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Before Midterm 2

# Before Midterm 2 - Answer key No.2 to selected homework...

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Answer key No.2 to selected homework problems: Math. 110B 7.6 1. aK = K means ae = k K ; so, a K . Conversely, if a K , then ak K for all k K because K is a subgroup. This shows aK K . Pick any k K , then k = ek = ( aa - 1 ) k = a ( a - 1 k ) aK , because K is subgroup. Thus K aK . This combined the earlier reverse inclusion shows that K = aK . 2. (a) (1) Every left coset aK is a subset of G ; so, a aK G . For each g G , g = ge gK ; so, G a aK . (2) Define f : K aK by f ( k ) = ak . To show f is injective, assume that f ( x ) = f ( y ). Then we need to show that x = y . f ( x ) = f ( y ) implies ax = ay ; then, by left-cancellation, we find x = y . x aK implies x = ak for k K ; so, x = f ( k ); therefore, f is surjective. Thus f is a bijection. In particular | aK | = | K | if | K | is finite. (b) We need to show that there are only two possibilities: aK = bK or aK bK = . Thus we need to show: aK bK = ∅ ⇒ aK = bK . Pick x = ak = bk in the intersection. Then b - 1 a = k k - 1 K . Thus a - 1 bk K for all k K ; in particular, bk = ebk = ( aa - 1 ) bk = a ( a - 1 bk ) aK ; thus aK bK . Interchanging the role of a and b in the above argument, we find the reverse inclusion aK bK ; so, aK = bK as desired. We are going to prove that G is a union of left cosets c 1 K , c 2 K ,... without overlapping (such a union is written as G = c 1 K c 2 K · · · c m K (here m = [ G : K ]) using the disjoint union symbol “ ”. Now start with K = eK , and write c 1 = e . Pick c 2 outside c i K = K if any. If there is none, we have G = K . Otherwise, c 1 K = c 2 K because c 2 is not in c 1 K ; so, we have c 1 K c 2 K = . Suppose we have found c 1 , . . ., c n so that c i K c j K = for any pairs ( i, j ) with i = j . By induction on n , we try to find c n +1 with the same disjointness property. Take c = c n +1 outside c 1 K c 2 K ∪ · · · ∪ c n K if any. If there is none, we have G = c 1 K c 2 K · · · c n K ; so, we are done, getting a disjoint union expression of G by left cosets. Otherwise, we have c j K = cK for all j = 1 , 2 , . . ., n because c c j K . Thus only possibility is to have cK c j K = ; so, by induction on n , we can keep going. Since G is a finite group, the sequence c 1 , c 2 , . . . has to end at some point (exhausting all elements), so at the end, we find the disjoint union expression. Then | G | = m j =1 | c j K | . By (a), | c j K | = | K | , we find | G | = m | K | ; so, we are done. (c) By (b), the number of left cosets is | G | / | K | . The number of right cosets is also | G | / | K | by Lagrange theorem; so, they are equal to each other. 4. Note that e is made of single element e . Then xex - 1 = xx - 1 = e ; so, x e x - 1 e ; so, by Theorem 7.34 (3), e is normal. Since G is a group xGy G ; in particular, xGx - 1 G ; so, again by Theorem 7.34 (3), it is normal. 5. (a) Hint: Compute the product of two upper triangular matrices to show it is again upper triangular. Also inverse of an upper triangular matrix is again upper triangular by computation.

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