Problems section 23

: 136.330 Intro to Algebra: Problems section 23
   By Anonymous on Saturday, December 07, 2002 - 09:49 pm: Edit Post

23.16 Prove that if q is a homomorphism of G onto H, B Ñ H, and A = {g element of G : q(g) element of B}, then A ÑG.

   By Sasho on Saturday, December 07, 2002 - 09:55 pm: Edit Post

23.16. take a in A and g in G. Show gag-1 is in A. This means, show q(gag-1) is in B. But q(gag-1)=q(g)q(a)q(g-1) and since q(a) is in B and since B is normal, q(g)q(a)q(g-1) is in B.

   By Anonymous on Sunday, December 08, 2002 - 01:41 pm: Edit Post

23.15 Find all homomorphic images of Zn (n a positive integer).

23.20 Assume that H and K are subgroups of a group G and K Ñ G.
a) Prove that HK = {hk : h in H and k in K } is a subgroup of G, and that K Ñ HK.
b) Define q : H ® HK/K by q(h) = Kh for h in H. Verify that q is a homomorphism of H onto HK/K.

   By Sasho on Sunday, December 08, 2002 - 05:29 pm: Edit Post

23.15 The homomorphic images of Zn are isomorphic to the corresponding factor groups of Zn. So, find all subgroups of Zn (check certain theorems on cyclic groups), then consider the corresponding factor groups and you are done.

23.20 (a) I will only show HK is closed under multiplication. Take x,y in HK. So x=h1k1, where h1 is in H and k1 is in K. Similarly, y= h2k2, where h2 is in H and k2 is in K. Now: xy=h1k1h2k2=h1h2h2-1k1h2k2=(h1h2)(h2-1k1h2)k2 where I put parenthesis just for emphasis. Now (h1h2) is in H since H is subgroup; (h2-1k1h2) is in K since K is normal, so (h2-1k1h2)k2 is also in K since K is a subgroup. So, to summarize, (h1h2) is in H, and the rest ((h2-1k1h2)k2) is in K, so that xy is in HK.

p.s. You supply neked statements of the problems as if you are communicating with an answers-generating machine.

   By Anonymous on Sunday, December 08, 2002 - 06:23 pm: Edit Post

I just used the special characters formulas given in the Formatting folder.

   By Sasho on Sunday, December 08, 2002 - 06:34 pm: Edit Post

Your formatting is fine (and your posts are easy to read). Thanks for taking your time to use it. That was not my point though. But - never mind my point; forget it.

Posting is currently disabled in this topic. Contact your discussion moderator for more information.