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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.

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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.

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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.

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23.15 The homomorphic images of Z_{n} are isomorphic to the corresponding factor groups of Z_{n}. So, find all subgroups of Z_{n} (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=h_{1}k_{1}, where h_{1} is in H and k_{1} is in K. Similarly, y= h_{2}k_{2}, where h_{2} is in H and k_{2} is in K. Now: xy=h_{1}k_{1}h_{2}k_{2}=h_{1}h_{2}h_{2}^{-1}k_{1}h_{2}k_{2}=(h_{1}h_{2})(h_{2}^{-1}k_{1}h_{2})k_{2} where I put parenthesis just for emphasis. Now (h_{1}h_{2}) is in H since H is subgroup; (h_{2}^{-1}k_{1}h_{2}) is in K since K is normal, so (h_{2}^{-1}k_{1}h_{2})k_{2} is also in K since K is a subgroup. So, to summarize, (h_{1}h_{2}) is in H, and the rest ((h_{2}^{-1}k_{1}h_{2})k_{2}) 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.

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

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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.