H) ⊕ (2 + H) = 3 + H 1 + 6 = 7 =⇒ (1 + H) ⊕ (2 + H) = 7 + H = 3 + H 1 + 10 = 11 =⇒ (1 + H) ⊕ (2 + H) = 11 + H = 3 + H 5 + 2 = 7 =⇒ (1 + H) ⊕ (2 + H) = 7 + H = 3 + H 5 + 6 = 11 =⇒ (1 + H) ⊕ (2 + H) = 11 + H = 3 + H 5 + 10 = 15 =⇒ (1 + H) ⊕ (2 + H) = 15 + H = 3 + H 9 + 2 = 11 =⇒ (1 + H) ⊕ (2 + H) = 11 + H = 3 + H 9 + 6 = 15 =⇒ (1 + H) ⊕ (2 + H) = 15 + H = 3 + H 9 + 10 = 19 =⇒ (1 + H) ⊕ (2 + H) = 19 + H = 3 + H Thankfully all nine results are the same! The other possibilities for adding cosets can be similarly checked (we’ll do this in general in a moment). It is not hard to produce a table for the binary . operation ⊕ on the set Z12 : H ⊕ H 1 + H 2 + H 3 + H H H 1 + H 2 + H 3 + H 1 + H 1 + H 2 + H 3 + H H 2 + H 2 + H 3 + H H 1 + H 3 + H 3 + H H 1 + H 2 + H This table should look very familiar: if you delete all the H expressions (so that H = 0 + H becomes 0), we recover precisely the Cayley table for the cyclic group Z4. Indeed we have proved the following: . Proposition 6.8. The set of left cosets Z12 = {H, 1 + H, 2 + H, 3 + H} forms a group under the H operation ⊕, which is moreover isomorphic to Z4. We will shortly refer to this as a factor group. It would be very nice if this sort of behavior were universal. Unfortunately it isn’t.

When cosets fail to form a group Let us repeat the process with the subgroup H = {e, µ1} ≤ D3. Recall that its left cosets are

eH = µ1 H = H = {e, µ1}, ρ1 H = µ3 H = {ρ1, µ3}, ρ2 H = µ2 H = {ρ2, µ2} . In the same way as above, we deﬁne the ‘natural’ operation on the set D3 of left cosets:7 H aH ⊗ bH := (ab)H . . . 6Strictly speaking we require that ⊕ : Z12 × Z12 → Z12 be a well-deﬁned function. H H H 7 Again, the ‘operation’ on the set of left cosets comes directly from the group operation on D3.

9 8 This time there is a problem. Suppose we want to compute ρ1 H ⊗ ρ1 H. There are four choices:

2 ρ1 H ⊗ ρ1 H = ρ1 H = ρ2 H ρ1 H ⊗ µ3 H = ρ1µ3 H = µ2 H = ρ2 H

µ3 H ⊗ ρ1 H = µ3ρ1 H = µ1 H = H 2 µ3 H ⊗ µ3 H = µ3 H = eH = H

Exercising the freedom of choice in the definition of ⊗ leads to different outcomes: it follows that ⊗ is . not well-defined. There is nothing stopping us from defining a group structure on D3 , but any such H definition must involve a choice on our part: we would be imposing structure rather that observing structure that occured naturally.

Well-deﬁnition of the natural group structure The above discussion can be generalized. Clearly . some subgroups H ≤ G behave better than others when considering the set G of left cosets. How H can we tell which subgroups? . Let H be a subgroup of G and deﬁne the natural operation on G : H aH · bH := (ab)H

This is well-deﬁned if and only if

∀a, b ∈ G, ∀x ∈ aH, y ∈ bH, we have aH · bH = xH · yH

Let us trace through what this means for the subgroup H.

x ∈ aH ⇐⇒ ∃h ∈ H such that x = ah . The operation on G is therefore well-deﬁned if and only if H

∀a, b ∈ G, ∀h1, h2 ∈ H, (ah1bh2)H = (ab)H

⇐⇒ ∀a, b ∈ G, ∀h ∈ H, (ahb)H = (ab)H (since h2 H = H for all h2 ∈ H) ⇐⇒ ∀a, b ∈ G, ∀h ∈ H, (ab)−1(ahb) ∈ H ⇐⇒ ∀b ∈ G, ∀h ∈ H, b−1hb ∈ H ⇐⇒ H / G (recall Lemma 6.4)

We’ve therefore proved the critical part of the following: . Theorem 6.9. Suppose that H is a subgroup of G. The set of (left) cosets G forms a group under the H natural operation

aH · bH := (ab)H if and only if H is a normal subgroup of G. 8 Write these as cycles if you’re having trouble computing: e.g. ρ1 = (123), µ1 = (23), etc.

10 . Proof. The above discussion proves the ⇒ direction: if G forms a group then the operation is H certainly well-defined, whence H is normal. For the converse, we check the group axioms. Closure By the above discussion, H / G =⇒ the natural operation is well-defined. Certainly . aH · bH = (ab)H is a coset, whence G , · is closed. X H Associativity aH · (bH · cH) = aH · (bc)H = a(bc)H. Similarly (aH · bH) · cH = (ab)cH. By the associativity of G these are identical. X Identity eH · aH = (ea)H = aH therefore eH = H is the identity. X − − − − Inversea 1 H · aH = (a 1a)H = eH = H, therefore (aH) 1 = a 1 H. X . Definition 6.10. If H is a normal subgroup of G, then G is called a factor group. H . Warning! Because the group structure on a factor group G comes naturally from the group struc- H ture on G, we typically use the same notation for the operation. Thus if (G, +) is a group, we also use + . . for the operation on G . Similarly (G, ·) will have factor groups G , · written multiplicatively. H H Make sure you know to which group an operation refers! The symbols ⊕, ⊗ in the above examples were used only to help you keep the operations separate.

6.5 Factor Groups: Examples and Calculations There are many nice examples of factor groups. The main question we are concerned with is how to . identify a given factor group G : by this we mean that we want to ﬁnd a well-understood group K . H which is isomorphic to G . We start with a general example which is important enough to merits H a full discussion. This should, however, be a revision of material from a previous class.

Factor Groups of Inﬁnite Cyclic Groups: The Group Zm For each integer m, its integer multiples mZ = hmi form a subgroup of Z. Since Z is abelian, mZ is a normal subgroup. Observe that

x + mZ = y + mZ ⇐⇒ x − y ∈ mZ ⇐⇒ x ≡ y mod m whence there is precisely one coset for each remainder modulo m. We may therefore write the cosets of mZ as Z. = {0 + mZ, 1 + mZ,..., (m − 1) + mZ} mZ Addition in the factor group is natural: x + mZ + y + mZ = (x + y) + mZ

This looks suspiciously like addition modulo m! Indeed we have the following:

11 Z. Theorem 6.11. µ : → Zm : x + mZ 7→ x (mod m) is an isomorphism. mZ Proof. We’ve already seen most of the relevant calculations!

Well-deﬁnition and Injectivity Note that

x + mZ = y + mZ ⇐⇒ x − y ∈ mZ ⇐⇒ x − y ≡ 0 (mod m)

⇐⇒ x = y (in Zm) ⇐⇒ µ(x + mZ) = µ(y + mZ)

Surjectivity Given any x ∈ Zm, we have x = µ(x + mZ). . Homomorphism For all x + mZ, y + mZ ∈ Z , we have mZ µ (x + mZ) + (y + mZ) = µ (x + y) + mZ = x + y (mod m) = µ x + mZ + µ y + mZ (mod m)

Z. By this argument, algebraists often take the factor group to be the deﬁnition of the group Zm. mZ

Factor Groups of Finite Cyclic Groups Z . ∼ We’ve already seen the example 12 = Z4 (Proposition 6.8). We can, in fact, identify all factor h4i groups of ﬁnite cyclic groups.

Z n 9 Recall that n has a single subgroup of order d for each divisor d of n, namely n n o hdi = 0, d, 2d,..., − 1 d d . Theorem 6.12. If d | n, then µ : Zn → Z : x + hdi 7→ x (mod d) is an isomorphism. hdi d Try proving this yourself: it should be very similar to the proof of Theorem 6.11.

Example h5i = {0, 5, 10, 15} ≤ Z20 has factor group . Z20 = {0 + h5i , 1 + h5i , 2 + h5i , 3 + h5i , 4 + h5i} h5i

This is isomorphic to Z5 under the isomorphism . Z20 µ : → Z5 : x + h5i 7→ x (mod 5) h5i 9Recall that hsi = hdi ⇐⇒ gcd(s, n) = gcd(d, n) = d.

12 Factor Groups of Finite Abelian Groups . If G is a ﬁnite abelian group with subgroup H, then G is also a ﬁnite abelian group. According to H the Fundamental Theorem of Finitely Generated Abelian Groups, the means that

G. ∼ = Zm × Zm H 1 k

∈ N ··· = ( ) = |G| for some m1,..., mk which satisfy m1 mk G : H |H| . Our goal in the following . examples is to identify G by ﬁnding the relevant m . H k

Examples . . 1. Identify the factor group G = (Z4 × Z8) in terms of the Fundamental Theorem of H h(0, 1)i Finitely Generated Abelian Groups. First consider the elements of the cyclic subgroup:

H = h(0, 1)i = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7)}

4·8 This has order 8 whence the number of cosets is index of H in G, namely (G : H) = 8 . We can easily compute the cosets: recall that

(x, y) + H = (v, w) + H ⇐⇒ (x, y) − (v, w) = (x − v, y − w) ∈ H ⇐⇒ x = v

It follows that there is a unique element in each coset of the form (x, 0), where x ∈ Z4. Indeed the factor group may be written . n o G = H, (1, 0) + H, (2, 0) + H, (3, 0) + H H G. ∼ The factor group is abelian of order 4. We therefore have two possibilities: = Z4 or H G. ∼ = Z2 × Z2. Which is it? H A straightforward answer comes by observing that the factor group is generated by the coset (1, 0) + H, indeed . h(1, 0) + Hi = {H, (1, 0) + H, (2, 0) + H, (3, 0) + H} = G H

The factor group is cyclic and is therefore isomorphic to Z4. If you want to be more explicit, observe that the function

G. ψ : Z4 → : x 7→ (x, 0) + H H is an isomorphism.

13 . . 2. We can do something similar for G = (Z4 × Z8) . This time the cyclic subgroup H h(0, 2)i H = {0, 0), (0, 2), (0, 4), (0, 6)} has order 4. As with the previous example, the elements of of the cyclic subgroup only inﬂuce the second factor Z8 of G, and we therefore expect to see . . (Z4 × Z8) ∼ Z8 ∼ = Z4 × = Z4 × Z2 h(0, 2)i h2i Indeed this is what we ﬁnd, for the function G. ψ : Z4 × Z2 → : (x, y) 7→ (x, y) + H H is an isomorphism. . If the above seems too, fast, try counting cosets. Clearly G = 4·8 = 8, whence the quotient H 4 group is abelian of order 8. There are three non-isomorphic possibilities:

G. ∼ = Z8, Z4 × Z2, Z2 × Z2 × Z2 H To distinguish these, consider the possible orders of elements in each group:

Group Possible orders of elements Z8 1, 2, 4, 8 Z4 × Z2 1, 2, 4 Z2 × Z2 × Z2 1, 2

Observe:

• (1, 0) + H has order 4, since h(1, 0) + Hi = {H, (1, 0) + H, (2, 0) + H, (3, 0) + H} • All elements of the factor group have order dividing 4: 4 (x, y) + H = (4x, 4y) + H = (0, 4y) + H = 2y (0, 2) + H = H

It follows that the factor group must be isomorphic to Z4 × Z2. . . 3. This time we identify10 G = (Z4 × Z8) . As ever, ﬁrst compute the subgroup: H h(2, 4)i

H = {(0, 0), (2, 4)}

4·8 has order 2. The factor group is therefore abelian of order 2 = 16. The Fundamental Theorem gives ﬁve distinct options for the factor group, namely

Z16, Z2 × Z8, Z4 × Z4, Z2 × Z2 × Z4, Z2 × Z2 × Z2 × Z2

A little scratch work allows us to identify which is correct:

10The previous examples may have lulled you into a false sense of security: the answer to this question is not . . Z4 × Z8 =∼ Z × Z h2i h4i 2 2 Indeed by counting the sixteen cosets, we see immediately see that this is impossible!

14 • If x = 2n is even, then (x, y) + H = (2n, y) + H = (0, y − 4n) + H. • If x = 2n + 1 is odd, then (x, y) + H = 1, y − 4n) + H. • It follows that there is precisely one representative of each coset whose ﬁrst entry is either 0 or 1, whence the following sixteen elements lie in distinct cosets of h(2, 4)i:

(0, 0), (0, 1),..., (0, 7), (1, 0),..., (1, 7)

• It now seems reasonable to conjecture that the factor group is isomorphic to Z2 × Z8. Here are two possible ways to prove this:

(a) Observe that the coset (0, 1) + H has order 8 in the factor group. This narrows our choice to Z16 or Z2 × Z8. Now check that all cosets have order at most 8: 8 (x, y) + H = (8i, 8j) + H = (0, 0) + H

There are no elements of order 16 whence we can rule out Z16 as a candidate. The only possibility remaining is Z2 × Z8. (b) We can deﬁne an explicit isomorphism: . Z4 × Z8 ψ : Z2 × Z8 → : (x, y) 7→ (x mod 2, y − 4x) + h(2, 4)i h(2, 4)i

We leave it as an exercise to check that ψ is a well-deﬁned isomorphism: it requires some creativity to invent such a function from nothing! In practice, method (a) is generally easier for us.

Other Examples There are many other examples of factor groups. Often these have to be considered individually.

1. Let H = {2πn : n ∈ Z}. This is a subgroup of the abelian group of (R, +). Can we identify the . factor group R ? H It should be clear that in any given coset there is a unique element x such that 0 ≤ x < 2π (this is like taking the remainder of x modulo 2π!). It follows that

R. = {x + H : x ∈ [0, 2π)} H Indeed, you can check that the function

R. ψ : → S1 : x + H 7→ eix H is a well-deﬁned group isomorphism! The factor group construction therefore corresponds to wrapping the real line inﬁnitely many times around a circle of circumference 2π.

15 2. Recall the description of the alternating group A4. It can be (tediously) checked that V = {e, (12)(34), (13)(24), (14)(23)}

is a normal subgroup of A4 which is moreover isomorphic to the Klein 4-group. In this case, the . A 12 factor group 4 has order = 3 and so must be isomorphic to Z3: can you ﬁnd an explicit V 4 isomorphism? S . ∼ 3. A similar analysis may be performed to see that 4 = S3. See the homework. V

Aside: An factor group of an infinite cyclic group As a final example we do something far more difficult: nothing this tricky is examinable! . . Identify G = (Z10 × Z6 × Z) . H h(4, 2, 3)i The challenge is that both G and H are infinite, whence we cannot simply use the index formula to count the number of cosets. Here is a way round the problem. • Let z = 3q + r be the result of the division algorithm applied to z ∈ Z, where r = 0, 1 or 2. Then (x, y, z) + H = (x, y, z) − q(4, 2, 3) + H = (x − 4q, y − 2q, r) + H It follows that in every coset (x, y, z) + H, there is precisely one element (x, y, z) with z = 0, 1 or 2.

• The elements (x, y, 0) where x ∈ Z10 and y ∈ Z6 all lie in different cosets: if two such were in the same coset, then

(x1, y1, 0) − (x2, y2, 0) ∈ H ⇐⇒ (x1 − x2, y1 − y2, 0) ∈ H  x ≡ x (mod 10)  1 2 ⇐⇒ and  y1 ≡ y2 (mod 6) Similarly all the elements (x, y, 1) and (x, y, 2) lie in different cosets. • It follows that the elements (x, y, 0), (x, y, 1) and (x, y, 2) are representatives of different cosets, and that all cosets have one such representative. There are therefore 10 · 6 · 3 = 180 distinct . cosets whence the factor group G is abelian of order 180. H • The prime decomposition of 180 is 22 · 32 · 5: there are, up to isomorphism, four abelian groups of order 180, namely ∼ Z4 × Z9 × Z5 = Z180 ∼ Z2 × Z2 × Z9 × Z5 = Z2 × Z90 ∼ Z4 × Z3 × Z3 × Z5 = Z3 × Z60 ∼ Z2 × Z2 × Z3 × Z3 × Z5 = Z6 × Z30 How do we distinguish between these? As before, we look for elements of a particular order (lots of scratch work may be required in general!).

16 • Compute the order of the coset (1, 1, 1) + H. Certainly its order must be divisible by 3, since the third entry of

n(1, 1, 1) = (n, n, n) ∈ H

requires 3 | n. Thus let n = 3k. Now

3k(1, 1, 1) + H = (3k, 3k, 3k) + H = (3k, 3k, 3k) − k(4, 2, 3) + H (since (4, 2, 3) ∈ H) = (−k, k, 0) + H = H ⇐⇒ 10 | k and 6 | k ⇐⇒ 30 | k ⇐⇒ 90 | n

G. Since contains an element of order 90, our choices are narrowed to Z180 and Z2 × Z90. H . • We show that every element of G has order dividing 90: H 90 (x, y, z) + H = (90x, 90y, 90z) − 30z(4, 2, 3) + H = (90x − 120z, 90y − 60z, 0) + H = (0, 0, 0) + H

G. ∼ We conclude that = Z2 × Z90. Phew! H You can check (it’s hard!) that the function . (Z10 × Z6 × Z) ψ : Z2 × Z90 → : (x, y) 7→ (y, 3x + y, y) + H h(4, 2, 3)i is an isomorphism, though you’d have to be truly inspired to conjecture this out of thin air!