# Math 345 Sp 07 Day 7. b. Prove that the image of a homomorphism is a subring.

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1 Math 345 Sp 07 Day 7 1. Last time we proved: a. Prove that the kernel of a homomorphism is a subring. b. Prove that the image of a homomorphism is a subring. c. Let R and S be rings. Suppose R and S are isomorphic. Prove that if R is commutative S is commutative. d. Let R and S be rings. Suppose R and S are isomorphic. Prove that if R has a unity (multiplicative Identity) then S has a unity. e. Let R and S be rings. Suppose R and S are isomorphic. Prove that if R has no zero divisors then S has a no zero divisors (and if R has a zero divisor then S has a zero divisor). f. Let R and S be rings. Suppose R and S are isomorphic. Prove that each nonzero element of R has an inverse then each nonzero element of S has an inverse. 2. Note that c, d, and e imply that if R and S are isomorphic then R is an I.D. if and only if S is an I.D. while c, d, and f imply that if R and S are isomorphic then R is a field if and only if S is a field. 3. Review of quotient groups a. What is a quotient group? What do the elements look like? Answer: A quotient group consists of the cosets of a normal subgroup under the operation ahbh = abh. b. Find all possible quotient groups of D 6, the symmetries of an equilateral triangle {I, R, R 2, F, FR, FR 2 }. By the way the operation table for D 6 is: I R R 2 F FR FR 2 I I R R 2 F FR FR 2 R R R 2 I FR 2 F FR R 2 R 2 I R FR FR 2 F F F FR FR 2 I R R 2 FR FR FR 2 F R 2 I R FR 2 FR 2 F FR R R 2 I

2 Answer: There is only one. Use the subgroup H = {I, R, R 2 }. The only other coset is FH = {F, FR, FR 2 }. The table is: H FH H H FH FH FH H c. Why can t H = {I, F} be used to construct a quotient group? What is wrong with the following? Cosets are IH ={I, F}, RH = {R, FR 2 }, and R 2 H = {R 2, FR} Using ahbh = abh, The operation table for D 6 /H is IH RH R 2 H IH IH RH R 2 H RH RH R 2 H IH R 2 H R 2 H IH RH So what is wrong with that? I ll tell you what is wrong. This operation is not well defined! Note that IH and FH are the same but IH RH = RH and FH RH = FRH. This is very bad since RH and FRH are not the same! d. What has to happen for the operation ahbh = abh to be well defined so that we actually get a group? (To be revealed below) e. In 344 we figured out that everything would work out if gh = Hg for every g G. We defined a normal subgroup to be one that has this property. We concluded with the fact that a subgroup can be used to make a quotient group if and only if it is normal. f. There is an alternative (equivalent) definition of normal subgroup. To figure out what it might be, let s focus on getting the operation to be well defined. Here is what we need to have happen: If ah = ch and bh = dh, we need abh = cdh (Just as we need 2/3 + 1/5 to be the same as 4/6 + 3/15) g. To make things easier, lets focus on a simpler situation: Let H be a subgroup of a group G and let g G and h H. Clearly hh = eh where e is the identity in G. So if our quotient group operation is going to be well defined, we need to get the same answer when we multiply these two versions of this coset by gh. So we need hgh = gh.

3 So what is necessary to make this true? h. Quick break to prove a little lemma. Let H be a subgroup of a group G. Let a, b G. Prove that ah = bh if and only if b -1 a H. (This is a handy tool for proving that cosets are equal.) Proof: Let a, b G. First, we suppose that ah = bh. We note that clearly a ah. But then since ah = bh we also have a bh. This means that a = bh for some h H. Then b -1 a = h. Therefore, b -1 a H. Second, we suppose that b -1 a H. We note that the inverse of this must also be in H since H is a subgroup. Thus (b -1 a) -1 = a -1 b H as well. Now suppose x ah. Then x = ah for some h H. Also since b -1 a H, b -1 a = h 1 for some h 1 H. Thus, a = bh 1. Substituting, we get x = bh 1 h which is clearly in bh. Similarly, if we suppose x bh. Then x = bh for some h H. Also since a 1 b H, a -1 b = h 2 for some h 2 H. Thus, b = ah 2. Substituting, we get x = ah 2 h which is clearly in ah. Thus ah = bh. So we conclude that ah = bh if and only if b -1 a H. i. So how what condition will ensure that hgh = gh? Answer: Let s answer a different question first. If the operation is well defined then from above (part G) we know that given any g G and h H we must have hgh = gh. So given our little lemma, it must be the case that given any g G and h H, we must have g -1 hg H. So for the coset multiplication to be well defined it is necessary that g -1 hg H for all g G and h H. But is this condition sufficient? j. Definition. A subgroup H of a group G is normal if g -1 hg H for every g G and h H. (Note this definition is equivalent the books definition simply substitute g -1 for g.) k. Prove that for a normal subgroup, G/H forms a group under the operation ahbh = abh. Answer: Here we are showing that the condition g -1 hg H for all g G and h H is sufficient to ensure our coset multiplication is well defined. Here we go!

4 Proof: Suppose that H is a normal subgroup of G. Let a, b, c, d in G be such that ah = ch and bh = dh. We need to show that ah bh = ch dh. Thus (by applying the operation) we need to show abh = cdh. By our lemma above, it is sufficient to show that (cd) -1 ab H. Also by our lemma, we note that c -1 a and d -1 b are both elements of H (since ah = ch and bh = dh). Let c -1 a= h 1 and d -1 b= h 2. Now, (cd) -1 ab = d -1 c -1 ab. Substituting, we get (cd) -1 ab = d -1 h 1 b. Since d -1 b= h 2, we have b= dh 2. Substituting again, we get (cd) -1 ab = d -1 h 1 dh 2. Since H is normal, we know that d -1 h 1 d is an element of H, say d -1 h 1 d = h 3. With one last substitution, we have (cd) -1 ab = h 3 h 2 so (cd) -1 ab is an element of H. This is what we needed to show. So we conclude that ah bh = ch dh. So the coset operation is well defined. Proving that G/H is a group is then trivial. By definition ah bh = abh which is clearly a coset (by closure of G) so G/H is closed under the operation. H = eh is clearly an identity element. Given any coset gh, g -1 H is clearly its inverse. Finally it is a routine exercise to show that this operation is associative. l. In conclusion, G/H forms a group if and only if H is a normal subgroup of G (i.e. ghg -1 H for every g G and h H). 4. Now before we move on to rings, lets consider the special case where our groups are additive and abelian. a. In this case, what does the normality condition look like? Answer: It looks like h H for all h H since then ghg -1 = hgg -1 = he = h. Of course it is always true that h H for all h H! b. In this case, which subgroups are normal? Answer: from above it is clear that all subgroups of an abelian group are normal!

5 5. Now lets figure out how to construct a quotient ring. How should we do that? a. How do we get our elements? What do they look like? Answer: We need a subring S of our ring R. The elements will be additive cosets and look like r + S where r is in R. We choose additive cosets because we want our quotient ring to also be a quotient group with respect to addition. b. What are our operations? Addition: (a + S) + (b + S) = (a + b) + S. This one is guaranteed to be well defined since our ring is an abelian group under addition! Multiplication: (a + S) (b + S) = (ab) + S. This operation may or may not be well defined. We need to figure out a necessary and sufficient condition for this to work. c. Try these two examples. For each see if you can verify that both operations (+ and ) on the cosets are well defined (provide a counterexample if an operation is not well defined.): i. Z/4Z. We would expect this to work since this is just the ring Z 4! (We will do this in class on Day 8 for practice with mod arithmetic proofs.) /Z. This one doesn t work! Note that 0 + Z = 1 + Z (both are equal to Z). Now pick any other coset, say ½ + Z. Then: (0 + Z) (½ + Z) = 0 + Z but (1 + Z) (½ + Z) = ½ + Z and these answers are not the same! d. Given a ring and a subgring, what is a necessary and sufficient condition for coset multiplication to be well defined? Hint: Look at the counterexample for /Z. Use something like it to find a necessary condition for coset multiplication to be well defined. Then prove the necessary condition is also sufficient. (For next time!)

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