Dihedral Groups 2 Keith Conrad PDF

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Introduction to Dihedral Groups: properties, generators, subgroups....

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DIHEDRAL GROUPS II KEITH CONRAD

We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of Dn , including the normal subgroups. We will also introduce an infinite group that resembles the dihedral groups and has all of them as quotient groups. 1. Abstract characterization of Dn The group Dn has two generators r and s with orders n and 2 such that srs−1 = r −1 . We will show any group with a pair of generators like r and s (except for their order) admits a homomorphism onto it from Dn , and is isomorphic to Dn if it has the same size as Dn . Theorem 1.1. Let G be generated by elements a and b where an = 1 for some n ≥ 3, b2 = 1, and bab−1 = a−1 . There is a surjective homomorphism Dn → G, and if G has order 2n then this homomorphism is an isomorphism. The hypotheses an = 1 and b2 = 1 do not mean a has order n and b has order 2, but only that their orders divide n and divide 2. For instance, the trivial group has the form ha, bi where an = 1, b2 = 1, and bab−1 = a−1 (take a and b to be the identity). When n = 4, the group (Z/(8))× has generators 3 and 5 with 34 = 1, 52 = 1, and 5 · 3 · 5−1 = 3 = 3−1 . Proof. The equation bab−1 = a−1 implies baj b−1 = a−j for any j ∈ Z (raise both sides to the jth power). Since b2 = 1, we have for any k ∈ Z kj

bk aj b−k = a(−1) by considering even and odd k separately. Thus k

bk aj = a(−1) j bk .

(1.1)

Thus, any product of a’s and b’s can have all the a’s brought to the left and all the b’s brought to the right. Taking into account that an = 1 and b2 = 1, we get G = ha, bi (1.2)

= {aj , aj b : j ∈ Z} = {1, a, a2 , . . . , an−1 , b, ab, a2 b, . . . , an−1 b}.

Thus G is a finite group with #G ≤ 2n. To write down an explicit homomorphism from Dn onto G, the equations an = 1, b2 = 1, and bab−1 = a−1 suggest we should be able send r to a and s to b by a homomorphism. This suggests the function f : Dn → G defined by f (r j sk ) = aj bk . This function makes sense, since the only ambiguity in writing an element of Dn as r j sk is that j can change modulo n and k can change modulo 2, which has no effect on the right side since an = 1 and b2 = 1. 1

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To check f is a homomorphism, we use (1.1): ′

′

′

f (r j sk )f (r j sk ) = aj bk aj bk

′

k ′

′

= aj a(−1) j bk bk k ′ ′ = aj +(−1) j bk+k and ′

′

kj ′

sk sk )

kj ′

sk+k )

f ((r j sk )(r j sk )) = f (r j r (−1)

= f (r j +(−1)

kj ′

= aj +(−1)

′

′

′

bk+k .

The results agree, so f is a homomorphism from Dn to G. It is onto since every element of G has the form aj bk and these are all values of f by the definition of f . If #G = 2n then surjectivity of f implies injectivity, so f is an isomorphism.  Remark 1.2. The homomorphism f : Dn → G constructed in the proof is the only one where f (r) = a and f (s) = b: if there is any such homomorphism then f (r j sk ) = f (r)j f (s)k = aj bk . So a more precise formulation of Theorem 1.1 is this: for any group G = ha, bi where an = 1 for some n ≥ 3, b2 = 1, and bab −1 = a−1 , there is a unique homomorphism Dn → G sending r to a and s to b. Mathematicians describe this state of affairs by saying Dn with its generators r and s is “universal” as a group with two generators satisfying the three equations in Theorem 1.1. As an application of Theorem 1.1, we can write down a matrix group over Z/(n) that is isomorphic to Dn when n ≥ 3. Set    ±1 c e (1.3) Dn = : c ∈ Z/(n) 0 1 e n has order 2n (since 1 6≡ −1 mod n for n ≥ 3). Inside inside GL2 (Z/(n)). The group D 1 1 −1 0 e e n, ( D 0 1 ) has order 2 and ( 0 1 ) has order n. A typical element of Dn is      ±1 c 1 c ±1 0 = 0 1 0 1 0 1  c   1 1 ±1 0 = , 0 1 0 1 e n . Moreover, ( 1 1 ) and ( 1 1 )−1 are conjugate by ( −1 1 ): so ( 01 11 ) and ( −10 01 ) generate D 01 01 0 1      −1 −1 0 1 −1 1 1 −1 0 = 0 1 0 1 0 1 0 1  −1 1 1 . = 0 1 e n is isomorphic to Dn , using ( 1 1 ) in the role of r and ( −1 0 ) in Thus, by Theorem 1.1, D 0 1 01 the role of s. This realization of Dn inside GL2 (Z/(n)) should not be confused with the geometric cos(2π/n) − sin(2π/n) 0 realization of Dn in GL2 (R) using real matrices: r = ( sin(2π/n) cos(2π/n) ) and s = ( 10 −1 ). ∼D Corollary 1.3. If n ≥ 6 is twice an odd number then Dn = n/2 × Z/(2).

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= D3 × Z/(2) and D10 ∼ For example, D6 ∼ = D5 × Z/(2). Proof. Let H = hr 2 , si, where r and s are taken from Dn . Then (r 2 )n/2 = 1, s2 = 1, and sr 2 s−1 = r −2 , so Theorem 1.1 tells us there is a surjective homomorphism Dn/2 → H. Since r 2 has order n/2, #H = 2(n/2) = n = #Dn/2 , so Dn/2 ∼ = H. n/2 Set Z = {1, r }, the center of Dn . The elements of H commute with the elements of Z, so the function f : H × Z → Dn by f (h, z) = hz is a homomorphism. Writing n = 2k where k = 2ℓ + 1 is odd, we get f ((r 2 )−ℓ , rn/2 ) = r −2ℓ+k = r and f (s, 1) = s, so the image of f contains hr, si = Dn . Thus f is surjective. Both H × Z and Dn have the same size, so f is injective too and thus is an isomorphism.  There is no isomorphism as in Corollary 1.3 between Dn and Dn/2 × Z/(2) when n is divisible by 4: since n and n/2 are even the center of Dn has order 2 and the center of Dn/2 × Z/(2) has order 2 · 2 = 4. Therefore the centers of Dn and Dn/2 × Z/(2) are not isomorphic, so these groups are not isomorphic. 2. Dihedral groups and generating elements of order 2 In Dn , we can obtain r from s and rs (just multiply: rs · s = rs2 = r), so we can use the reflections rs and s as generators for Dn : Dn = hr, si = hrs, si. In group-theoretic terms, Dn is generated by two elements of order 2. They do not commute: rs · s = r and s · rs = srs = r −1 ss = r −1 . What finite groups besides Dn for n ≥ 3 can be generated by two elements of order 2? Suppose G = hx, yi, where x2 = 1 and y2 = 1. If x and y commute, then G = {1, x, y, xy}. This has size 4 provided x 6= y. Then we see G behaves just like the group Z/(2) × Z/(2), where x corresponds to (1, 0) and y corresponds to (0, 1). If x = y, then G = {1, x} = hxi is cyclic of size 2. If x and y do not commute, it turns out that G is essentially a dihedral group! Theorem 2.1. Let G be a finite non-abelian group generated by two elements of order 2. Then G is isomorphic to a dihedral group. Proof. Let the two elements be x and y, so each has order 2 and G = hx, yi. Since G is non-abelian and x and y generate G, x and y do not commute: xy 6= yx. The product xy has some finite order, since we are told that G is a finite group. Let the order of xy be denoted n. Set a = xy and b = y. (If we secretly expect x is like rs and y is like s in Dn , then this choice of a and b is understandable, since it makes a look like r and b look like s.) Then G = hx, yi = hxy, yi is generated by a and b, where an = 1 and b2 = 1. Since a has order n, n | #G. Since b 6∈ hai, #G > n, so #G ≥ 2n. The order n of a is greater than 2. Indeed, if n ≤ 2 then a2 = 1, so xyxy = 1. Since x and y have order 2, we get xy = y−1 x−1 = yx, but x and y do not commute. Therefore n ≥ 3. Since (2.1)

bab−1 = yxyy = yx,

a−1 = y−1 x−1 = yx,

where the last equation is due to x and y having order 2, we obtain bab−1 = a−1 . By Theorem 1.1, there is a surjective homomorphism Dn → G, so #G ≤ 2n. We saw before  that #G ≥ 2n, so #G = 2n and G ∼ = Dn .

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Theorem 2.1 says we know all the finite non-abelian groups generated by two elements of order 2. What about the finite abelian groups generated by two elements of order 2? We discussed this before Theorem 2.1. Such a group is isomorphic to Z/(2) × Z/(2) or (in the degenerate case that the two generators are the same element) to Z/(2). So we can define new dihedral groups D1 = Z/(2), D2 = Z/(2) × Z/(2). In terms of generators, D1 = hr, si where r = 1 and s has order 2, and D2 = hr, si where r and s have order 2 and they commute. With these definitions, • #Dn = 2n for every n ≥ 1, • the dihedral groups are precisely the finite groups generated by two elements of order 2, • the description of the commutators in Dn for n > 2 (namely, they are the powers of r 2 ) is true for n ≥ 1 (commutators are trivial in D1 and D2 , and so is r 2 in these cases), • for even n ≥ 1, Corollary 1.3 is true when n is twice an odd number (including n = 2) and false when n is a multiple of 4, • the model for Dn as a subgroup of GL2 (R) when n ≥ 3 is valid for all n ≥ 1. However, D1 and D2 don’t satisfy all properties of dihedral groups when n > 2. For example, • Dn is non-abelian for n > 2 but not for n ≤ 2, • the description of the center of Dn when n > 2 (trivial for odd n and of order 2 for even n) is false when n ≤ 2, • the matrix model for Dn over Z/(n) doesn’t work when n ≤ 2, Remark 2.2. Unlike finite groups generated by two elements of order 2, there is no elementary description of all finite groups generated by two elements with equal order > 2. Theorem 2.3. If N is a proper normal subgroup of Dn then Dn /N is a dihedral group. Therefore any nontrivial homomorphic image of a dihedral group is a dihedral group. Proof. The group Dn /N is generated by rs and s, which both square to the identity, so they have order 1 or 2 and they are not both trivial since Dn /N is not trivial. If rs and s both have order 2 then Dn /N is a dihedral group by Theorem 2.1 if Dn /N if nonabelian or it is isomorphic to Z/(2) or Z/(2) × Z/(2) if Dn /N is abelian, which are also dihedral groups by our convention on the meaning of D1 and D2 . If rs or s have order 1 then only one of them has order 1, which makes Dn /N ∼  = Z/(2) = D1 . We will see what the proper normal subgroups of Dn are in Theorem 3.8; aside from subgroups of index 2 (which are normal in any group) they turn out to be the subgroups of hr i. 3. Subgroups of Dn We will list all subgroups of Dn and then collect them into conjugacy classes of subgroups. Our results will be valid even for n = 1 and n = 2. Recall D1 = hr, si where r = 1 and s has order 2, and D2 = hr, si where r and s have order 2 and commute. Theorem 3.1. Every subgroup of Dn is cyclic or dihedral. A complete listing of the subgroups is as follows: (1) hr d i, where d | n, with index 2d ,

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(2) hr d , r i si, where d | n and 0 ≤ i ≤ d − 1, with index d . Every subgroup of Dn occurs exactly once in this listing. In this theorem, subgroups of the first type are cyclic and subgroups of the second type ∼ Dn/d. are dihedral: hr d i ∼ = Z/(n/d) and hr d , r i si = Proof. It is left to the reader to check n = 1 and n = 2 separately. We now assume n ≥ 3. Let H be a subgroup of Dn . The composite homomorphism H ֒→ Dn → Dn /hri to a group of order 2 is either trivial or onto. Its kernel is H ∩ hri. If the homomorphism is trivial then H = H ∩ hri, so H ⊂ hri, which means H = hr d i for a unique d | n. The order of hr d i is n/d and its index in Dn is 2n/(n/d) = 2d . If the homomorphism H → Dn /hri is onto then H/(H ∩ hri) has order 2, so H ∩ hri has index 2 in H. Set H ∩ hri = hr d i, so [H : hr d i] = 2. Since hr d i has order n/d, #H = 2n/d and [Dn : H] = 2n/#H = d. Choosing h ∈ H with h 6∈ hr d i, we know h is not a power of r since hr d i = H ∩ hri, so h is a reflection. Write h = r i s. Then H contains o n n r dk , r dk+i s : 0 ≤ k ≤ − 1 , d which is already 2(n/d) terms, so H = hr d , r i si. Multiplying r i s by an appropriate power of r d will produce an r j s where 0 ≤ j ≤ d − 1, and we can replace r i s with this r j s in the generating set. So we may assume 0 ≤ i ≤ d − 1. The subgroup hr d , r i si is nontrivial and generated by two elements of order 2 (r i s and r d · r i s), so it is isomorphic to a dihedral group. Since r d has order n/d, the order of hr d , r i si is 2(n/d) = 2n/d, whose index in Dn is d . To check the two lists of subgroups in the theorem have no duplications, first we show the lists are disjoint. The only dihedral groups that are cyclic are groups of order 2, and hr d , r i si has order 2 only when d = n. The subgroup hr n , r i si = hr i si has order 2 and r i s is not a power of r, so this subgroup is not on the first list. The first list of subgroups has no duplications since the order of hr d i changes when we change d (among positive divisors of n). If the second list of subgroups has a duplication, say hr d , r i si = hr e , rj si, then computing the index in Dn shows d = e. The reflections in hr d , r i si are all r dk+i s, so r j s = r dk+i s for some k. Therefore j ≡ dk + i mod n, and from d | n we further get j ≡ i mod d. That forces j = i, since 0 ≤ i, j ≤ d − 1.  Corollary 3.2. Let n be odd and m | 2n. If m is odd then there are m subgroups of Dn with index m. If m is even then there is one subgroup of Dn with index m. Let n be even and m | 2n. • If m is odd then there are m subgroups of Dn with index m. • If m is even and m doesn’t divide n then there is one subgroup of Dn with index m. • If m is even and m | n then there are m + 1 subgroups of Dn with index m. Proof. Check n = 1 and n = 2 separately first. We now assume n ≥ 3. If n is odd then the odd divisors of 2n are the divisors of n and the even divisors of 2n are of the form 2d, where d | n. From the list of subgroups of Dn in Theorem 3.1, any subgroup with odd index is dihedral and any subgroup with even index is inside hri. A subgroup with odd index m is hr m , r i si for a unique i from 0 to m − 1, so there are m such subgroups. The only subgroup with even index m is hr m/2 i by Theorem 3.1. If n is even and m is an odd divisor of 2n, so m | n, the subgroups of Dn with index m are hr m , r i si where 0 ≤ i ≤ m − 1. When m is an even divisor of 2n, so (m/2) | n, hr m/2 i

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has index m. If m does not divide n then hr m/2 i is the only subgroup of index m. If m divides n then the other subgroups of index m are hr m , r i si where 0 ≤ i ≤ m − 1.  From knowledge of all subgroups of Dn we can count conjugacy classes of subgroups. Theorem 3.3. Let n be odd and m | 2n. If m is odd then all m subgroups of Dn with index m are conjugate to hr m , si. If m is even then the only subgroup of Dn with index m is hr m/2 i. In particular, all subgroups of Dn with the same index are conjugate to each other. Let n be even and m | 2n. • If m is odd then all m subgroups of Dn with index m are conjugate to hr m , si. • If m is even and m doesn’t divide n then the only subgroup of Dn with index m is hr m/2 i. • If m is even and m | n then any subgroup of Dn with index m is hr m/2 i or is conjugate to exactly one of hr m , si or hr m , rsi. In particular, the number of conjugacy classes of subgroups of Dn with index m is 1 when m is odd, 1 when m is even and m doesn’t divide n, and 3 when m is even and m | n. Proof. As usual, check n = 1 and n = 2 separately first. We now assume n ≥ 3. When n is odd and m is odd, m | n and any subgroup of Dn with index m is some hr m , r i si. Since n is odd, r i s is conjugate to s in Dn . The only conjugates of r m in Dn are r ±m , and any conjugation sending s to r i s turns hr m , si into hr ±m , r i si = hr m , r i si. When n is odd and m is even, the only subgroup of Dn with even index m is hr m/2 i by Theorem 3.1. If n is even and m is an odd divisor of 2n, so m | n, a subgroup of Dn with index m is some hr m , r i si where 0 ≤ i ≤ m − 1. Since r i s is conjugate to s or rs (depending on the parity of i), and the only conjugates of r m are r ±m , hr m , r i si is conjugate to hr m , si or hr m , rsi. Note hr m , si = hr m , r m si and r m s is conjugate to rs (because m is odd), Any conjugation sending r m s to rs turns hr m , si into hr m , rsi. When m is an even divisor of 2n, so (m/2) | n, Theorem 3.1 tells us hr m/2 i has index m. Any other subgroup of index m is hr m , r i si for some i, and this occurs only when m | n, in which case hr m , r i si is conjugate to one of hr m , si and hr m , rsi. It remains to show hr m , si and hr m , rsi are nonconjugate subgroups of Dn . Since m is even, the reflections in hr m , si are of the form r i s with even i and the reflections in hr m , rsi are of the form r i s with odd i. Therefore no reflection in one of these subgroups has a conjugate in the other subgroup, so the two subgroups are not conjugate.  Example 3.4. For odd prime p, the only subgroup of Dp with index 2 is hri and all p subgroups with index p (hence order 2) are conjugate to hr p , si = hsi. Example 3.5. In D6 , the subgroups of index 2 are hri, hr 2 , si, and hr 2 , rsi, which are nonconjugate to each other. All 3 subgroups of index 3 are conjugate to hr 3 , si. The only subgroup of index 4 is hr 2 i. A subgroup of index 6 is hr 3 i or is conjugate to hsi or hrsi. Example 3.6. In D10 the subgroups of index 2 are hri, hr 2 , si, and hr 2 , rsi, which are nonconjugate. The only subgroup of index 4 is hr 2 i, all 5 subgroups with index 5 are conjugate to hr 5 , si, and a subgroup with index 10 is hr 5 i or is conjugate to hr 10 , si or hr 10 , rsi. Example 3.7. When k ≥ 3, the dihedral group D2k has three conjugacy classes of subgroups with each index 2, 4, . . . , 2k−1 .

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We now classify the normal subgroups of Dn , using a method that does not rely on our listing of all subgroups or all conjugacy classes of subgroups. Theorem 3.8. In Dn , every subgroup of hri is a normal subgroup of Dn ; these are the subgroups hr d i for d | n and have index 2d. This describes all proper normal subgroups of Dn when n is odd, and the only additional proper normal subgroups when n is even are hr 2 , si and hr 2 , rsi with index 2. In particular, there is at most one normal subgroup per index in Dn except for three normal subgroups hri, hr 2 , si, and hr 2 , rsi of index 2 when n is even. Proof. We leave the cases n = 1 and n = 2 to the reader, and take n ≥ 3. Since hri is a cyclic normal subgroup of Dn all of its subgroups are normal in Dn , and by the structure of subgroups of cyclic groups these have the form hr d i where d | n. It remains to find the proper normal subgroups of Dn that are not inside hri. Any subgroup of Dn not in hri must contain a reflection. First suppose n is odd. All the reflections in Dn are conjugate, so a normal subgroup containing one reflection must contain all n reflections, which is half of Dn . The subgroup also contains the identity, so its size is over half of the size of Dn , and thus the subgroup is Dn . So every proper normal subgroup of Dn is contained in hri. Next suppose n is even. The reflections in Dn fall into two conjugacy classes of size n/2, represented by r and rs, so a proper normal subgroup N of Dn containing a reflection will contain half the reflections or all the reflections. A proper subgroup of Dn can’t contain all the reflections, so N contains exactly n/2 reflections. Since N contains the identity, |N | > n/2, so [Dn : N ] < (2n)/(n/2) = 4. A reflection in Dn lying outside of N has order 2 in Dn /N , so [Dn : N ] is even. Thus [Dn : N ] = 2, and conversely every subgroup of index 2 is normal. Since Dn /N has order 2 we have r 2 ∈ N . The subgroup hr 2 i in Dn is normal with index 4, so the subgroups of index 2 in Dn are obtained by taking the inverse image in Dn of subgroups of index 2 in Dn /hr 2 i = {1, r, s, rs} ∼ = Z/(2) × Z/(2): • the inverse image of {1, r} is hr i, • the inverse image of {1, s} is hr 2 , si, • the inverse image of {1, rs} is hr 2 , rsh.  Example 3.9. For an odd prime p, the only nontrivial proper normal subgroup of Dp is hri, with index 2. Example 3.10. In D6 , the normal subgroups of index 2 are hri, hr 2 , si, and hr 2 , rsi. The normal subgroup of index 4 is hr 2 i and of index 6 is hr 3 i. There is no normal subgroup of index 3. Example 3.11. The normal subgroups of D10 of index 2 are hri, hr 2 , si, and hr 2 , rsi. The normal subgroup of index 4 is hr 2 i and of index 10 is hr 5 i. There is no normal subgroup of index 5. Example 3.12. When k ≥ 3, the dihedral group D2k has one normal subgroup of each index except for...