Questions for test # 3

: 136.330 Intro to Algebra: Questions for test # 3
   By waiteduntillthelastminute on Tuesday, November 26, 2002 - 06:31 pm: Edit Post

There is some questions from the text book that cause me problems. The first one is 15.27, then 17.13 has "there is five" for answer can you tell me what are they? And also, 17.18 says find [G:H], what is it?

   By Anonymous on Tuesday, November 26, 2002 - 08:28 pm: Edit Post

Could you please help me out with a few questions.
14.29 Prove that a group is Abelian if each of its nonidentity elements have order 2.

15.22 b)List the elements in the subgroup
X of Z4 X Z8.

17.4 Find[Z40 : ]

17.25 Prove that if H is a subgroup of G,
[G:H]= 2, a,b in G and a,b not in H, then ab in H.

   By Sasho on Tuesday, November 26, 2002 - 08:49 pm: Edit Post

Regarding the first post above: I need to see the statements of the questions that bother you (no text book around me).

14.29: Want to show that ab=ba for any two elements. Start with (ab)2=1 (since the order of any element is 2). So (ab)(ab)=1. So, After multiplying first by b-1 and then by a-1) ab=b-1a-1. But since a2=1 it follows that a-1=a; similarly b-1=b; so ab=ba as wanted.

15.22 the question is not stated clearly (what is X?)

17.4. The question is not stated clearly ([Z40: ??].)

17.25 (I will be using below the following easy exercise: x is in H only if Hx=H) Since [G:H]=2, there are two right cosets of H in G, one of them is H and the other is the rest of G (call it K). Since a is not in H, Ha is not H. So Ha=K. Since b is not in H, neither is b-1. So Hb-1 is also not H, so it must be K. So Ha=Hb-1. After multiplying both sides by b we get Hab=H. So ab is in H.

   By Anonymous on Tuesday, November 26, 2002 - 10:00 pm: Edit Post

Just a few questions

18.13 Use the mapping theta([a]6) = ([a]2,[a]3) to show that Z6 equivalent Z2 X Z3.

19.12 Is there a noncyclic group of order 39?

21.34 define theta: Z X Z -> Z by theta((a,b)) = a + b. Verify that theta is a homomorphism and determine ker(theta)


   By Sasho on Tuesday, November 26, 2002 - 10:35 pm: Edit Post

18.13. I am assuming you mean isomorphic (rather than equiavalent) and that q:Z6 -> Z2xZ3 is defined by q([a]6)=([a]2, [a]3).

I will only prove that q is onto; perhaps that will help you see what is to be done. Compute: q([0]6)=([0]2, [0]3), q([1]6)=([1]2, [1]3), q([2]6)=([2]2, [2]3)=([0]2, [2]3), q([3]6)=([3]2, [3]3)=([1]2, [0]3) ... etc going through all of the elements of Z6 and checking that the images will cover all of the elements of Z2xZ3.

19.12 Are you sure this is the question 19.12? Is not the word "abelian" somewhere there?

21.34. q((a,b)+(c,d))=q((a+c,b+d))=a+c+b+d. On the other hand q((a,b))+q((c,d))=(a+b)+(c+d) and we see we get equal numbers. That showed that q is a homomorphism. The kernel is obviously the set of all pairs of the type (a,-a) (only these pairs are sent to 0).

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