EXAMPLES CLASS 2 MORE COSETS, FIRST INTRODUCTION TO FACTOR GROUPS Let (G, ) be a group, H G, and [G : H] the set of right cosets of H in G. We define a new binary operation on [G : H] by (1) (Hx 1 ) (Hx 2 ) = H(x 1 x 2 ). This may look familiar to some of you: when H is a normal subgroup of G, this is exactly the binary operation on the factor group G/H. Since we haven t defined normal subgroups yet, the point of these questions is to get a feel for what it means to define a binary operation on cosets, and to see what can go wrong when H is not a normal subgroup. 1. G = (Z, +), H = nz, n 2. 1. Questions (a) If you didn t do this last week, work out the cosets for H in G (for a fixed n). Note that they have the form H + x rather than Hx since the binary operation on G is +. (b) Translating the definition in equation (1) to this setting, we have (H + x 1 ) (H + x 2 ) := H + (x 1 + x 2 ). Apply this definition to the cosets you ve worked out in (a). How does this compare to the group (Z n, +) that you re familiar with? 2. G = D 3, H = r G. (a) Write down the (right) cosets of H in G. (b) With the binary operation on [G : H] defined in (1), what group is [G : H] isomorphic to? 3. G = D 3, H = {e, s}. Find an example to show that the multiplication of cosets (1) is not well defined on [G : H], i.e. cosets Hx 1 = Hx 2 and Hy 1 = Hy 2 such that (Hx 1 ) (Hx 2 ) (Hy 1 ) (Hy 2 ). 1
2 EXAMPLE CLASS 2 2. Hints To work out the right cosets, follow the same pattern as in Example Class 1: find an element of G that hasn t appeared in any cosets so far, work out its coset, and keep doing this until you ve written down the whole group. Remember that the cosets partition G (viewed as a set rather than a group), so all we re doing is breaking up G into smaller pieces in a well-defined way. In 2(b), it may help to think about the order of [G : H]: we know G : H = G / H, and there aren t many different groups of this size (up to isomorphism). It s a lot easier to write down an isomorphism once you know what group you re trying to identify [G : H] with!
1. G = (Z, +), H = nz, n 2. EXAMPLE CLASS 2 3 3. Answers (a) If you didn t do this last week, work out the cosets for H in G (for a fixed n). Note that they have the form H + x rather than Hx since the binary operation on G is +. (b) Translating the definition in equation (1) to this setting, we have (H + x 1 ) (H + x 2 ) := H + (x 1 + x 2 ). Apply this definition to the cosets you ve worked out in (a). How does this compare to the group (Z n, +) that you re familiar with? Answer: See the writeup of Example Class 1 for a description of the cosets of nz in Z. They are: nz + 0 = {..., n, 0, n, 2n,... } nz + 1 = {..., n + 1, 1, n + 1, 2n + 1,... }. nz + (n 1) = {..., 1, n 1, 2n 1, 3n 1,... } The transversal set is a set T consisting of exactly one element from each coset. Here we have chosen T = {0, 1, 2,..., n 1} but there are many other choices. The chosen element from each coset is representing that coset: in this example, since H = nz consists of multiples of n, a coset nz + x consists of all integers which give a remainder of x when divided by n. So 1 and n + 1 represent the same coset because they tell us the same piece of information: all elements in nz + 1 = nz + (n + 1) give a remainder of 1 when divided by n. We decide that we might as well think of 1 and n + 1 as equal here, because their cosets are equal. Now consider [Z : nz], the set of cosets of nz in Z (here we don t need to specify right/left cosets because Z is abelian) and define the binary operation (1) on it. We ll take T = {0, 1, 2,..., n 1} for simplicity. If nz + x, nz + y are two cosets in [Z : nz], then their sum is defined to be the coset nz + (x + y). This makes sense: for any an + x nz + x, any bn+y nz+y, their sum is an+x+bn+y = (a+b)n+(x+y) nz+(x+y). What is nz + (x + y)? If (x + y) T then we will leave it in that form; otherwise, x + y n and we have an equality of cosets nz + (x + y) = nz + (x + y n). All we have done is reduced x + y modulo n. In other words, the set of cosets [Z : nz] with the binary operation in (1) behaves in exactly the same way as (Z n, +), except that the elements of [Z : nz] are cosets and the elements of Z n are integers. In fact, this is often
4 EXAMPLE CLASS 2 how Z n is defined: we fix the transversal T = {0, 1, 2,..., n 1} for nz in Z and write x to mean the coset nz + x. So if you re happy working with integers modulo n, you ve already mastered one example of working with cosets! 2. G = D 3, H = r G. (a) Write down the (right) cosets of H in G. (b) With the binary operation on [G : H] defined in (1), what group is [G : H] isomorphic to? Answer: Again, see Example Class 1 for the process of working out the cosets. We ll work with right cosets here, but we ll see in Week 4 that H is a normal subgroup of G, which means that the left cosets and right cosets of H in G are equal. We have Hr = {e, r, r 2 } Hs = {s, rs, r 2 s} Here we re working with the transversal T = {r, s}, although again there are many other choices. The following approach may help you understand the group structure on the cosets: we are essentially defining a group structure on the elements of T using the binary operation on G, and we force this binary operation to be closed on T by declaring that anything in the coset Hx is equal to x, for each x T. The binary operation on the set of cosets [G : H] is Hx 1 Hx 2 = H(x 1 x 2 ), where the product x 1 x 2 on the RHS is the binary operation in G. In our example, we have (Hr) (Hr) = H(r 2 ) = Hr, (Hr) (Hs) = H(rs) = Hs, (Hs) (Hr) = H(sr) = Hs, (Hs) (Hs) = H(s 2 ) = Hr, since Hr = Hr 2 as cosets, since Hrs = Hs as cosets, since sr = r 1 s = r 2 s and Hr 2 s = Hs, since s 2 = e and He = Hr. Equally, we could have declared that from now on we were going to assume that e = r = r 2 and s = rs = r 2 s, and worked out what structure this defined on our transversal set T = {r, s} when we multiply them using the binary operation from G. We get r r = r, r s = s, s r = s, s s = r, which is exactly what we had in the coset computation above as well. Notice that r is acting as the identity in this new group (which makes sense, it s in the same coset as e), and s is the only non-trivial element. There is only one group with this property: (Z 2, +). We obtain an isomorphism [D 3 : r ] = (Z 2, +).
EXAMPLE CLASS 2 5 3. G = D 3, H = {e, s}. Find an example to show that the multiplication of cosets (1) is not well defined on [G : H], i.e. cosets Hx 1 = Hx 2 and Hy 1 = Hy 2 such that (Hx 1 ) (Hx 2 ) (Hy 1 ) (Hy 2 ). Answer: This question illustrates why we only ever consider the binary operation (1) when H is a normal subgroup of G. The right cosets of H in G are He = {e, s} Hr = {r, sr} Hr 2 = {r 2, sr 2 } We have Hsr = Hr and Hr 2 = Hsr 2, since any element of a coset represents that coset. However, if we try to multiply them: (Hr 2 ) (Hsr) = H(r 2 sr) = H(r 1 sr) = H(sr 2 ) (Hsr 2 ) (Hr) = H(sr 3 ) = Hs But Hsr 2 and Hs are completely different cosets! The binary operation from (1) is not well defined, because we get different answers to the same question depending on how we look at it. Thinking about this in the same terms as question 2: we ve tried to declare that e = s, r = sr, and r 2 = sr 2 (and no other equalities), so we would expect that r 2 sr = sr 2 r. But r 2 sr = sr 2, while sr 2 r = s. We decided that sr 2 and s were not equal to each other, which means that our multiplication does not make sense.