Classifying groups of order 1-11, p, and 2p. Part 2: Enter Lagrange!

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1 Classifying groups of order 1-11, p, and 2p Part 2: Enter Lagrange! Intro: This is the story of how we classified the groups of order 1-11, and all groups of order p, 2p where p is prime. Like we did in class and homework, I ll introduce new tools as needed. Tool 4: Theorem of Lagrange The Theorem of Lagrange states that the order of any subgroup, H, of any finite group G, must divide the order of G. An immediate consequence of this theorem is that the order of any element of a finite group must divide the order of the group (because the order of any element is the same as the order of the subgroup it generates). Our next tool will actually be the proof of the Theorem of Lagrange. But for now, let s see what we can get from the statement alone. Suppose that the order of G = p where p is prime (from now on, always assume that p is a prime number). Then since G has only one identity element, and only the identity has order 1, we know that G must have an element of order p, which in turn implies that G is cyclic. So, we can use Lagrange to knock off infinitely many values of N! Any group of prime order, p, is isomorphic to Zp. This could have saved a lot of trouble doing N = 5! Tool 5: Proof of Theorem of Lagrange The Theorem of Lagrange is proved by partitioning the group G into the cosets of the subgroup H. A number of facts come together to give the result. First, is the observation that all cosets ah of G in H have the same number of elements as H. (This follows from the cancellation law.) Second, is the observation that a bh implies that ah = bh. (This says that cosets are either equal or do not intersect at all.) Finally, there is the observation that any element is in at least one coset (a is in ah). Using these facts, you can partition G into the cosets of H in G. The union is the whole thing because every element is in a coset. And they are pairwise disjoint if you only list each coset once. The number of cosets of H in G is denoted by [G: H] and from the above, it is clear that G = H [G: H. This idea of listing the elements of G as cosets of a subgroup turns out to be very powerful!

2 Tool #6: If Every Element of a Group is a Self- Inverse, the Group is Abelian Proof: Suppose every element of G is a self- inverse. Now, let a, b be elements of G. Then (ab)(ab) = e. Multiplying both sides by a on the left we get bab = a. Then, multiplying by b on the left, we get ab =ba. So, G is abelian. Note on FTFGAG: In class, we used FTFGAG to make things a bit easier. But people seemed uncomfortable with that (not being sure why it is true) so I will try to avoid that here. N = 6: If G has an element of order 6, then it is cyclic an isomorphic to Z6. So, suppose that G doesn t have an element of order 6. If all of the non- identity elements are of order 2, then G is abelian. Let a be an element of order 2. Then we have the elements of the subgroup {e, a} in G. Now, let b be an element not in {e, a}. Then we can list the elements of the coset {b, ab} giving us a total of 4 elements. Note that ba = ab, so we can t generate any more elements using just a and b. So, now let c be an element of G not in {e, a} or {b, ab}. We may then list the elements of the coset {c, ac}. Now we have listed 6 elements that must be in the group. But what of the product abc? This must be in the group. Is it already in our list or do we have an illegal 7 th element? abc = e implies that c = ab which is a contradiction since c is not in {b, ab}. abc = a implies that bc = e which implies b = c (contradiction). abc = b implies that ac = e which implies a = c (contradiction). abc = c implies that ab = e which implies a = b (contradiction). abc = ab implies that c = e (contradiction). abc = ac implies that b = e (contradiction). So, it isn t possible for all elements to be of order 2. Actually one could show that in an abelian group, if all elements are of order 2, then the order of G must be 2 n for some n. I might do this later if I think it will save some time! Back to work! Now we know that G must have an element, a, of order 3. Let s start by listing the elements of the subgroup generated by a: {e, a, a 2 }. Now let b be an element of G not in <a>. Then the three remaining elements of G are from the coset {b, ab, a 2 b}. Since all elements can be written as a power of a or a power of a times b, we can completely characterize multiplication in this group by figuring out how to write ba in this form. We know that ba is not in {e, a, a 2 }, since b is not equal to a or e. We also know that ba is not b, since a is not e. This gives us two choices for ba. Either ba = ab or ba = a 2 b.

3 Before considering these cases, let s consider the order of b. If b is order- 3, then we get the coset {b 2, ab 2, a 2 b 2 }. Now, b 2 cannot be in <a> because if it were then so b. Can b 2 be in {b, ab, a 2 b}? If not, then we have a brand new coset and our group has 9 elements! b 2 is not b because b is not e. b 2 is not ab because b is not a. b 2 is not a 2 b because b is not a 2. Thus we know that b must be of order 2. Case 1: ba = ab. I claim that the order of ab is 6. It isn t 1 since a is not b. Suppose that abab = 1, then aabb = 1 which implies that aa = 1 (since b is order 2). But this isn t true since a is order 3. So, the order of ab isn t 2 either. Suppose (ab) 3 = e. Then since ab = ba we have a 3 b 3 = e. This implies that b 3 = e since a is order 3. But b 2 = e, so this means that b = e (contradiction). This guarantees ab is order 3 since it can t be anything else. Just for fun, note also that (ab) 6 = a 6 b 6 =( a 3 ) 2 ( b 2 ) 3 = 3. So, this Case only happens if we have an element of order 6 (which we already handled). So, we know that if G is not isomorphic to Z6, then it can be generated by an order- 3 element a and an order- 2 element b with the property that ba = a 2 b. This is our old buddy D6 (the group of symmetries of an equilateral triangle)! LOOKING INTO THE FUTURE! We just did a semi- direct product. Since we proved we had an order- 3 element, we know we have a normal subgroup of order 3. We then proved we also had an order- 2 element (else we would have at least 9 elements). This gave us a normal subgroup N = <a> and a subgroup H = <b> such that N H = G and N H = {e}. This means that G = NH and we know how to multiply any two elements of NH as long as we know how to express any product hn as n h for some n in N. Normality of N guarantees that this can always be done, but the trick is figuring out what the n will be. In the future we will learn that for any given h, the map taking n to n (where hn = n h) will be an automorphism on N. Further, we will learn that the map associates each h with the automorphism that tells how to commute h with elements of n is a homomorphism. Now, a group of order 3 has two automorphisms (a can go to a or a 2 ). So Aut N is isomorphic to Z2. H is also isomorphic to Z2. There are two viable homomorphisms from a group of order 2 to a group of order 2. One is the trivial map. With this map, for every h, the rule for commuting with elements of h is given by the identity automorphisms. In other words, for every h in H, hn = nh for every n in N. So, with this homomorphism, we have a group isomorphic to the direct product Z2 Z3 which is isomorphic to Z6. The second will take e in H to the idendity automorphism and b in H to the nontrivial automorphism given by a a 2. This means that ba = a 2 b. So, in summary for N= 6 we have two groups (up to isomorphism) Z6 and D6.

4 N= 7: (Done! 7 is prime) N= 8: If G has an element of order 8 then it is isomorphic to Z8. So, let s suppose that it doesn t. Case 1: (All non- identity elements of G are of order 2.) Then we can pick out three different elements a, b, c and note that e, a, b, c, ab, ac, bc, and abc are all different and all must be elements of G. In fact, because G must be commutative, this tells us that each element can be uniquely expressed in the form a j b k c l where j, k, and l are each 0 or 1. a j b k c l a i b m c n = a j+i b k+m c l+n. This observation makes it clear that G is isomorphic to <a> <b> <c>, which is isomorphic to Z2 Z2 Z2. Case 2: (G contains an element, a, of order 4.) We start by listing the elements of <a>, {e, a, a 2, a 3 }. Now pick an element, b, of G not in <a>, and list the elements of <a>b, {b, ab a 2 b, a 3 b}. This is all of the elements of G. Again, we see that all elements are expressed in terms of two generators. We simply need to determine the order of b and how to express ba in the form a k b l where 0 k 3 and 0 l 1 (or in other words, express ba as an element of <a> or <a>b). So, we break things into cases: Case 1: b = 2. We can quickly rule out ba as an element of <a> because b is not in <a>. Or, if we recall that a subgroup of index two must always be normal, then we know that in this case our left coset and right cosets must be equal. Since ba is in b<a> it must be in <a>b. Of course ba can t be b, so we are left with three choices, ba = ab, ba = a 2 b, and ba = a 3 b. If ba = ab, then a k b l a m b n = a k+m b l+n. This makes it clear that this group is isomorphic to <a> <b> which is in turn clearly isomorphic to Z4 Z2. Now, if ba = a 2 b, then (ba) 2 = a 2 bba = a 3. But a 3 is an element of order 4, which means that ba is NOT order 2 or 4. And if ba is order 8 the group is cyclic which implies it is abelian which makes it hard to accept that ba = a 2 b! So this can t happen. If ba = a 3 b. Then we have the same generators and relations as D8, so G is isomorphic to D8. Case 2: b = 4. Again we can rule out ba as an element of <a> because b is not in <a> (or because <a> is a normal subgroup. And of course ba is not b. So, we have again three choices ba = ab, ba = a 2 b, and ba = a 3 b. Suppose ba = ab. Note that the order of b 2 is 2. Let s see where b 2 lives. We know b 2 must be in either {e, a, a 2, a 3 } or {b, ab a 2 b, a 3 b}. If b 2 is in {b, ab a 2 b, a 3 b}, then we have an order- 2 element not in <a> that commutes with a. This is exactly the situation that gave us Z4 Z2.

5 So, suppose that b 2 is in <a>, then it must equal a 2. Now, check out order of ab. (ab) 2 = abab = a 2 b 2 = a 4 = e. So, again we have an order- 2 element not in <a> that commutes with a. So, again our group is isomorphic to Z4 Z2. Suppose ba = a 2 b. Then (ba) 2 = a 2 ba 2 b = a 2 a 2 bab = bab = a 2 b 2. If a 2 b 2 = e, then a 2 = b 2. Then substituting into ba = a 2 b, we get ba = b 3 which would imply that a = b 2. So, (ba) 2 is not e. (ba) 4 = a 2 b 2 a 2 b 2 = a 10 b 4 = a 2. This means the order of ba is 8. But, as before, a non- abelian group of order- 8 cannot have an element of order- 8. So this can t happen! Finally, suppose that ba = a 3 b. Then we have the generators and relations for the quaternion group. Let s show that. First I ll use my rules for a and b to make my operation table for G. Then I ll define a mapping to change the names and see if I get the quaternion table. It will help to observe that b 2 cannot be in <a>b (easy check). This means that b 2 = a 2 since that is the only order 2 element in <a>. e a a 2 a 3 b ab a 2 b a 3 b e e a a 2 a 3 b ab a 2 b a 3 b a a a 2 a 3 e ab a 2 b a 3 b b a 2 a 2 a 3 e a a 2 b a 3 b b ab a 3 a 3 e a a 2 a 3 b b ab a 2 b b b a 3 b a 2 b ab a 2 a e a 3 ab ab b a 3 b a 2 b a 3 a 2 a e a 2 b a 2 b ab b a 3 b e a 3 a 2 a a 3 b a 3 b a 2 b ab b a e a 3 a 2 Now, let e = 1, a = i, b = j, and let a 2 = - 1 since it is an order 2 element. Let s give ab the name k. Then ba = a 3 b becomes ji = - ij = - k. Let s change the names and see what we have. 1 i i j k - j - k 1 1 i i j k - j - k i i i 1 k - j - k j i 1 i - j - k j k - i - i 1 i k j k - j j j - k - j k - 1 i 1 - i k k j - k - j - i - 1 i 1 - j - j k j - k 1 - i - 1 i - k - k - j k j i 1 - i - 1 Hey, that looks just like the Quaternion group! So for N=8 we found FIVE groups: Z8, Z2 Z2 Z2, Z4 Z2, D8, and Q8.

6 Tool #7: The Handy Little Coset Lemma Let G be a group and H a subgroup of G. Then for any a, b G, ah = bh iff b - 1 a is in H. Proof: Suppose that ah = bh, then since a is clearly in ah, we know that a is in bh. Then a = bh where h is in H. This means that b - 1 a = h is in H. (By symmetry of set equality, we also get that a - 1 b is in H. Now suppose that b - 1 a is in H. This means that b - 1 a = h for some h in H. Let x ah. Then x = ah for some h in H. Solving the above equation for a, we get a = bh. So, x = bh h is an element of bh. Now let x bh. Then x = bh for some h in H. Solving the above equation for b, we get b = ah - 1. So, x = ah - 1 h is an element of ah. Thus ah = bh. Warning! If H is not normal, then it is NOT true that ah = bh if and only if ab - 1 is in H. For example, in D8, R{I, F} = FR 3 {I, F} but RFR 3 = FR 2 is not in {I, F}. (Here a = R and b = FR 3 and b is a self- inverse.) So, when you use HLCL on LEFT COSETS make sure you do the inverse first in the product! Since we have been mostly working with right cosets, let s see how this converts. I claim it works backwards : Ha = Hb if and only if ab - 1 is in H. I ll verify one direction and leave the rest to the reader (just mimic the above). If Ha = Hb, then since a is in Ha, a is in Hb. This means a = hb for some h in H. So that ab - 1 is in H. When you use HLCL on RIGHT COSETS make sure you do the inverse second in the product! N = 9: If G has an element of order 9, then G is isomorphic to Z9. If not, then Lagrange assures us that all of the non- identity elements are of order 3. Let a be an order- 3 element. List the elements of <a> = {e, a, a 2 }. Now let b be an order 3 element not in <a>. Observe that by the HLCL we can t have <a>b = <a>b 2 since b 2 b 2 = b is not in <a>. So our group consists of the three cosets <a>, a>b, and <a>b 2. This means that each element can be expressed as a product of a power of a (between 0 and 2) and a power of b (between 0 and 2). So, let s find out how to express ba this way and this will completely determine the group. Clearly ba is not in <a> and ba is neither b nor b 2. Four guys left to check. ba = a 2 b would imply that (ba) 2 = b 2, which would make ba an order- 6 element (nope!). ba = a 2 b 2 would imply that (ba) 2 = e, which would imply ba = e (nope!) or ba = 2 (nope!). ba = ab 2 would imply that (ba) 2 = a 2, which would make ba an order- 6 element (nope!). So, we must have ab = ba. Then a n b m a k b l = a n+k b m+l, so that G is isomorphic to Z3 Z3. So, for N = 9, there are only two groups: Z9 and Z3 Z3.

7 Tool #8 If G is finite and contains only elements of order 2 (or 1), then G = 2 n for some positive integer n. Actually we can do better than that! G must be isomorphic to a direct product of n copies of Z2 for some positive integer n. (Informal) Proof: Suppose G is finite and contains only elements of order 2 (or 1). Then we know that G is abelian (Tool #6). Now, start grabbing elements of G and forming all possible products with them until you generate all elements of G. Let n be the size of the smallest generating set of G, {a1, a2,, an}. Then every element of G can be expressed uniquely in the form a1 k1 a2 k2 an kn where all those superscripts are either 0 or 1. And because the group is abelian we also know that (a1 k1 a2 k2 an kn )(a1 l1 a2 l2 an ln ) = a1 k1+l1 a2 k2+l2 an kn+ln. So this group is clearly isomorphic to Z2 Z2 Z2 (where there are n copies of Z2). Tool #9 In a group of order 2p (p a prime bigger than 2), the product of an element of order 2 is either an order- 2p element or an order- 2 element. Our proof will show more than this! If we let these elements be a and b, we will see that either ab = ba (and ab = 2p) or ba = a p- 1 b (and ab = ba = 2) Proof: Let G = 2p, and a, b in G with a = p and b = 2. Clearly ab and ba are not 1. Now since <a> is normal in G, ba is in <a>b. Since ba clearly can t be b, we know that ba = a k b for some k between 1 and p- 1. If ba = ab then (ab) 2 = (ba) 2 = a 2 b 2 = a 2 and (ab) p = (ba) p = a p b p = b. So ab and ba can t have order 1, 2, or p, so they must have order 2p. Now, if ba = a k b where k is bigger than 1, we see that (ab) p = a something b p = a something b. You can see this by imagining using the rule ba = a k b repeatedly to write the (ab) p as a product of a power of a times a power of b. This rule leaves b unchanged, so you will have to end up with b p in your final expression. No matter what the power of a is, this will never be e. Note that if ba = a k b where k is bigger than 1, we can t have an order- 2p element (if we did, we would be looking at a cyclic, and therefore abelian, group). This means that in this case the order of ab and ba must both be 2 (the only option left). How can this happen? If (ba) 2 = e, then a k bba = a k+1 = e. This means that k must be p- 1.

8 N = 10: If G has an element of order 10, then it is isomorphic to Z10. Now suppose that it does not. By Tool #8, G must have an element, a, of order 5. We also know it has at least one element, b, of order 2 (Tool #2). We can list the elements of G as <a> = {e, a, a 2, a 3, a 4 } and <a>b = {b, ab, a 2 b, a 3 b, a 4 b}. So all we have to do now is figure out which element of <a>b is ba. We ll since we are in the non- cylic case, we know that our only choice is ba = a 4 b. So the group must be isomorphic to D10. So for N=10 we only get two groups: Z10 and D10. N = 2p: These all work the same way! Either G is Z2p or we must have an element, a, of order p (using Tool #8). It also has an element, b, of order 2 (Tool #2). And since we are now in the non- cyclic case we must have ba = a p- 1 b. This means we are looking at D2p. So a group of order 2p is either isomorphic to Z2p or D2p. The 100 s chart below shows how much progress we have made in classifying groups of order <= 100: Blue stands for victory! The number in parentheses is how many groups there are of that order (5)